Sequences and Series: Theory and Practice [2024 ed.] 3031672011, 9783031672019

Citation preview

Ana Alves de Sá · Bento Louro

Sequences and Series Theory and Practice

Sequences and Series

Ana Alves de S´a • Bento Louro

Sequences and Series Theory and Practice

Ana Alves de S´ a Department of Mathematics Nova School of Science and Technology-NOVA University Lisbon (Faculdade de Ciˆencias e Tecnologia-Universidade Nova de Lisboa) Lisboa, Portugal

Bento Louro Department of Mathematics Nova School of Science and Technology-NOVA University Lisbon (Faculdade de Ciˆencias e Tecnologia-Universidade Nova de Lisboa) Lisboa, Portugal

ISBN 978-3-031-67201-9 ISBN 978-3-031-67202-6 https://doi.org/10.1007/978-3-031-67202-6

(eBook)

Mathematics Subject Classification: 97I30, 40A05, 40, 40A30, 40D15, 40C15 The translation was done with the help of an artificial intelligence machine translation tool. A subsequent human revision was done primarily in terms of content. Translation from the Portuguese language edition: “Sucess˜ oes e S´ eries: Teoria e Pr´ atica” by Ana Alves de S´ a and Bento Louro, © Ana Maria de Souza Alves de S´ a and Bento Jos´ e Carrilho Miguens Louro 2014. Published by Escolar Editora. All Rights Reserved. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland If disposing of this product, please recycle the paper.

Preface This book is primarily intended for students of Science and Engineering. Although the topics were not always addressed on the same subject, we decided to group them here due to their scientific coherence. We have strived to create a self-contained text, presenting theoretical exposition, illustrated with various examples, solved exercises of different natures and degrees of difficulty, and proposed exercises. Thus, the readers can test their knowledge and level of understanding of the material. We rigorously present the statements of definitions and results without delving too much into the theoretical exposition and without demonstrating a large part of the Theorems. Many examples highlight aspects that could go unnoticed by readers, mainly those inexperienced in reading mathematical texts, for whom this book is primarily intended. Any reader who has completed High School should be able to comprehend Chap. 1, and the same holds true for Chap. 2, with the possible exception of the Integral Test. For those unfamiliar with integral calculus, skipping this section will not detract from their understanding of the rest of the chapter. To understand Chap. 3, the reader must have basic knowledge of differential and integral calculus. We go beyond the typical introduction to function series by delving into the study of Fourier series. This topic will prove to be especially useful when studying Differential Equations, but no additional knowledge is required beyond what is already necessary for a general understanding of function series. At the end of each chapter, there are two lists of exercises: one titled ”Solved Exercises” followed by the respective solutions, and another titled ”Proposed Exercises,” which are intended for the reader’s practice and whose answers can be found at the end of the book. v

vi We strongly encourage students to attempt the “Solved Exercises” on their own before referring to the provided solution. It is important to note that the written solution does not include the reasoning, drafts, or false starts made in arriving at this point. By comparing their solution to the one presented in the book, readers can more accurately assess their understanding. Reviewing the solution without first attempting the problem independently may not effectively facilitate learning. The exercises featured in this book were used during practical classes and exams for the subjects we taught. We extend our sincere appreciation to the students whose inquiries aided in enhancing this publication, as well as to our fellow educators who assisted us in teaching Calculus courses at Nova School of Science & Technology. Their thorough review of the text and copious suggestions were invaluable contributions. In the margins of the text, we have provided concise biographies of the mentioned mathematicians. For those interested in learning more about these individuals, among many others, we suggest consulting the following web page: https://mathshistory.st-andrews.ac.uk/Biographies/. Despite our efforts toward meticulous writing and thorough proofreading, we know some typos persisted. We extend our apologies to our readership for any such errors and express our gratitude in advance for their assistance in identifying and correcting them for future editions. Lisboa, Portugal Lisboa, Portugal 2024

Ana Alves de S´a Bento Louro

Notes to the Reader This section consists of a list of notations. They are quite common. Our aim is to avoid ambiguity and to help the reader who, here or there, may be used to a different notation. N = {1, 2, 3, 4, . . . }: the set of natural numbers

.

N0 = N ∪ {0}

.

Z = {. . . , −4, −3, −2, −1, 0, 1, 2, 3, 4, . . . }: the set of integers

.

Q = {x ∈ R : x = p/q, p, q ∈ Z, q = 0}: the set of rationals

.

R : the set of real numbers (remark that .N ⊂ Z ⊂ Q ⊂ R)

.

R+ = {x ∈ R : x > 0} : the set of positive real numbers

.

+ R+ 0 = R ∪ {0}

.

triangle inequality: .|x + y| ≤ |x| + |y|, ∀x, y ∈ R intervals on .R: (suppose .a, b ∈ R, with .a < b) are the following sets:

.

[a, b] = {x ∈ R : a ≤ x ≤ b};

] − ∞, b] = {x ∈ R : x ≤ b};

]a, b[ = {x ∈ R : a < x < b};

] − ∞, b[ = {x ∈ R : x < b};

[a, b[ = {x ∈ R : a ≤ x < b};

[a, +∞[ = {x ∈ R : a ≤ x};

]a, b] = {x ∈ R : a < x ≤ b};

]a, +∞[ = {x ∈ R : a < x};

[a, a] = {a};

] − ∞, +∞[ = R

ε neighborhood of a: is the set .Vε (x) = ]a − ε, a + ε[ (where .a ∈ R and .ε > 0)

.

V is a neighborhood of a: there is an .ε > 0 such that .Vε (a) ⊂ V vii

viii n  .

i=n n  .

ai = an , .an ∈ R and .n ∈ N ai = am + am+1 + · · · + an , .ai ∈ R, i = m, m + 1, . . . , n and .m, n ∈ N with .n > m

i=m

factorial of .n ∈ N0 : .0! = 1, 1! = 1 and .n! = n × (n − 1)! (in a suggestive way, .n! = n × (n − 1) × · · · × 3 × 2 × 1)   n n! number of p-combinations of n distinct objects (.n, p ∈ N0 , .n ≥ p): . = p p! × (n − p)! n    n k n−k Newton Binomial: if .a, b ∈ R, .n ∈ N then .(a + b)n = a b k k=0

Method of Mathematical Induction: to prove that a proposition .P (n) is true for all .n ∈ N proceed as follows: 1) prove that .P (1) is true 2) prove hereditary, that is, assume that the Induction Hypothesis .P (n) is true, and then prove the Induction Thesis .P (n + 1) is true integer part of .x ∈ R+ : .Int(x), the largest integer less than or equal to x upper bound of .X ⊂ R: a real number z such that .x ≤ z, ∀x ∈ X X ⊂ R is bounded from above: there exists an upper bound of X

.

supremum of .X ⊂ R: a real number w such that w is an upper bound of X and .w ≤ z for all z upper bound of X; we write .w = sup(X); if X is bounded from above, .sup(X) exists maximum of .X ⊂ R: if .sup(X) ∈ X it is called the maximum of X and is denoted by .max(X) lower bound of .X ⊂ R: a real number y such that .y ≤ x, ∀x ∈ X X ⊂ R is bounded from below: there exists a lower bound of X

.

infimum of .X ⊂ R: a real number u such that u is a lower bound of X and .u ≥ y for all y lower bound of X; we write .u = inf(X); if X is bounded from below, .inf(X) exists minimum of .X ⊂ R: if .inf(X) ∈ X it is called the minimum of X and is denoted by .min(X) X ⊂ R is bounded: X is bounded from below and from above

.

a function of the set X into the set Y is a relation between X and Y such that, to each element .x ∈ X corresponds an unique element .y ∈ Y , and we will write .f : X → Y . X is the domain of f and the set .{y ∈ Y : ∃x ∈ X, y = f (x)} is the range of f

ix if .a ∈ R+ , .f : [−a, a] → R is an odd function if .f (x) = −f (−x), ∀x ∈ [−a, a] and is an even function if .f (x) = f (−x), ∀x ∈ [−a, a] function of class .C 1 on .X ⊂ R: function with continuous derivative on X function of class .C p on .X ⊂ R: function with all continuous derivatives until order p on X indeterminate forms:

.

0 ∞ , , ∞ − ∞, 0 × ∞, 1∞ , 00 ; ∞0 0 ∞

L’Hˆ opital’s Rule: Let I be an interval, .a ∈ I, f and g two differentiable functions on I such that: a) .g  (x) = 0, ∀x ∈ I \ {a}. b) . lim f (x) = lim g(x) = 0 or . lim f (x) = lim g(x) = +∞. x→a

x→a

x→a

x→a

f  (x) f  (x) f (x) = lim . , then . lim   x→a g (x) x→a g (x) x→a g(x)

If there exists (real or .∞) . lim

L’Hˆ opital’s Rule (another version): Let .b ∈ R, f and g two differentiable functions on .]b, +∞[ such that: a) .g  (x) = 0, ∀x ∈ ]b, +∞[. b)

.

lim f (x) = lim g(x) = 0 or . lim f (x) = lim g(x) = +∞.

x→+∞

x→+∞

x→+∞

x→+∞

f  (x) f  (x) f (x) = lim . , then . lim x→+∞ g  (x) x→+∞ g  (x) x→+∞ g(x)

If there exists (real or .∞) . lim

Contents

Preface

v

Notes to the Reader

vii

1 Sequences of Real Numbers 1.1 Sequences of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Solved Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Proposed Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 38 64

2 Numerical Series 2.1 Generalization of the Addition Operation . . . . . . . . 2.2 Definition of Series: Convergence – General Properties 2.3 Alternating Series . . . . . . . . . . . . . . . . . . . . 2.4 Absolute Convergence . . . . . . . . . . . . . . . . . . 2.5 Series of Nonnegative Terms . . . . . . . . . . . . . . 2.6 Products of Series . . . . . . . . . . . . . . . . . . . . 2.7 Solved Exercises . . . . . . . . . . . . . . . . . . . . . 2.8 Proposed Exercises . . . . . . . . . . . . . . . . . . . 3 Series of Functions 3.1 Introduction: Sequences of Functions . 3.2 Pointwise and Uniform Convergence of 3.3 Power Series . . . . . . . . . . . . . . 3.4 Taylor Series and Maclaurin Series . . 3.5 Introduction to Fourier Series . . . . . 3.6 Solved Exercises . . . . . . . . . . . . 3.7 Proposed Exercises . . . . . . . . . .

. . . . . Series of . . . . . . . . . . . . . . . . . . . . . . . . . x

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

67 . 67 . 69 . 82 . 87 . 90 . 113 . 116 . 161

. . . . . . Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . . .

167 167 172 186 193 201 229 320

Contents

xi

Answers to Proposed Exercises

327

Bibliography

337

Index

339

Sequences of Real Numbers 1

In everyday language, the word “sequence” refers to a collection of objects or events arranged in a specific order. For instance, the daily stock quotes of a particular company on the stock exchange market can be considered a sequence. In this chapter, we shall explore sequences of real numbers and establish precise definitions for the terms “sequence” and “convergence.”

1.1

Sequences of Real Numbers

Definition 1.1.1 A sequence of real numbers is a function of N into R. Note: The elements of the range of a sequence u are called terms of the sequence, and they are usually denoted by un , instead of the notation u(n) which is generally used for functions. A sequence u is represented by (un )n∈N or, more simply, (un ). As a function, its graph is the set formed by ordered pairs of the form (n, un ), n ∈ N. For example, in Fig. 1.1 is represented the graph of the sequence un = n/(n + 1). Of course, it is only possible to represent part of a sequence graph. Definition 1.1.2 The designatory expression that defines the sequence is called general term of the sequence.

Example 1.1.1 The sequences of general terms an = n2 and bn = cos(n) Figure 1.1: The graph of the sequence un = n/(n + 1) are illustrated in Fig. 1.2.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 A. Alves de S´a, B. Louro, Sequences and Series, https://doi.org/10.1007/978-3-031-67202-6 1

1

2

Sequences of Real Numbers Notes: 1. We can define a sequence without specifying a general term. This is the case for sequences defined recursively. A recursive relation is an expression that relates the nth element of a sequence to the previous element or elements. For example, Leonardo Fibonacci, also known as Leonardo of Pisa (1170–1250), was born in Pisa, Italy and received his education in North Africa. He is primarily known for his work on number theory and algebra. There are copies of some of his works, one of the most important being the Liber Abaci, which introduced Arabic numerals to Europe. It is also in this book that the famous Fibonacci sequence appears: 1, 1, 2, 3, 5, 8, 13, 21, 34, ..., where each number is the sum of the two previous numbers. (Source of image: I benefattori dell’umanit` a; vol. VI, Firenze, Ducci, 1850)

u1 = 1, u2 = 2, un+2 = un+1 + un , ∀n ∈ N (Fibonacci sequence). In this sequence, we obtain the term un as the sum of the two immediately preceding terms: u3 = u2 + u1 = 3, u4 = u3 + u2 = 5, u5 = u4 + u3 = 8, . . . 2. Sometimes only a few terms of the sequence are given, leading the reader to “infer” the rest, for example, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, . . .

.

3. There are sequences that are not defined for a finite number of values 1 of n ∈ N. For instance, the sequence with general term un = n−3 is only defined for n = 3.

Definition 1.1.3 A sequence is bounded from above (or bounded above) if the set of its terms has an upper bound; it is bounded from below (or bounded below) if the set of its terms has a lower bound; and it is bounded if the set of its terms is bounded, that is, if the set has an upper bound and a lower bound.

Example 1.1.2 The sequence an = n2 is bounded below but not above (see Fig. 1.2). Specifically, x = 1 is a lower bound; however, there are no upper bounds since there is no number L such that n2 ≤ L, ∀n ∈ N. Example 1.1.3 The sequence bn = cos(n) is bounded (see Fig. 1.2). It is enough to note that −1 ≤ cos(x) ≤ 1, ∀x ∈ R. Example 1.1.4 The sequence cn = −n is bounded above but not below (see Fig. 1.3). In fact, x = −1 is an upper bound, but there are no lower bounds because there is no number L such that −n ≥ L, ∀n ∈ N.

1.1. Sequences of Real Numbers

(a) The graph of an = n2

3

(b) The graph of bn = cos(n)

Figure 1.2: A bounded below sequence (a) and a bounded sequence (b)

Example 1.1.5 The sequence dn = (−1)n n is not bounded above or below (see Fig. 1.3).

(a) The graph of cn = −n

(b) The graph of dn = (−1)n n

Figure 1.3: A bounded above and unbounded below sequence (a) and a sequence not bounded above or below (b)

Theorem 1.1.1 A sequence u is bounded if and only if there exists M ∈ R+ such that |un | ≤ M , ∀n ∈ N. Proof: Let u be a bounded sequence. According to Definition 1.1.3, u has upper and lower bounds. Suppose α is a lower bound and β is an upper bound. This means that α ≤ un ≤ β, ∀n ∈ N. Let us consider M = max (|α|, |β|). Then |un | ≤ M, ∀n ∈ N.

.

4

Sequences of Real Numbers Conversely, suppose there exists M ∈ R+ such that |un | ≤ M, ∀n ∈ N.

.

We have −M ≤ un ≤ M , ∀n ∈ N. Thus, −M is a lower bound and M is an upper bound of the sequence. By Definition 1.1.3, u is a bounded sequence. n+2 is bounded (see Fig. 1.4): Example 1.1.6 The sequence un = (−1)n n     2 n n + 2  = 1 + ≤ 3, ∀n ∈ N. .|un | = (−1)  n  n

Figure 1.4: A bounded sequence

Definition 1.1.4 Given two sequences of real numbers, u and v, we call sum, difference, and product of u and v to the sequences u + v, u − v, and uv with general terms, respectively, un + vn , un − vn , and un vn . If vn = 0, ∀n ∈ N, we define the quotient sequence of u and v as the sequence u/v with general term un /vn . Definition 1.1.5 A sequence u is increasing if un ≤ un+1 , ∀n ∈ N; it is strictly increasing if un < un+1 , ∀n ∈ N; it is decreasing if un ≥ un+1 , ∀n ∈ N; it is strictly decreasing if un > un+1 , ∀n ∈ N; it is monotonic if it is increasing or decreasing; and it is strictly monotonic if it is strictly increasing or decreasing.

1.1. Sequences of Real Numbers

5

2n is strictly increasing. We can 3n + 7 prove this by computing and simplifying the expression un+1 − un : Example 1.1.7 The sequence un =

2n (2n + 2)(3n + 7) − 2n(3n + 10) 2(n + 1) − = 3(n + 1) + 7 3n + 7 (3n + 10)(3n + 7) 14 > 0, ∀n ∈ N. = (3n + 10)(3n + 7)

un+1 − un = .

5n is strictly decreasing. In fact, n2 + 1 5n 5(n + 1) − un+1 − un = (n + 1)2 + 1 n2 + 1

Example 1.1.8 The sequence un =

.

=

5(n + 1)(n2 + 1) − 5n(n2 + 2n + 2) (n2 + 1)(n2 + 2n + 2)

= −

5(n2 + n − 1) < 0, ∀n ∈ N. (n2 + 1)(n2 + 2n + 2)

Example 1.1.9 The sequence un = n2 is strictly increasing because u

. n+1

− un = (n + 1)2 − n2 = n2 + 2n + 1 − n2 = 2n + 1 > 0, ∀n ∈ N.

Example 1.1.10 The sequence un = −n strictly decreases because u

. n+1

− un = −(n + 1) + n = −n − 1 + n = −1 < 0, ∀n ∈ N.

Example 1.1.11 The sequence un = (−n)n is not monotonic. Indeed,    n+1 .un+1 − un = − (n + 1) − (−n)n = (−1)n+1 (n + 1)n+1 + nn , and this difference is positive if n is odd and negative if n is even. Example 1.1.12 The sequence un = Fig. 1.5).

n + (−1)n is not monotonic (see n2

⎧ n+1−1 n+1 n3 − (n + 1)3 ⎪ ⎪ − = < 0, ⎪ ⎨ (n + 1)2 n2 n2 (n + 1)2 .un+1 − un = ⎪ n+1+1 n−1 n2 + n + 1 ⎪ ⎪ ⎩ − = 2 > 0, 2 2 (n + 1) n n (n + 1)2

if n is even if n is odd.

6

Sequences of Real Numbers

Figure 1.5: The sequence un =

n + (−1)n n2

• Arithmetic and Geometric Sequences We will now discuss two distinct types of sequences that hold significant importance due to their properties: arithmetic sequences (also known as arithmetic progressions) and geometric sequences (also known as geometric progressions). An arithmetic sequence .(an ) is a sequence defined by the recursive relation: a1 = k, .

an+1 = an + r,

∀n ∈ N,

where .k, r ∈ R, and r is called the common difference of the arithmetic sequence. According to this expression, it is easy to conclude that the arithmetic sequence is: • Strictly increasing if .r > 0 • Strictly decreasing if .r < 0 • Constant if .r = 0 As an example, .1, 4, 7, 10, 13, . . . is an arithmetic sequence with .r = 3. Let .(an ) be an arithmetic sequence with common difference r. Then

1.1. Sequences of Real Numbers

.

7

a2 = a1 + r a3 = a2 + r = a1 + 2r a4 = a3 + r = a1 + 3r .. . an = an−1 + r = a1 + (n − 1) r .. .

The general term of the arithmetic sequence with common difference r is given by the expression a = a1 + (n − 1) r, ∀n ∈ N.

. n

At first glance, the sum of the first terms of an arithmetic sequence may seem like a time-consuming problem to solve. However, according to legend, Gauss, one of the greatest mathematicians of all time, solved the question of calculating the sum of the first 100 natural numbers in a few minutes when he was still a child. What reasoning would have allowed him to arrive so quickly at the result? If we write the integers from 1 to 100, we find that: • The sum of the first and last numbers is 101. • The sum of the second and penultimate numbers is 101. • The sum of the third and third-to-last numbers is 101, and so on. There are precisely 50 pairs of numbers whose sum is a constant of value 101, so it is concluded that the sum of the first 100 natural numbers is .101 × 50 = 5050. In general, .Sn , the sum of the first n consecutive terms of an arithmetic sequence .(an ) with common difference r, is given by the formula Sn =

.

a1 + an × n. 2

Let us prove this using the method of induction. For .n = 1, the equality is trivial. The property is proven if it is hereditary. If we admit that equality is valid for n, then

Carl Friedrich Gauss (1777–1855) was a German mathematician. He worked in a wide variety of areas in both Mathematics and Physics. He was a gifted child. His first mathematical discoveries were made in his teens. His magnum opus, Disquisitiones Arithmeticae, was completed in 1798, when Gauss was only 21 years old. (Source of image: Oil portrait by Gottlieb Wilhelm Emil Biermann)

8

Sequences of Real Numbers

a1 + an × n + an+1 2 a1 + a1 + (n − 1)r × n + a1 + nr 2 (n − 1)r n a1 + × n + a1 + nr 2

 n−1 (n + 1)a1 + + 1 × nr 2 n+1 (n + 1)a1 + × nr 2 nr (n + 1) a1 + 2 a1 + a1 + n r (n + 1) 2 a1 + an+1 , (n + 1) 2

Sn+1 = Sn + an+1 = = = = .

= = = =

which confirms that the equality holds for .n + 1. A geometric sequence .(an ) is the sequence defined by the recursive relation .

a1 = k an+1 = an × r,

∀n ∈ N,

where .k, r ∈ R\{0}, and r is known as the common ratio of the geometric sequence. For example, .

1 1 1 1 , , , . . . is a geometric sequence with .r = . 10 100 1000 10

Similar to arithmetic sequences, it is possible to calculate any term of the sequence knowing the first term and the common ratio. Considering the definition of a geometric sequence with common ratio r, we obtain the following: a2 = a1 × r a 3 = a2 × r = a 1 × r 2 .

a4 = a3 × r = a 1 × r 3 .. . an = a1 × rn−1 .. .

1.1. Sequences of Real Numbers

9

The general term of a geometric sequence with ratio r is expressed as follows: n−1 .an = a1 × r , ∀n ∈ N. According to this expression, it is easy to conclude that the monotonicity of a geometric sequence is: • Increasing if .a1 > 0 and .r > 1 or if .a1 < 0 and .0 < r < 1 • Decreasing if .a1 > 0 and .0 < r < 1 or if .a1 < 0 and .r > 1 • Constant if .r = 1 • Not monotonic if .r < 0 There is a well-known legend that a Hindu prince, fascinated with the chess game, decided to reward its inventor, who asked for this reward to be given in wheat, receiving one grain for the first square of the chessboard, two grains for the second square, four grains for the third, and so on, doubling the number of grains from the last square considered. We conclude that the number of wheat grains is represented by the sum of the first 64 terms of a geometric sequence with common ratio 2. S64 = 1 + 2 + 22 + 23 + . . . + 263 .

.

How can we determine the value of this expression? We can calculate the sum in the more general case of a geometric sequence, which involves finding the sum of the first n terms. Let Sn = a1 + a 2 + a 3 + . . . + a n

.

(1.1)

be the sum of the first n terms. Multiplying .Sn by the ratio r, we obtain r Sn = r a1 + r a2 + r a3 + . . . + r an = a2 + a3 + a4 + . . . + an+1 . (1.2)

.

Subtracting (1.2) from (1.1), we obtain Sn − r Sn = (a1 + a2 + a3 + . . . + an ) − (a2 + a3 + a4 + . . . + an+1 ),

.

that is, Sn =

.

1 − rn a1 − a1 r n a1 − an+1 = = a1 . 1−r 1−r 1−r

10

Sequences of Real Numbers Returning to the inventor of the game of chess, the requested reward is S64 = 264 − 1 = 18 446 744 073 709 551 615

.

wheat grains (which not even the richest prince would be able to pay).

Example 1.1.13 Let us consider Fig. 1.6, and let .(An ) be the sequence of the areas of the semicircles delimited by the lines .C1 , .C2 , .C3 , .. . .. We show that .(An ) is a geometric sequence and express its general term as follows:

(a)

(b)

(c)

Figure 1.6: Semicircles

1 Each semicircle has an area . π r2 , where r is the radius of the respective 2 circumference. In Fig. 1.6a, the radius of the circumference is r; in Fig. 1.6b, r each semicircle has a radius of . ; in Fig. 1.6c, each semicircle has a radius 2 r of . ; and so on. Therefore, we can write the first terms of the sequence 4 .An . 1 π r2 2 1 r 2 A2 = π ×2=π 2 2 1 r 2 ×4=π A3 = π 2 4 A1 = .

r2 4 r2 . 8

1 By induction, it can be proven that .An+1 = ×An , that is, it is a geometric 2 1 sequence with common ratio . , so the general term of the sequence is 2 r2 .An = π . 2n

1.1. Sequences of Real Numbers

11

• Limits of Sequences Definition 1.1.6 A sequence u tends to +∞ (and we write un → +∞ or lim un = +∞ or lim un = +∞), if n→+∞

∀ L ∈ R+ ∃ p ∈ N : n > p ⇒ un > L.

.

A sequence u tends to −∞ (and we write un → −∞ or lim un = −∞ or lim un = −∞), if

n→+∞

∀ L ∈ R+ ∃ p ∈ N : n > p ⇒ un < −L.

.

Example 1.1.14 We prove that un = n2 → +∞. Given L > 0, there √ exists p ∈ N such that p > L; if n > p, then n2 > p2 > L. It is evident that p depends on L; for instance, if L = 100, it is sufficient to consider p = 10, but if L = 200, we must consider p = 14. This example is illustrated in Fig. 1.7.

Figure 1.7: The sequence un = n2 tends to +∞

Similarly, it can be shown that vn = nα , α ∈ R+ , tends to +∞, that yn = −n tends to −∞, and that if wn = (−n)n , then |wn | = nn → +∞. Notes: 1. Sometimes the following concept is also used: A sequence (un ) has general infinite limit if |un | → +∞. For example, the sequence

12

Sequences of Real Numbers un = (−1)n n has general infinite limit despite un → +∞ and un → −∞. 2. The fact that un → +∞ does not imply that u is increasing (nor that there is an order from which it is increasing). For instance, the sequence un = n + (−1)n , illustrated in Fig. 1.8, tends to +∞, but it is not monotonic.

Figure 1.8: The sequence un = n + (−1)n

3. If u is such that un → +∞, un → −∞, or |un | → +∞, then u is unbounded. The converse is not true. For example, the sequence ⎧ ⎨n, .un = ⎩1, n

if n is even if n is odd

is unbounded and un → +∞, un → −∞, |un | → +∞.

From Definition 1.1.6, the next result follows immediately. Theorem 1.1.2 Let u and v be sequences such that from a certain order, un ≤ vn . Then, a) un → +∞ ⇒ vn → +∞. b) vn → −∞ ⇒ un → −∞.

Example 1.1.15 Let us consider the sequence

u =

. n

Figure 1.9: An illustration of Theorem 1.1.2

n  1 1 1 1 √ = 1 + √ + √ + ··· + √ . n 2 3 k k=1

Because we have inequality √ 1 1 1 1 1 + √ + √ + ··· + √ ≥ n × √ = n n n 2 3

.

and

√ n → +∞, we can state that un → +∞ (see Fig. 1.9).

1.1. Sequences of Real Numbers

13

Theorem 1.1.3 Let un → +∞, vn → +∞, and wn → −∞. Then a) lim (un + vn ) = +∞. b) lim (un vn ) = +∞. c) lim (un wn ) = −∞. d) lim (un )p = +∞, ∀p ∈ N. e) lim |un | = lim |vn | = lim |wn | = +∞.

Definition 1.1.7 Let u be a sequence and a ∈ R. We say that u converges to a (or that tends to a or, still, that the limit of the sequence is a), and we write un → a or lim un = a or lim un = a, if n→+∞

∀ ε > 0 ∃ p ∈ N : n > p ⇒ |un − a| < ε.

.

We say that a sequence u is convergent if there is a ∈ R such that u converges to a; otherwise, u is divergent. In other words, u is convergent if for every ε > 0, we can choose p such that from the order p all terms un are in the interval ]a − ε, a + ε[. 5n , we illustrate this definition in Fig. 1.10 Taking the sequence un = 9n + 1 for two values of ε.

1 (a) If ε = 80 then a − whatever n > 4

1 80

< un < a +

1 80

(b) If ε = n > 30

1 500

Figure 1.10: The value of p varies with the value of ε

then a −

1 500

< un < a +

1 500

whatever

14

Sequences of Real Numbers 5 The limit of this sequence is a = . To prove this fact using the definition, 9 we proceed as follows: Let ε > 0; we must show that there exists a natural number, p, such that    5n 5   < ε. − .if n > p then  9n + 1 9  Note that .

      5n −5 5 5     = −  9n + 1 9   9(9n + 1)  = 9(9n + 1) .

Solving inequality .

5 p ⇒ n > 81ε 9n + 1 9

for n, we obtain n >

5 5 Note that if ε ≥ , we can choose p = 1. If ε < , we can choose 9

9

 5 − 9ε 5 − 9ε p = Int . This number p = Int is the smallest number 81ε 81ε that makes the statement true. 1 Example 1.1.16 Let us prove that un = → 0. Given an arbitrary ε > 0, n we must show that there exists a natural number, p, such that   1 1 .if n > p then   = < ε. n n 1 Let p be a natural number greater than or equal to . Then, for n > p, we ε 1 1 1 , this will be the smallest order have < < ε. If we choose p = Int n p ε from which the proposition is true for a given value of ε (see Fig. 1.11). 1 Similarly, we can prove that the sequence vn = α , α ∈ R+ , converges to n zero.

1.1. Sequences of Real Numbers

Figure 1.11: If ε = 0.1, then −ε
10

Notes: 1. In the language of neighborhoods, Definition 1.1.7 is equivalent to ∀ ε > 0 ∃ p ∈ N : n > p ⇒ un ∈ Vε (a).

.

2. We could also write equivalently ∀ ε > 0 ∃ p ∈ N : |un − a| < ε, ∀n > p.

.

3. Consider the set R = R ∪ {−∞, +∞}, where −∞ and +∞ are two distinct mathematical objects that are not real numbers. We can introduce an order relation in this set as follows: i) If x, y ∈ R, x < y on R if and only if x < y on R. ii) −∞ < x < +∞, ∀x ∈ R. The set R, with this order relation, is called extended real line. We can extend the notion of neighborhood to R. Let ε > 0. If a ∈ R, we call the ε neighborhood of a the set Vε (a) = ]a − ε, a + ε[ (which   coincides with the neighborhood in R). The set Vε (+∞) = 1ε , +∞   is called the ε neighborhood of +∞. The set Vε (−∞) = −∞, − 1ε is called the ε neighborhood of −∞. With the definitions given previously, we can unify from a formal viewpoint Definitions 1.1.6 and 1.1.7: If a ∈ R, u → a if and only if ∀ ε > 0 ∃ p ∈ N : n > p ⇒ un ∈ Vε (a).

. n

16

Sequences of Real Numbers Theorem 1.1.4 (Uniqueness of the limit) Let a, b ∈ R. If un → a and un → b, then a = b. Proof: (By contradiction)

|b − a| . By the definition of a convergent If a = b, we can consider ε = 3 sequence, .∃ p ∈ N : n > p ⇒ |un − a| < ε

and ∃ q ∈ N : n > q ⇒ |un − b| < ε.

.

Let m > max{p, q}. Then, |b − um + um − a| |b − um | + |um − a| |b − a| = ≤ 3 3 3 ε+ε 2ε < = 0, ∀n ∈ N, and the limits of the sequences are not both zero. g) lim |un | = | lim un |. h) (∃ p ∈ N ∀n ≥ p, un ≥ 0) ⇒ lim un ≥ 0. i) (∃ p ∈ N ∀n ≥ p, un ≥ vn ) ⇒ lim un ≥ lim vn .

Proof: We will prove only items a) and g).

1.1. Sequences of Real Numbers

17

a) Let a and b be the limits of sequences (un ) and (vn ), respectively. Let ε > 0 be arbitrary. By definition, ε .∃ p1 ∈ N : n > p1 ⇒ |un − a| < 2 and ∃ p2 ∈ N : n > p2 ⇒ |vn − b|
p, |(un + vn ) − (a + b)| = |un + vn − a − b| = |un − a + vn − b| .

≤ |un − a| + |vn − b|
p ⇒ |un − a| < ε. But ||un | − |a|| ≤ |un − a|, so ||un | − |a|| < ε, ∀n > p. We conclude that |un | → |a|. Definition 1.1.8 A sequence u is a null sequence if un → 0. Notes: 1. It is evident from the definitions that (un ) converges to a if and only if (un − a) is a null sequence. 2. Considering the definitions, it is easy to verify that if u is a null sequence, |1/un | → +∞, and, reciprocally, if |vn | → +∞, then 1/vn → 0. Theorem 1.1.6 If un → 0 and v is a bounded sequence, then un · vn → 0. Proof: Let M > 0 such that |vn | ≤ M, ∀n ∈ N. Given an arbitrary ε > 0, ε let p ∈ N, such that |un | < , ∀n > p. Then M |un · vn | < ε, ∀n > p.

.

18

Sequences of Real Numbers

Figure 1.12: An illustration of Theorem 1.1.6

Example 1.1.17 Let us calculate the limit of the sequence a =

. n

−2 + 4 cos(n) . n

We know that −1 ≤ cos(n) ≤ 1, ∀n ∈ N; therefore, .

− 6 ≤ −2 + 4 cos(n) ≤ 2,

∀n ∈ N.

In other words, the sequence bn = −2 + 4 cos(n) is bounded (see Fig. 1.12). 1 As Example 1.1.15 shows, the sequence with general term is a null sen quence. By Theorem 1.1.6 we can assert that .

lim

−2 + 4 cos(n) = 0. n

From Definition 1.1.7 it is easy to show the following theorem: Theorem 1.1.7 Every convergent sequence is bounded.

Note: The converse is not true. For example, the sequence u = cos(nπ) = (−1)n

. n

is bounded but does not converge.

1.1. Sequences of Real Numbers Theorem 1.1.8 (Squeeze Theorem) If un → a, vn → a and, from a certain order, un ≤ wn ≤ vn , then wn → a. Proof: Let ε be any positive real number. Then ∃ p1 ∈ N : n > p1 ⇒ a − ε < un < a + ε, .

∃ p2 ∈ N : n > p2 ⇒ a − ε < vn < a + ε, ∃ p3 ∈ N : n > p3 ⇒ u n ≤ w n ≤ v n .

Let p = max{p1 , p2 , p3 }. If n > p, then a − ε < un ≤ wn ≤ vn < a + ε.

.

Figure 1.13: An illustration of the Squeeze Theorem

Example 1.1.18 Let us calculate the limit of the sequence a =

. n

−2 + 4 cos(n) n

from the previous example, now applying Theorem 1.1.8. We know that −1 ≤ cos(n) ≤ 1, ∀n ∈ N, which implies that .

Considering that Fig. 1.13).

−

−2 + 4 cos(n) 2 6 ≤ ≤ , ∀n ∈ N. n n n

−2 + 4 cos(n) 1 → 0, we conclude that lim = 0 (see n n

19

20

Sequences of Real Numbers Theorem 1.1.9 Let P (x) = a0 xp +· · ·+ap and Q(x) = b0 xq +· · ·+bq be two polynomial functions with real coefficients, p, q ∈ N, a0 = 0, b0 = 0, and v be a sequence, vn → +∞. Then ⎧ a0 , ⎪ ⎪ ⎪ b0 ⎪ ⎪ ⎪ ⎨+∞,

P (vn ) = . lim Q(vn ) ⎪ ⎪ −∞, ⎪ ⎪ ⎪ ⎪ ⎩ 0,

if p = q a0 >0 b0 a0 if p > q ∧ q ∧

Proof: Let p = q. Dividing by vnp both members of the fraction .

a0 vnp + a1 vnp−1 + · · · + ap−1 vn + ap P (vn ) = , Q(vn ) b0 vnp + b1 vnp−1 + · · · + bp−1 vn + bp

we obtain P (vn ) = . Q(vn )

a1 ap−1 ap + · · · + p−1 + p vn vn vn . b1 bp−1 bp b0 + + · · · + p−1 + p vn vn vn

a0 +

By hypothesis, vn → +∞; then Theorem 1.1.5 that .

1 → 0 for all s ∈ N. We conclude by vns

a0 P (vn ) → . Q(vn ) b0

Let p = q. Putting vnp in evidence in the numerator and vnq in the denominator, we obtain the equalities

.

a0 vnp + a1 vnp−1 + · · · + ap−1 vn + ap P (vn ) = Q(vn ) b0 vnq + b1 vnq−1 + · · · + bq−1 vn + bq

 a1 ap−1 ap p + · · · + p−1 + p v n a0 + vn vn vn

 = b1 bq−1 bq vnq b0 + + · · · + q−1 + q vn vn vn a1 ap−1 ap a0 + + · · · + p−1 + p v v n v n n = vnp−q . b1 bq−1 bq b0 + + · · · + q−1 + q vn vn vn

1.1. Sequences of Real Numbers

21

We saw earlier that a1 + ··· + vn . b1 b0 + + ··· + vn a0 +

ap−1 ap p−1 + v p a0 n vn → . bq−1 bq b0 + q vn vnq−1

a0 P (vn ) will be +∞ if If p > q, vnp−q → +∞, so the limit of > 0 and Q(vn ) b0 a0 −∞ if < 0. b0 a0 P (vn ) If p < q, vnp−q → 0, so →0· = 0. Q(vn ) b0 Theorem 1.1.10 Let a ∈ R. Then ⎧ ⎪ ⎪ ⎨+∞, n . lim a = 0, ⎪ ⎪ ⎩1,

if a > 1 if |a| < 1 if a = 1.

If a ≤ −1, the limit does not exist. If a < −1, |an | → +∞. Proof: Let a ∈ R such that |a| > 1. We intend to prove that |a|n → +∞. Consider an arbitrary L > 0. Since log(|a|) is positive when |a| > 1, we have |a|n > L ⇔ en log(|a|) > elog(L) ⇔ n log(|a|) > log(L) ⇔ n >

.

log(L) . log(|a|)

Therefore, it is sufficient to consider in Definition 1.1.6, for example, log(L) p = Int log(|a|) + 1. Thus, we have shown that if |a| > 1, then |an | → +∞. If a > 1, |an | = an , so an → +∞. If a < −1, |an | → +∞ (note that, in this case, an is negative when n is odd and positive when n is even). If a = 0, it is obvious that an = 0 → 0. Let a be such that |a| < 1, a = 0, and take an arbitrary ε > 0. As log(|a|) < 0 because |a| < 1, we get |an | = |a|n < ε ⇔ n log(|a|) < log(ε) ⇔ n >

.

log(ε) . log(|a|)

22

Sequences of Real Numbers If ε < 1, it is enough to take in Definition 1.1.7, for instance, log(ε) + 1; if ε ≥ 1, we can consider p = 1. p = Int log(|a|) If a = 1, then an = 1, ∀n ∈ N, so an → 1. If a = −1, then an = (−1)n , and this sequence does not converge.

Example 1.1.19 Let us calculate the limit of the sequence

.

4n 2n + 1

n  . n∈N

Given n ∈ N, 3n ≥ 2n + 1 ⇒

.

4n 4n 4 = ≤ ⇒ 3 3n 2n + 1

n

n 4 4n ≤ . 3 2n + 1

n 4 = +∞. Considering the last inequality, 3

n 4n we conclude by Theorem 1.1.2 that lim = +∞. 2n + 1

But by Theorem 1.1.10, lim

Example 1.1.20 Let us evaluate the limit of the sequence of general term

n n .(an ) = . 5n + 1 n

The sequence (an ) is monotonically increasing and has a limit of n 1 1 ≤ ≤ ⇒ . 6 5n + 1 5

1 , so 5

n

n n 1 n 1 ≤ ≤ . 6 5n + 1 5

n

n 1 1 = lim = 0. From the previous 5 6

n n inequalities, we conclude by Theorem 1.1.8 that lim = 0. 5n + 1

But by Theorem 1.1.10, lim

1.1. Sequences of Real Numbers

23

Theorem 1.1.11 a) Let a ∈ R. If un → a, then b) If b ∈ R+ , then

√ n

u1 + · · · + un → a. n

b → 1.

c) Let c ∈ R. If un > 0, ∀n ∈ N, and

√ un+1 → c, then n un → c. un

Proof: a) Let a ∈ R and ε > 0 be arbitrary. We want to prove that     u1 + u2 + · · · + un  < ε, − a .∃ s ∈ N : n > s ⇒    n that is,

   u1 − a + u2 − a + · · · + un − a   < ε. ∃ s ∈ N : n > s ⇒   n

.

We know that ∃ p ∈ N : n > p ⇒ |un − a| < 2ε . Additionally,    u1 − a + u2 − a + · · · + up − a  ≤    n .

≤

|u1 − a| + |u2 − a| + · · · + |up − a| Mp ≤ , n n

where M = max{|u1 − a|, |u2 − a|, · · · , |up − a|}. Let q ∈ N such that n > q ⇒ Mn p < 2ε . Let s = max{p, q}. If n > s, we have    u1 − a + u2 − a + · · · + un − a  ≤    n .

≤

|u1 − a| + |u2 − a| + · · · + |up − a| |up+1 − a| + · · · + |un − a| + n n

≤

ε (n − p) ε + 2 ≤ ε. 2 n

Thus, if a ∈ R, the statement is proved. Now, we show the theorem for a = +∞. Let L be an arbitrary positive real number. We know that there exists p ∈ N : n > p ⇒ un > 2 L. Given that u1 + u2 + · · · + up → 0, . n then −L u1 + u2 + · · · + up > . .∃ q ∈ N : n > q ⇒ n 2

24

Sequences of Real Numbers Additionally,

2 L (n − p) up+1 + up+1 + · · · + un > . n n 3 → 1, ∃ r ∈ N : n > r ⇒ n−p n > 4 . Let s = max{p, q, r}. If n > s, .

As n−p n we have .

−L 2 L (n − p) −L 3 u1 + u2 + · · · + un > + > + 2 L = L. n 2 n 2 4

Thus, the statement is proved if a = +∞. For a = −∞, the procedure is similar but with obvious adaptations. √ √ b) Let b > 1. In this case n b > 1, so we can write n b = 1 + hn , where, for each n, hn > 0. Then, by Newton’s Binomial, n   n k n .b = (1 + hn ) = h > 1 + n hn , k n k=0

so n hn < √ b − 1, from which 0 < hn < that is, n b → 1. If b < 1, we take a =

If b = 1,

b−1 n ,

and we deduce that hn → 0,

1 b

> 1. Then  √ 1 1 n n 1 = √ → = 1. . b= n a 1 a

√ √ n b = n 1 = 1 for all n, which is the limit of a constant sequence.

c) Let c ∈ R+ and 0 < ε < c be arbitrary. We know that ∃p ∈ N : n > p ⇒ c −

.

un+1 ε ε < p, we obtain up+2 ε up+3 ε un ε ε ε ε < .c− < c+ , c− < < c+ , . . . , c− < < c+ , 2 up+1 2 2 up+2 2 2 un−1 2 and multiplying term by term these inequalities, un ε n−p−1 . c− < < c+ 2 up+1

we obtain ε n−p−1 , 2

which implies .



ε n ε n up+1 up+1 c − c + < u < .    n p+1 p+1 2 2 c − 2ε c + 2ε

1.1. Sequences of Real Numbers

25

Applying roots,   ε √ ε up+1 up+1 n n  c − c+ < . . n  < u   n p+1 p+1 2 2 c − 2ε c + 2ε By item b),  . n

up+1  p+1 → 1 and c − 2ε

 n

up+1 p+1 → 1; c + 2ε



therefore,   ε ε ε ε up+1 up+1 n  c − c + → c − and →c+ . . n    p+1 ε p+1 ε 2 2 2 2 c− 2 c+ 2 As c − 2ε = c + 2ε , we cannot apply the Squeeze Theorem. However, by applying the definition of limit, we know that there exists q ∈ N such that if n > q, then  ε ε ε up+1 ε ε c− − < n < c− + = c, .c − ε = c − ε p+1 2 2 (c − 2 ) 2 2 2 and there exists r ∈ N such that if n > r, then  ε ε ε up+1 ε ε c + − < n < c + + = c + ε; .c = c + 2 2 (c + 2ε )p+1 2 2 2 taking s = max{p, q, r}, if n > s, then   ε √ ε up+1 up+1 n n c− c + < un < n  < c + ε, .c − ε <  ε p+1 p+1 (c − 2 ) 2 2 c+ ε 2

and it follows from the definition of limit that lim

√ n

un = c.

If c = 0, we follow the same reasoning, considering that un > 0 and, √ therefore, n un > 0. We show that, from a certain order,  √ up+1 ε n .0 < un < n  p+1 . ε 2 2

From this, we deduce that there is an order from which 0 < √ as before, we conclude that lim n un = 0.

√ n u n < ε and,

Let c = +∞. Let L be an arbitrary positive number. We know that un+1 .∃ p ∈ N : n > p ⇒ > 2L. un

26

Sequences of Real Numbers Taking n > p, we have up+2 up+3 un . > 2L, > 2L, . . . , > 2L, up+1 up+2 un−1 and multiplying term by term these inequalities yields un . > (2L)n−p−1 , up+1 up+1 which implies un > up+1 (2L)n−p−1 = (2L)n . Applying roots, (2L)p+1  √ up+1 n . un > 2L n . (2L)p+1  up+1 By item b), n → 1; therefore, (2L)p+1  up+1 1 .∃ q ∈ N : n > q ⇒ n > . (2L)p+1 2 √ Let s = max{p, q}. If n > s, then n un > 2L 12 = L. Notes: 1. It is easy to see, using item c) of the previous theorem, that

√ n n → 1.

2. The converse of item c) of the previous theorem is not true, that is, √ un+1 n un → c does not imply that → c (un > 0, ∀n ∈ N). For un n example, consider the sequence un = e−n−(−1) : (−1)n √ n . un = e−1− n → e−1 , and n un+1 . = e−1+2(−1) = un

e−3 ,

if n is odd

e,

if n is even

has no limit. Example 1.1.21 Let us calculate the limit of the sequence with general √ term n 2n + 3n . The sequence un = 2n + 3n is positive and

n+1 2 +1 n+1 n+1 un+1 2 +3 3

 = 3. . lim = lim = lim n n→+∞ un n→+∞ n→+∞ 2n + 3n 2 1 1 · + 3 3 3

1.1. Sequences of Real Numbers

27

By item c) of Theorem 1.1.11, we can conclude that √ n . lim 2n + 3n = 3. n→+∞

Theorem 1.1.12 Every bounded monotonic sequence (un ) is convergent and sup{un : n ∈ N}, if (un ) is increasing . lim un = inf{un : n ∈ N}, if (un ) is decreasing. Proof: For example, suppose that the sequence (un ) is increasing. The set of terms of the sequence is bounded; therefore, it has a supremum, which we denote by S. Let ε > 0 be arbitrary. There is an element of the sequence, up , such that S − ε < up ≤ S. However, the sequence is increasing; therefore, up ≤ un , ∀n > p. We then have S − ε < un ≤ S, ∀n > p, so un → S. Note: The converse is not true; that is, there are convergent sequences that 1 are not monotonic. For example, the sequence un = (−1)n converges to n 0 and is not monotonic (Fig. 1.14).

Figure 1.14: The sequence un =

(−1)n n

28

Sequences of Real Numbers



• The Sequence

1+

1 n

n  n∈N

Let us consider the sequence with general term .un =



1 1+ n

n . According

to Newton’s Binomial, it can be written as follows:

n  n  p n 1 1 un = 1 + = p n n

 0 p=0

 1

 2

 n n n n n 1 1 1 1 = + + + ··· + 0 1 2 n n n n n 1 n! 1 1 n! n! · + · 2+ · 3+ = 1+ 1!(n − 1)! n 2!(n − 2)! n 3!(n − 3)! n 1 1 n! n! · · + ··· + + (n − 1)!(n − (n − 1))! nn−1 n!(n − n)! nn n! 1 1 n! 1 n! 1 1 1 · + · · 2+ · · 3+ = 1+ · 1! (n − 1)! n 2! (n − 2)! n 3! (n − 3)! n . n! 1 n! 1 1 1 · · · · + + ··· + (n − 1)! (n − (n − 1))! nn−1 n! (n − n)! nn 1 n(n − 1) 1 1 n(n − 1)(n − 2) = 1+ + · + · + 2 1! 2! n 3! n3 n! 1 n! 1 · · + + ··· + (n − 1)! nn−1 n! nn





 1 1 1 1 1 2 = 1+ + 1− + 1− 1− + 1! 2! n 3! n n





 1 1 2 n−1 + ··· + 1− 1− ··· 1 − . n! n n n Since .0 < 1 −

p < 1, for any .1 ≤ p ≤ n − 1, we can write n 1 1 1 + + ··· + . .2 ≤ un ≤ 1 + 1 + 2 3! n!

It is easy to prove by induction that .

1 1 < n−1 , ∀n ≥ 3, n! 2

which allows us to write, taking into account the formula for the sum of the terms of a geometric sequence,  n 1 − 12 1 1 1 1 + 2 + · · · + n−1 = 1 + .2 ≤ un ≤ 1 + 1 + < 1+ = 3. 2 2 2 1 − 12 1 − 12

1.1. Sequences of Real Numbers

29

We have just proved that the sequence is bounded. Let us see that it is monotonic: u

. n+1

− un = 



 1 1 1 2 1− + 1− 1− + n+1 3! n+1 n+1





 1 1 2 n−1 + ··· + 1− 1− ··· 1 − + n! n+1 n+1 n+1





 1 2 n 1 1− 1− ··· 1 − − + (n + 1)! n+1 n+1 n+1 





 1 1 1 1 1 2 − 1+ + 1− + 1− 1− + 1! 2! n 3! n n





 1 2 n−1 1 1− 1− ··· 1 − . + ··· + n! n n n

1 1 = 1+ + 1! 2!

.



By associating terms, we obtain 

 1 1 1− − 1− + n+1 n 







 1 1 2 1 2 + 1− 1− − 1− 1− + 3! n+1 n+1 n n 





 1 1 2 n−1 +··· + 1− 1− ··· 1 − − n! n+1 n+1 n+1 





2 n−1 1 1− ··· 1 − + − 1− n n n





 1 1 2 n + 1− 1− ··· 1 − . (n + 1)! n+1 n+1 n+1

1 un+1 − un = 2!

.



p p > 1− > 0 and the last term is positive, we have n+1 n .un+1 − un > 0, for all .n ∈ N. The sequence

n 1 .un = 1+ n Since .1 −

is bounded and monotonic. By Theorem 1.1.12 we conclude that it is convergent. The limit of this sequence is an irrational number denoted by

Leonhard Euler (1707– 1783) was a Swiss mathematician who studied philosophy and mathematics at the University of Basel. He was a professor in St. Petersburg (Russia) and Berlin. His contributions are important in almost all areas of mathematics, namely, number theory, differential equations, analysis, calculus of variations, and rational mechanics, as well as in other scientific areas, such as calculation of planetary orbits, artillery and ballistics, shipbuilding, and navigation. In 1771, he became blind, but this did not prevent him from continuing an extraordinary scientific production. After his death, St. Petersburg Academy continued to publish his unpublished works for 50 years. Euler introduced the notation .f (x) for a function, e for the base of natural logarithms, i for the root  for summaof .−1, .π, . tion, .Δy for finite differences, and many others. He was the most prolific mathematician of all time. (Source of image: Oil portrait by Jakob Emanuel Handmann (1718–1781), Deutsches Museum)

30

Sequences of Real Numbers

the letter e:

1 . lim 1+ n

n = e.

This number is usually referred to as Neper’s number or Euler’s constant.

John Napier (or Jhone Neper) (1550–1617) was a Scottish mathematician. He has dedicated himself to studies in theology and mathematics. He is essentially known as the inventor of logarithms, although he also has works on spherical triangles. He invented a calculating rule, known as “Napier’s bones” because it was made of ivory, which allowed multiplication, division, and the mechanical calculation of square and cubic roots. (Source of image: Engraving by Samuel Freeman (1773–1857))

x n Theorem 1.1.13 Let .x ∈ R. Then .lim 1 + = ex . n

Theorem 1.1.14 Let .x ∈ R

and u bea sequence such that .lim un = +∞ un x or .lim un = −∞. Then .lim 1 + = ex . un

Example 1.1.22 We know that .xn = n + 5 → +∞. For every .n ∈ N,

.

an =

n+3 n+5



2n+1 =

2 1− n+5



2n+1 =

2 1− n+5

n+5  2n+1 n+5 .

Using the previous theorem, we obtain the following:

2 . lim 1− n+5 Since .

n+5

= e−2 .

2n + 1 → 2, then n+5 .

lim an = (e−2 )2 = e−4 .

• Subsequences Let f and g be two functions. The composition .f ◦ g is only possible if the range of g is contained in the domain of f . Sequences are functions of a natural variable; thus, the composition of two sequences, .u ◦ v, is only possible if the range of v is a subset of .N. Consider two sequences u and v, where v is a sequence of natural numbers. The composition .u ◦ v is still a sequence, of general term .uvn . For example,

1.1. Sequences of Real Numbers if u is the sequence 1, 2, 1, 3, 1, 4, . . . and .vn = 2n − 1, then .uvn = 1; if z = 2n, then .uzn = n + 1; and if .sn = 4, then .usn = 3.

. n

A subsequence of a sequence can be obtained by omitting some of its terms and keeping the remaining terms in the original order. Let us now consider a more formal definition. Definition 1.1.9 Given two sequences u and w, we say that w is a subsequence of u if there exists a strictly increasing sequence of natural numbers v, such that .w = u ◦ v. Example 1.1.23 From the sequences considered earlier, .u ◦ v and .u ◦ z are subsequences of u, but .u ◦ s is not a subsequence of u.

Notes: 1. Every subsequence of a bounded sequence is bounded. 2. An unbounded sequence may have bounded subsequences. For example, ⎧ ⎨n, if n is even .un = ⎩ 1 , if n is odd. n 3. Every subsequence of a monotonic sequence is monotonic. 4. A non-monotonic sequence can have monotonic subsequences, as shown by the sequence of Example 1.1.12. The same can be verified in the sequence whose graph is illustrated in Fig. 1.14.

Theorem 1.1.15 Every subsequence of a convergent sequence converges to the same limit. Proof: Let u be a sequence that converges to a. Let .ε > 0. By definition, we know that ∃ p ∈ N : n > p ⇒ |un − a| < ε.

.

31

32

Sequences of Real Numbers Let w be a subsequence of u. By Definition 1.1.9, there exists a sequence of natural numbers, strictly increasing, v, such that .w = u ◦ v. Since .vn ≥ n, ∀n ∈ N, then if .n > p, also .vn > p and |wn − a| = |uvn − a| < ε,

.

which proves that .wn → a. Note: This theorem is mainly used as a negative test: If two subsequences with different limits are found, the sequence diverges. Theorem 1.1.16 Let u be a sequence. Let .(uvn ) and .(uzn ) be two subsequences such that the union of their indices is .N. If .(uvn ) and .(uzn ) converge to the same limit a, then u converges to a. Proof: Let .ε > 0 be arbitrary. By definition, we know that ∃p1 ∈ N : vn > p1 ⇒ |uvn − a| < ε

.

and ∃p2 ∈ N : zn > p2 ⇒ |uzn − a| < ε.

.

Let .p = max{p1 , p2 }. Observing that every natural number is a term of one of the sequences .(vn ) or .(zn ), then .|un − a| < ε if .n > p. Note: Theorem 1.1.16 remains valid for a finite number of subsequences with the same limit, provided that the union of the indices is .N. Lemma 1.1.1 Every sequence of real numbers has monotonic subsequences. Proof: Let M = {p ∈ N : up < un , ∀n > p}.1

.

If .M is infinite, that is, if there exists .p1 < p2 < · · · < pn < · · · belonging to .M, .up1 < up2 < up3 < · · · < upn < · · · , for example, .(un ) is strictly increasing, .M = N, if .(un ) is decreasing, .M = ∅, 1 and if .un = (−1)n , .M is the set of odd numbers. n 1 If,

1.1. Sequences of Real Numbers

33

and the subsequence .(upn ) is monotonic (increasing). If .M is finite or empty, we will show that there exists a decreasing subsequence .(uqn ). If .M is empty, we consider .q1 = 1. If .M is finite, let .q1 = max(M) + 1. It is clear that .q1 ∈ / M. By the definition of .M, there exists .q2 > q1 such that .uq2 ≤ uq1 . Again, by the definition of .M, there exists .q3 > q2 such that .uq3 ≤ uq2 . Similarly, for each .n ∈ N, there exists .qn > qn−1 such that .uqn ≤ uqn−1 . Therefore, there exists a subsequence .(uqn ) satisfying .uq1 ≥ uq2 · · · ≥ uqn ≥ uqn+1 ≥ · · · , that is, the subsequence .(uqn ) is decreasing. Theorem 1.1.17 Every bounded sequence has convergent subsequences. Proof: By Lemma 1.1.1, the sequence .(un ) admits at least one monotonic subsequence .(upn ). Because .(upn ) is monotonic and bounded, then by Theorem 1.1.12 it is convergent. Definition 1.1.10 A number .a ∈ R is a sublimit of the sequence u if there exists a subsequence of u that converges to a.

1 . The sublimits n 1 of .(un ) are .−1 and 1 because 1 is the limit of subsequence .u2n = 1 + 2n 1 and .−1 is the limit of subsequence .u2n−1 = −1 + (see Fig. 1.15). 2n − 1 Example 1.1.24 Consider the sequence .un = (−1)n +

Figure 1.15: Sublimits of sequence .un = (−1)n +

1 n

34

Sequences of Real Numbers Notes: Let S be the set of sublimits of the sequence u. 1. By Theorem 1.1.17, if u is bounded, .S = ∅. 2. S can be empty. As an example, .un = n, which is not bounded, does not have convergent subsequences. 3. If u is convergent, then S is a unit set, that is, it has only one element. 4. S can be a unit set, and u can be divergent; for example, we have ⎧ ⎨ 1 , if n is even n .un = ⎩n, if n is odd. 5. S can be an infinite set; for example, given the sequence 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, . . . ,

.

then .S = N. Theorem 1.1.18 The set of sublimits of a bounded sequence has a maximum and a minimum. Proof: Let .M > 0, such that .−M ≤ un ≤ M, ∀n ∈ N. Let S be the set of sublimits of sequence u, which is not empty because the sequence is bounded. By Theorem 1.1.5, if .a ∈ S, then .−M ≤ a ≤ M , so S is bounded. Therefore, S has an infimum, .α, and a supremum, .β. We will show that .β ∈ S, that is, .β = max S. By definition of supremum, there exists .a1 ∈ S such that .β − 1/2 < a1 ≤ β; by definition of sublimit, there exists .up1 such that .a1 − 1/2 < up1 < a1 + 1/2; and then, β − 1 < a1 −

.

1 1 < up1 < a1 + < β + 1. 2 2

Again, by definition of supremum, there exists .a2 ∈ S such that β − 1/4 < a2 ≤ β;

.

by definition of sublimit, there exists .up2 such that a − 1/4 < up2 < a2 + 1/4;

. 2

and then, β−

.

1 1 1 1 < a2 − < u p2 < a2 + < β + . 2 4 4 2

1.1. Sequences of Real Numbers Repeating this reasoning, for each .n ∈ N, there exists .upn such that 1 1 < u pn < β + . n n We have constructed a subsequence .(upn ) that converges to .β, so .β ∈ S. Similarly, it can be shown that .α ∈ S. β−

.

Definition 1.1.11 Let u be a bounded sequence and S be the set of sublimits of u. The maximum of S is called the superior limit or upper limit of u and is represented by .lim un = lim sup un = max(S). The minimum of S is called the inferior limit or lower limit of u and is represented by .lim un = lim inf un = min(S). If u is not bounded above, we define .lim un = +∞. If u is not bounded below, we define .lim un = −∞. If .un → +∞, then we define .lim un = lim un = +∞. If .un → −∞, we define .lim un = lim un = −∞. Theorem 1.1.19 A bounded sequence is convergent if and only if lim un = lim un .

.

Proof: If u is convergent, with limit a, every subsequence of u has limit a (Theorem 1.1.15), which implies that .S = {a} and lim un = min(S) = a = max(S) = lim un .

.

Suppose u is bounded and .lim un = lim un = a. If .un → a, then ∃ ε > 0 ∀ p ∈ N ∃ n > p : |un − a| ≥ ε.

.

Thus, there exists .n1 such that .|un1 − a| ≥ ε; there exists .n2 > n1 such that .|un2 − a| ≥ ε; and there exists .nm > nm−1 such that .|unm − a| ≥ ε. Because it is bounded, the subsequence .(unm ) has a convergent subsequence .(unm ): .unm → b = a, since .|unm − a| ≥ ε for all terms. Therefore, we s s s have shown that S has an element b different from a, which is incompatible with the hypothesis of .lim un = lim un = a.

Definition 1.1.12 A sequence u is a Cauchy sequence if ∀ε > 0 ∃ p ∈ N : m, n > p ⇒ |un − um | < ε.

.

35

36

Sequences of Real Numbers

1 Example 1.1.25 We prove that the sequence .un = is a Cauchy sen quence. Let .m, n > p; then   1 1 1  1 1 1 2 − . n m ≤ n + m < p + p = p.

Augustin Louis Cauchy (1789–1857) was a French engineer and mathematician. He was a pioneer of Mathematical Analysis and initiated a project to express and rigorously prove the theorems of calculus, making significant contributions to Complex Analysis. He wrote about 800 works covering all areas of Mathematics and Mathematical Physics. (Source of image: Charles H. Reutlinger, Smithsonian Institution Libraries)

Let .ε denote an arbitrary positive number; to conclude, it is enough to take 2 .p > . ε Note: In the definition of a convergent sequence, we introduced an element external to the sequence, the limit. The sequence converges if, from a certain order, all elements of the sequence “are close” to the limit. In the definition of Cauchy sequence, we compare only the elements of the sequence with each other. We say that the sequence is Cauchy if, from a certain order, all elements of the sequence “are close” to each other. Theorem 1.1.20 A real sequence is convergent if and only if it is a Cauchy sequence. Proof: Let u be a convergent sequence and a be its limit. Let .ε > 0 be arbitrary. By definition, ∃ p ∈ N : |un − a|
p. 2

Let .m, n > p be arbitrary. Then, |um − un | = |um − a + a − un | ≤ |um − a| + |a − un | ≤

.

ε ε + = ε, 2 2

so u is a Cauchy sequence. We must now prove the converse, that is, every Cauchy sequence is convergent. Let u be a Cauchy sequence. We begin by demonstrating that u is bounded. In the definition, we can take, for example, .ε = 1: There exists .p ∈ N such that .|un − um | < 1, ∀m, n > p; in particular, |un | − |up+1 | ≤ |un − up+1 | < 1, ∀n > p,

.

so |un | < |up+1 | + 1, ∀n > p.

.

1.1. Sequences of Real Numbers

37

Considering .M = max{|u1 |, |u2 |, . . . , |up |, |up+1 | + 1}, we obtain |un | ≤ M, ∀n ∈ N.

.

Because sequence u is bounded, it has by Theorem 1.1.17 a convergent subsequence .(uvn ), with limit a. Let .ε > 0 be arbitrary. There exists .q ∈ N such that ε , ∀ vn > q. .|uvn − a| < 2 By definition of Cauchy sequence, there exists .r ∈ N such that |un − um |
r. 2

Let .p = max{q, r}. Let .n > p and .vn > p; then, |un − a| = |un − uvn + uvn − a| ≤ |un − uvn | + |uvn − a|
m; thus, we obtain    1 1 1   |un − um | =  + + ··· + 2 (m + 1)2 (m + 2)2 n 1 1 1 = + + ··· + 2 (m + 1)2 (m + 2)2 n 1 1 1 . ≤ + + ··· + m(m + 1) (m + 1)(m + 2) (n − 1)n





 1 1 1 1 1 1 − − − = + + ··· + m m+1 m+1 m+2 n−1 n 1 1 1 − ≤ . = m n m 1 If .p > and .n ≥ m > p, we obtain .|un − um | < ε, so the sequence is ε Cauchy and therefore converges.

38

Sequences of Real Numbers

1.2

Solved Exercises 

1. Evaluate the limits of the following sequences and provide an argument for the calculations: √ n + nn +3 n+1 b) 2n + 5 a)

c) d) e) f) g) h) i) j) k)

n2

h)

2 n · e1/n √ (−1)n + n2 + 5

2. Calculate the limits of the following sequences: n  2 n −1 a) 2 n  n  n 4 −5 2 b) 4n + 3   n + 2 n+1 c) n+4   2+n n d) 5 + 5n   3n + 1 n e) 3n + 2   2n + 1 4n−2 f) 3 − n   2n + 5 n+4 g) 2n + 1

n earctan(n)

3. Find the limits of the following sequences and justify the calculations: a)

n3 + 3n2 + 5 n2 + 2 35n7 + 1 71n7 + 5n6 + 1 √ 3 n2 + n + n √ √ 4 2 n4 + 1 + n √ n3 + 2n4 + 1 − n √ −2n2 + 3 n2 + 3 √ √ (1 + 2 n ) n + 1 √ n+ 3n √ 3 1 − 27n3 1 + 4n  √  n (−1)n + n √ 2 + n3 + 1 √ n 3 n2 + 2 n2 + (−1)n n

n2 + 3 2n2 + 1

b) c) d) e) f) g) h) i)

3n sin(23n + 1) 23n + 1 1 · cos(n + 1) · log(n) n   n2 + 3 √ n3 + 2 · cos 3 n n +2 1 √ n n! n  n4  √ n4 n n2 e−n − n4 + 1 n sin(n) √ 2n 5n3 + 1  n2 + 2n − n     (n + 1)n+2 1 n − · sin (n + 2)n+1 3 n 3n − 5 5n + 3

4. Find the limits of the following sequences and provide a clear explanation of the reasoning behind each calculation: √ n 3 n−1   sin2 (n) 2n √ c) a) 3 4+k n n2 + 3k2 k=1 k=1

b)

n  k=1

√

5n n4 + k

5. Evaluate the supremum and infimum of the set of terms, and the limit of the sequence .an

6.

=n+

 1 − n2 + 1. 2n

a) Find the limit of the sequence √  √ .an = 2n + 1 − 2n · cos2 (n) and give a clear explanation of the answer. b) Determine the set of sublimits of the sequence nπ · arctan(n) and give a justifibn = sin 2 cation for the answer provided.

1.2. Solved Exercises

39 a) By induction show that xn > 0, ∀n ∈ N.

7. Consider the sequence  n

.un

=

n·(−1)n

1+2

b) Show that the sequence is decreasing. .

a) Write the subsequence of even-indexed terms and calculate its limit. b) Write the subsequence of odd-indexed terms and calculate its limit. c) Calculate lim un and lim un . d) Given the previous items, what can we conclude about the convergence of the sequence? 8. Consider the sequence defined recursively: .

√

u1 = un+1

2 √ = 2 un ,

∀n ∈ N.

a) By induction prove that 0 < un < 2, ∀n ∈ N. b) Prove that the sequence is increasing. c) Prove that the sequence is convergent. d) Evaluate the limit of the sequence. 9. Consider the sequence ⎧ ⎨a1 = √2 . ⎩an+1 = √2 an ,

c) Show that the sequence is convergent and determine its limit. 11. Let a ∈ R be a positive number. Consider the sequence of real numbers defined recursively .

x0 = 0, x1 = a xn+1 = xn + x2n−1 ,

∀n ∈ N.

a) Show that the sequence is increasing. b) Show that xn > 0, ∀n ∈ N. c) Show that if there exists b ∈ R such that lim xn = b, then b = 0. d) Given the previous items, calculate if it exists, lim xn . 12. Consider the sequence of real numbers defined recursively ⎧ ⎨ x1 = 2 . 1 x ⎩xn+1 = n + , ∀n ∈ N. 2 xn a) Prove that xn > 0, ∀n ∈ N. √ 2, ∀n ∈ N.

b) Prove that xn > ∀n ∈ N.

c) Prove that (xn ) is monotonic. d) Show that (xn ) converges.

a) By induction show that √ .

e) Evaluate the limit of (xn ).

2 ≤ an < 2, ∀n ∈ N.

b) By induction show that the sequence (an ) is increasing. c) Show that there exists a ≤ 2 such that an → a. 10. Let a ∈ R be a positive number. Consider the sequence of real numbers defined recursively ⎧ ⎪ ⎨x1 = a .

⎪ ⎩xn+1 =

xn , 2 + xn

13. Consider the sequence of real numbers defined recursively ⎧ ⎪ ⎨ x1 = 3 . x2 + 3 ⎪ , ∀n ∈ N. ⎩xn+1 = n 2 xn a) Show by induction that .xn

−

√

3 > 0, ∀n ∈ N.

b) Show that the sequence (xn ) is decreasing. ∀n ∈ N.

c) Show that the sequence (xn ) converges. d) Find the limit of the sequence (xn ).

40

Sequences of Real Numbers

14. Prove by definition the following limits: a) b) c) d)

25 25n = lim 3n + 1 3 √ n =1 lim √ n+1 √ 2 n + 5−n lim √ = 2 n+1   sin(n) lim 1+ =1 n

e) lim n3 = +∞ n2 = +∞ f) lim n+1 15. The Koch curve is created from a line segment of length 1. This segment is divided into three equal parts. We then remove the middle segment. Next, we replace the removed segment with two segments of length 13 . The resulting shape can be seen in the figure below.

The middle segment of each of the four segments is removed and replaced by two segments of the same . length. The resulting curve has a length of 16 9

Continuing this process ad infinitum, the result is the Koch curve.

What is the length of the Koch curve?

1.2. Solved Exercises

41 SOLUTIONS

1.

n . It is not possible to directly apply Theorem 1.1.5, as the indeterminate form n2 + 3 To solve this, we can rewrite an as follows:

a) Let an =

n .an = = n2 + 3

We know, from Example 1.1.16, that

∞ ∞

occurs.

1 1 n2 · n n   = . 3 3 n2 1 + 2 1+ 2 n n

  1 3 is a null sequence, which by Theorem 1.1.5, implies that 2 → 0. n n

Again by Theorem 1.1.5, . lim an

Since lim

√ n

= 0.

n = 1 (see Note 1 after Theorem 1.1.11), we can conclude that   √ n n n . lim + = 1. n2 + 3

Note: The procedure described here is equivalent to dividing both terms of the fraction by the power of n with the highest exponent that appears in the expression. Alternatively, we could use Theorem 1.1.9, with vn = n, keeping in mind that p = 1 and q = 2. n+1 . Taking into account the note from the previous item, we b) The following sequence is given by an = 2n + 5 divide both members of the fraction that defines the sequence (an ) by n, which is the highest power in the expression. n+1 1 1+ n+1 n n = . = .an = 5 2n + 5 2n + 5 2+ n n   1 Since is a null sequence, we can apply Theorem 1.1.5 and conclude that n . lim an

=

1 . 2

n3 + 3n2 + 5 . Since n3 is the highest power of n that appears in the fraction that defines the n2 + 2 sequence (an ), we can divide the numerator and denominator by n3 :

c) Let an =

.an

=

n3

n3 + 3n2 + 5 3 5 1+ + 3 + +5 n3 n n = = . n2 + 2 1 2 n2 + 2 + n3 n n3 3n2

Using Theorem 1.1.5, we can conclude that . lim an = +∞,   1 because is a null sequence and an > 0, ∀n ∈ N. n

42

Sequences of Real Numbers 35n7 + 1 . If we divide both the numerator and denominator of the fraction that defines 71n7 + 5n6 + 1 the sequence (an ) by n raised to the highest power, n7 , we obtain

d) Let an =

35n7 + 1 n7

1 35 + 7 +1 n .an = = = . 1 5 71n7 + 5n6 + 1 71n7 + 5n6 + 1 71 + + 7 n n n7 35n7

Knowing that

  1 is a null sequence, we can use Theorem 1.1.5, to conclude that n . lim an

=

35 . 71

√ 3

n2 + n + n √ , we can simplify the expression by dividing both the numerator and denomi2n4 + 1 + n nator by n as shown below:   √ 2 3 1 n2 + n + n 3 n +n 3 1 + 2 +1 +1 √ 3 n3 n n n2 + n + n n   . .an = √ =  √ = √ √ =  4 4 4 2n4 + 1 + n 2n4 + 1 + n n 1 1 4 2n + 1 4 + 2+ 4 + n n4 n2 n n

e) Given an = √ 4

Using Theorem 1.1.5, the conclusion is . lim an

√

1 . = √ 4 2

n3 + 2n4 + 1 − n √ . We can rewrite an using the following equalities: −2n2 + 3 n2 + 3   √ 1 1 1 1 n3 + 2n4 + 1 n3 + 2n4 + 1 − n +2+ 4 − √ − 4 3 4 2 n n n n n n + 2n + 1 − n n √ √   .an = = = . = −2n2 + 3 n2 + 3 −2n2 + 3 n2 + 3 2+3 3 1 n 3 3 −2 + + 6 −2 + n2 n4 n n6

f) Let an =

By applying Theorem 1.1.5 to the above expression, we conclude that the limit is given by √ 2 . lim an = − . 2  √ √ 1+2 n n+1 . To simplify the expression of (an ), we can divide both the numerator and g) Let an = √ n+ 3n denominator of the fraction that defines it by n. This gives us   √ √ √ √   1+2 n n+1 1+2 n n+1 1 1  √ √ 1+ √ √ √ +2 n+1 1+2 n n n n n n √   = .an = = . = √ n+ 3n n n+ 3n 1 3 1+ 3 3 1+ n n n2 By applying Theorem 1.1.5, we can determine that . lim an

= 2.

1.2. Solved Exercises √ 3 h) Let an =

43

1 − 27n3 . To find the limit of this sequence, we will manipulate the expression for an to 1 + 4n

simplify it. First, we will divide both the numerator and denominator of the fraction that defines the sequence (an ) by n, and next we simplify the expression. This gives us   √ 3 3 1 − 27n3 1 3 1 − 27n 3 √ − 27 3 n3 1 − 27n3 n n3 = = . = .an = 1 + 4n 1 1 1 + 4n +4 +4 n n n Therefore, by Theorem 1.1.5, 3 =− . 4  √  n (−1)n + n √ i) Consider the sequence (an ) given by an = . In this case, by dividing both the numerator 2 + n3 + 1 and denominator of the fraction that defines the sequence (an ) by n3/2 , we can simplify the expression as follows:  √  √ n (−1)n + n (−1)n (−1)n + n   √ +1 √ 3/2 1/2 n (−1)n + n n n n   √ √ .an = = = . = 2 + n3 + 1 2 + n3 + 1 2 1 2 n3 + 1 √ + 1 + √ + n3/2 n3 n3 n3 n3     1 is a null sequence, we can apply Theorem 1.1.6 to conclude that Since (−1)n is bounded and √ n n (−1) → 0. Moreover, applying Theorem 1.1.5, we can compute the limit of the sequence and obtain √ n . lim an

. lim an

= 1.

√ 3

n n2 + 2 . By dividing both the numerator and the denominator of the fraction that defines n2 + (−1)n n 2 sequence (an ) by n , we obtain   √ √ 3 2 n2 + 2 2 n 3 n2 + 2 3 n +2 3 1 √ + 3 3 n3 n n2 + 2 n2 n n n = = . = 2 .an = = n + (−1)n n (−1)n (−1)n n2 + (−1)n n (−1)n 1+ 1+ 1+ 2 n n n n

j) Let an =

Therefore, according to Theorem 1.1.5, we arrive at the following result: . lim an

= 0.

2 n e1/n 2n √ √ = · e1/n = bn · e1/n . By dividing both the numerator and + n2 + 5 (−1)n + n2 + 5 denominator of the fraction that defines the sequence (bn ) by n, we obtain

k) Let an =

.bn

(−1)n

=

2n √

(−1)n +

n2 + 5

=

2 2 2   √ = = . n (−1)n + n2 + 5 (−1) 5 n 2 (−1) n +5 + 1 + + n n n2 n n2

44

Sequences of Real Numbers Hence, . lim bn

Since lim e

1/n

= 1, we can conclude that . lim

2.

= 2.

2 n e1/n √ = 2. + n2 + 5

(−1)n

a) Consider the sequence (an ) defined as  .an

=

n2 − 1 n2



n

1 n2

1−

=



n

1−

=

1 n2

n2 1/n , ∀n ∈ N.

Using Theorem 1.1.14, we can show that  . lim

Since lim

1−

1 n2

 n2

= e−1 .

1 = 0, we can conclude by Theorem 1.1.5 that n . lim an

= (e−1 )0 = 1.

b) Denoting by an the general term of the sequence, we can write  .an

=

4n − 5 4n + 3

2n

⎡   ⎤2 n 5 1− n ⎢ 1− ⎥ ⎢ 4 ⎥ =⎢   =⎢ ⎢ ⎣ 3 ⎦ ⎣ 4n 1 + n 1+ 4 ⎡

4n

5 4n 3 4n

22n ⎤1/2n ⎥ ⎥ 22n ⎥ ⎦

, ∀n ∈ N.

Using Theorem 1.1.14, we can find the limit of the sequence:  . lim

and

1− 

. lim

But by Theorem 1.1.10, lim

1+

5 4n

3 4n

22n

 n  5 4 = lim 1 − n = e−5 4

22n

 n  3 4 = lim 1 + n = e3 . 4

1 = 0; therefore, 2n  . lim an

=

e−5 e3

0 = 1.

c) If an is the general term of the sequence being discussed, we have  .an

=

n+2 n+4

n+1

 ⎤n+1 ⎡ ⎡  2 n 1+ 1+ ⎢ ⎢ n ⎥ ⎥ =⎢ =⎢ ⎦ ⎣  ⎣ 4 n 1+ 1+ n

  ⎤n+1 2 2 n 1+ 1+ n⎥ n ⎥ n · =  ⎦ 4 4 1+ 1+ n n

2 n , ∀n ∈ N. 4 n

1.2. Solved Exercises

45

Using Theorem 1.1.13, we arrive at the following results:   2 n . lim 1 + = e2 n   4 n 1+ = e4 ; n

and . lim

therefore, . lim an

=

e2 · 1 = e−2 . e4

d) To find the limit of the sequence (an ), we can simplify the expression defining it by putting n in evidence in the numerator of the expression that defines (an ) and 5n in the denominator. This gives us the following expression:   ⎤n  ⎡  2 2 n  n n 1+ 1+ ⎢ 1 n ⎥ n  ⎥ =  , ∀n ∈ N. .an = ⎢ · ⎣ 1 ⎦ 1 n 5 5n 1 + 1+ n n  n 1 We know from Theorem 1.1.10 that lim = 0. Additionally, by Theorem 1.1.13, we can find that 5   2 n . lim 1 + = e2 n 

and . lim

Therefore,

1+

1 n

n = e.

  2 n  n 1+ e2 1 n  =0· . lim an = lim · lim  = 0. 1 n 5 e 1+ n

e) If we consider an as the general term of the sequence, we can write it as follows:  .an

=

3n + 1 3n + 2

n

 ⎤n ⎡  1 3n 1 + 1+ ⎢ ⎢ 3n ⎥  ⎥ =⎢ =⎢ ⎣ ⎣ ⎦ 2 3n 1 + 1+ 3n ⎡

  ⎤n 1/3 n 1 1+ 3n ⎥ n ⎥ =   , ∀n ∈ N. 2/3 n 2 ⎦ 1+ n 3n

According to Theorem 1.1.13, we know that  . lim

1+

1/3 n

1+

2/3 n



and . lim

n = e1/3 n = e2/3 .

Therefore, . lim an

=

e1/3 = e−1/3 . e2/3

46

Sequences of Real Numbers f) We have  .an

=

2n + 1 n

3−

4n−2

     1 4n 1 n 4 3−2− 1− n n =  , ∀n ∈ N.  =   1 2 1 2 3−2− 1− n n

By Theorem 1.1.13, we know that



1 n

1−

. lim

Thus, we can conclude that

n

= e−1 .

4  = e−1 = e−4 .

. lim an

g) We have  .an

=

2n + 5 2n + 1

n+4

  ⎤n+4 ⎡⎛ 5 2n 1 + ⎢⎜ 1 + ⎢ 2n ⎥ ⎢ ⎥  =⎢ = ⎢⎜ ⎣ ⎦ 1 ⎣⎝ 2n 1 + 1+ 2n ⎡

⎞2n ⎤1/2 ⎡ 5 1+ ⎥ ⎢ 2n ⎟ ⎟ ⎥ ·⎢ ⎥ ⎣ ⎠ ⎦ 1 1+ 2n

⎤4 5 2n ⎥ ⎥ , ∀n ∈ N. 1 ⎦ 2n

Using Theorem 1.1.14, we can obtain  . lim

and

1+ 

. lim

1+

Therefore, we can find that

 . lim an

=

5 2n 1 2n e5 e

2n = e5 2n = e. 1/2 = e2 .

π h) Knowing that lim arctan(x) = , we can conclude1 that lim earctan(n) = eπ/2 . Now, let us consider x→+∞ 2  2 n n +3 2 . Putting n in evidence in this expression yields an = 2n2 + 1  .an

=

n2

+3 2n2 + 1

n

   ⎤n  1/n 3 n2 3 n 3  n  n 1 + 1 + ⎥ ⎢ 1 1 n2 n2 n2  ⎥ n = =⎢ · · . ⎦ = 2 ⎣

2n2 1/2n 1 1 2 1 2n2 1 + 1 + 1+ 2n2 2n2 2n2 

⎡

n2

1+

Applying Theorem 1.1.14, we obtain . lim

and

 . lim

1 We

are using the following result: If

  2 3 n 1+ 2 = e3 n

1+

1 2n2

2n2 = e.

lim f (x) = L and f (n) = un , n ∈ N, then lim un = L.

x→+∞

1.2. Solved Exercises Therefore, as lim

47 1 1 = lim = lim n 2n

 n 1 = 0, we can conclude that 2 . lim an

Thus, we get

 . lim

3.

a) Let an =

  3n sin 23n + 1 23n + 1

=

=0·

n2 + 3 2n2 + 1

(e3 )0 = 0. e0

n · earctan(n) = 0.

  3n · sin 23n + 1 . We know that 23n + 1

. |sin(x)| ≤ 1, ∀x ∈ R; %  3n %   % % therefore, sin 2 + 1 ≤ 1, ∀n ∈ N, that is, the sequence with general term sin 23n + 1 is a bounded n 3 sequence. We prove that the sequence with general term 3n is a null sequence. 2 +1  n 3 n n 3 3 8  n = 0. = lim n = lim . lim 1 23n + 1 8 +1 1+ 8

We conclude that the given sequence is a null sequence because it is the product of a null sequence and a bounded sequence (see Theorem 1.1.6). 1 b) Let an = · cos(n + 1) · log(n). We know that −1 ≤ cos(x) ≤ 1, ∀x ∈ R. Then n .

− 1 ≤ cos(n + 1) ≤ 1, ∀n ∈ N.

Therefore, √  √  1 log(n) log(n) n ≤ · cos(n + 1) · log(n) ≤ = log n n , ∀n ∈ N. n =− n n n √ Because we know that lim n n = 1, we have √  n . lim log n = 0. .

− log

It follows from the Squeeze Theorem that . lim an

= 0.

  n2 + 3 n3 + 2 . For all n ∈ N, we have · cos c) Let an = √ n n3 + 2 %  %% % .0 ≤ %cos n3 + 2 % ≤ 1. n2 + 3 √ . By dividing the numerator and denominator of n n3 + 2 the fraction that defines this sequence by the highest power of n, we obtain   6 1 9 n2 + 3 n2 + 3 (n2 + 3)2 + 3 + 5 5/2 5/2 5 n n n n n n  . lim √ = lim √ = lim  = lim = 0. 2 n n3 + 2 n3 + 2 n3 + 2 1+ 3 n n5/2 n3/2 n3

Let us consider the sequence of general term

We conclude that the sequence (an ) is a null sequence because it is the product of a null sequence and a bounded sequence (see Theorem 1.1.6).

48

Sequences of Real Numbers d) Let an =

1 √ n n! = n

 n

n! n! . Let bn = n . It is evident that bn > 0, ∀n ∈ N. nn n

(n + 1)!  n n (n + 1) nn bn+1 (n + 1)! nn 1 (n + 1)n+1 = lim . lim = lim = lim = lim = . n! bn (n + 1)n+1 n! (n + 1)n+1 n+1 e nn By Theorem 1.1.11 we can conclude that lim an = e) Let an =

√ n

 n2 e−n −

n4 4 n +1

 n4

bn+1 . lim = lim bn

1 . e

and bn = n2 e−n . It is evident that bn > 0, ∀n ∈ N.

(n + 1)2  2 2 n 1 en+1 = lim (n + 1) e = 1 lim n + 1 = . n2 en+1 n2 e n e en

By Theorem 1.1.11 we can conclude that lim

√ n

n2 e−n =

result:  . lim

n4 4 n +1

n4 = lim 

1 n4

+1 n4

1 . Using Theorem 1.1.14 we obtain the following e

n4 = lim 

1

1  n4 = e . 1 1+ 4 n

1 1 − = 0. e e  n 1 n sin(n) n √ = · sin(n). We know that ·√ f) Let an = 2 2n 5n3 + 1 5n3 + 1 We conclude that lim an =

. |sin(n)|

therefore, the sequence



≤ 1, ∀n ∈ N;

 sin(n) is a bounded sequence. We show that the sequence

 √

n 5n3 + 1

 is a

null sequence. 1 1 n √ √ 3/2 n n n . lim √ = lim  = lim  = 0. = lim √ 5n3 + 1 5n3 + 1 1 5n3 + 1 5 + n3 n3/2 n3 n

Since lim

 n 1 = 0, we have 2 . lim

 n 1 n ·√ = 0. 2 5n3 + 1

We conclude that the given sequence is a null sequence because it is the product of a null sequence by a bounded sequence (see Theorem 1.1.6).

1.2. Solved Exercises

49

g) It is not possible to directly apply Theorem 1.1.5, as the indeterminate form ∞ − ∞ occurs. Dividing and multiplying the general term of the sequence by its conjugate, we obtain √ √    n2 + 2n − n n2 + 2n + n n2 + 2n − n2 √ lim n2 + 2n − n = lim = lim √ 2 n + 2n + n n2 + 2n + n .

2n 2 2 n = lim √ = lim √ = lim  = 1. = lim  n2 + 2n + n n2 + 2n + n 2 2 n + 2n + 1 1 + +1 n n n2 2n

 h) Let an =

(n + 1)n+2 n − (n + 2)n+1 3

 · sin

  1 . Then n

        1 n + 1 n+2 1 (n + 1)n+2 n n · sin = lim · sin − · (n + 2) − n+1 (n + 2) 3 n n+2 3 n      n+2 1 1 n 1 = lim · (n + 2) · sin − · sin 1− n+2 n 3 n  n+2 n+2        1 1 n 1 1 1 . = lim 1− + 1− − · sin · n · sin · 2 · sin n+2 n n+2 n 3 n    ⎞ ⎛ 1 1     n+2 sin n+2 sin ⎜ 1 1 1 1 n n ⎟ ⎟. ⎜ + 1− · · 2 · sin = lim ⎝ 1 − − · ⎠ 1 1 n+2 n+2 n 3 n n 

lim

 n+2   1 1 1 By Theorem 1.1.14, lim 1 − = e−1 . Since lim = 0, we conclude that lim sin = 0 n+2 n n   1 sin 1 n = 1. Using Theorem 1.1.5, we arrive at the following result: lim an = e−1 − . and lim 1 3 n i) By dividing both the numerator and denominator of the fraction that defines the general term of the sequence by the power of the largest base and exponent, we obtain, by Theorem 1.1.10, 3n − 5 n 3 −5 5n = lim n . lim = lim 5 +3 5n + 3 5n

 n  n−1 3 1 − 5 5 = 0. 3 1+ n 5

4. In this exercise, we apply the Squeeze Theorem. When dealing with sequences defined by summations, a useful technique is to first identify the largest and smallest terms in the summation. Then, these terms can be used to bound the sequence. a) The given sequence has a general term an which is defined as the sum from k = 1 to k = n − 1 of sin2 (n) sin2 (n) . As we show below, the largest term is 2 (which corresponds to k = 1), and the smallest 2 2 n + 3k n +3 2 sin (n) is 2 (corresponding to k = n − 1). Clearly, n + 3(n − 1)2

50

Sequences of Real Numbers

.n

+ 3k2 > n2 , ∀ k, n ∈ N

2

and .n

2

+ 3k2 ≤ n2 + 3(n − 1)2 , ∀ k, n ∈ N such that k ≤ n − 1.

Therefore, for all n and 1 ≤ k ≤ n − 1, we obtain n2 < n2 + 3k2 ≤ n2 + 3(n − 1)2

⇒

1 1 1 > 2 ≥ 2 n2 n + 3k2 n + 3(n − 1)2

⇒

sin2 (n) sin2 (n) sin2 (n) > 2 ≥ 2 . 2 2 n n + 3k n + 3(n − 1)2

.

As (an ) is defined as the sum of n − 1 terms, (n − 1) ·

.

⇔ ⇔

n2

n2

n−1  sin2 (n) sin2 (n) sin2 (n) ≤ < (n − 1) · , ∀n ∈ N 2 + 3k 2 n2 + 3(n − 1)2 n n2 k=1

n−1  sin2 (n) n−1 n−1 · sin2 (n) ≤ < · sin2 (n), ∀n ∈ N 2 2 + 3k 2 2 + 3(n − 1) n n k=1

  n−1  sin2 (n) 1 n−1 1 − 2 · sin2 (n), ∀n ∈ N. · sin2 (n) ≤ < 2 2 2 + 3(n − 1) n + 3k n n k=1

n−1 n−1 . Dividing the numerator and denominator of the fraction that = n2 + 3(n − 1)2 4n2 − 6n + 3 2 defines the sequence by n , we get Let bn =

1 1 1 1 − 2 − 2 n−1 n n n n = lim . lim = lim = 0. 4n2 − 6n + 3 3 6 4n2 − 6n + 3 + 4 − n2 n n2 1 1 − 2 . It is evident that lim cn = 0. n n   As we know that | sin2 (n)| ≤ 1, ∀n ∈ N, the sequence sin2 (n) is bounded. Since the product of a null sequence by a bounded sequence is a null sequence, we can assert that the sequences of general terms Let cn =

.

n−1 · sin2 (n) n2 + 3(n − 1)2 

and .

1 1 − 2 n n

 · sin2 (n)

are null sequences. Finally, as the two limits are equal, the Squeeze Theorem allows us to conclude that . lim an

= 0.

b) The general term an of the sequence is defined as the sum from k = 1 to k = n of √

5n n4 + k

. The largest

5n 5n (which corresponds to k = 1), and the smallest is √ (corresponding to k = n), as n4 + 1 n4 + n shown below. We have  2 . n4 + k > n , ∀ k, n ∈ N, term is √

1.2. Solved Exercises

51

and

 .

n4 + k ≤



n4 + n, ∀ k, n ∈ N such that k ≤ n.

Therefore, for all n and 1 ≤ k ≤ n, we have n2
√ n2 n4 + k n4 + n

⇒

5n 5n 5n ≥ √ . > √ n2 n4 + k n4 + n

.

Because (an ) is defined as the sum of n terms, we obtain n· √ .

√

⇔ 

5n2 = 5· n4 + n 4 sequence by n , we obtain

Let bn = √

5n n4

+n

5n2 n4

≤

+n

≤

n 

√

k=1 n 

√

k=1

5n n4 + k

5n n4

+k

<

n4 , ∀ k, n ∈ N, and .

√ 3

n4 + k ≤

√ 3

n4 + n, ∀ k, n ∈ N such that k ≤ n.

Therefore, for all n and 1 ≤ k ≤ n, we obtain √ 3

n4
√ ≥ √ 3 3 3 n4 n4 + k n4 + n √ √ √ 3 3 3 2n 2n 2n ⇒ √ > √ ≥ √ . 3 3 3 n4 n4 + k n4 + n

n4 + n

.

Because (an ) is defined as the sum of n terms, we obtain √ 3

√ 3 2n , ∀n ∈ N 0, ∀n ∈ N, we can calculate

2n+1

1+ 1 + 2n

. lim

(−1)2k 2k

1+2

1 + 2n+1 1 1 + n+1 2n+1 2 = lim = lim = 2. 1 + 2n 1 1 + 2n+1 2 2n+1

The limit of the sequence (an ) is 2; therefore, all its subsequences have this limit. In particular, the evenindexed subsequence has limit 2. However, this subsequence is equal to the sequence (u2k ). We can affirm that lim u2k = 2. b) The odd-indexed subsequence of (un ) is given by 

2k+1

.u2k+1

and

lim

k→+∞

=

(−1)2k+1 (2k+1)

1+2



2k+1

=

−(2k+1)

1+2

 =

2k+1

1+

1 , k ∈ N, 22k+1

u2k+1 = 1.

c) Taking into account the results obtained in the previous items, .lim un

= 1 and lim un = 2.

d) Since lim un = 1 = lim un = 2, the sequence (un ) is divergent.

8.

a) We will use the Principle of Mathematical Induction to demonstrate that the sequence (un ) satisfies .0

< un < 2, ∀n ∈ N.

For the base case n = 1, we have the following trivial formula: .0

=

√

0
0. Therefore, the sequence is increasing. c) Since the sequence is bounded (as shown in item a)) and increasing (as shown in item b)), we can conclude that the sequence (un ) is convergent because by Theorem 1.1.12, every bounded monotonic sequence is convergent. d) Let b ∈ R be the limit of the sequence. Since every subsequence of a convergent sequence is convergent to the same limit, we have . lim un+1 = b. Hence, .b

= lim un+1 = lim

√

2 un =

√

2 b.

√

Therefore, b satisfies equation b = 2 b, which means that b2 − 2b = 0. Thus, b(b − 2) = 0, and we can exclude the solution b = 0 because by item b), we have .un

≥ u1 =

√

2, ∀n ∈ N.

Therefore, we conclude that the limit of u is b = 2. 9.

√ √ √ x a) Let f be the function defined by f (x) = 2 = ex·log( 2) . Since 2 > 1, f is continuous and increasing on R. √ √ For n = 1, the formula is trivial: 2 ≤ a1 = 2 < 2. √ Induction hypothesis: 2 ≤ an < 2.

Induction thesis:

√

2 ≤ an+1 < 2.

Proof: Assuming that the property is valid for n (induction hypothesis) and using the fact that f is an increasing function, we have √ .

2 ≤ an < 2 ⇒

 √ √2 √  = f 2 ≤ f (an ) = an+1 < f (2) = 2. 2

Using again the monotonicity of f , we obtain .1


0 by hypothesis, which implies that x1 > 0. Induction hypothesis: xn > 0. Induction thesis: xn+1 > 0. xn . Since xn > 0 by induction hypothesis, we know that 2 + xn > 0. Thus, 2 + xn is the quotient of two positive quantities, so xn+1 > 0.

Proof: We have xn+1 = xn+1

Therefore, by the Principle of Induction, we have established that xn > 0, ∀n ∈ N. b) We want to prove that xn+1 − xn < 0 for any n ∈ N. We can simplify this expression as follows: .xn+1

− xn =

xn xn − 2xn − x2n −xn − x2n xn + x2n − xn = = =− . 2 + xn 2 + xn 2 + xn 2 + xn

Since by item a), xn > 0 for all n ∈ N, we have xn + x2n > 0, which implies that − xn+1 − xn < 0.

xn + x2n < 0. Therefore, 2 + xn

c) Since for all n ∈ N we have xn > 0 (item a)) and (xn ) is a decreasing sequence (item b)), we have 0 < xn ≤ x1 , that is, 0 < xn ≤ a, for all n ∈ N. In other words, the sequence of general term xn is bounded. Because every monotone and bounded sequence is convergent, (xn ) is also convergent. Let us assume that lim xn = b. Note that b ≥ 0 because xn > 0 for all n ∈ N.

xn is convergent, 2 + xn since it is the quotient of two convergent sequences where the denominator never becomes zero and has a limit different from zero.

If the sequence converges to b, we also have lim xn+1 = b. Furthermore, the sequence

56

Sequences of Real Numbers Thus, we have lim xn+1 = lim .

b xn lim xn ⇔b= ⇔ lim xn+1 = 2 + xn lim(2 + xn ) 2+b

2b + b2 − b b2 + b b =0⇔ =0⇔ = 0 ⇔ b2 + b = 0 (b = −2) 2+b 2+b 2+b

⇔

b−

⇔

b(b + 1) = 0 ⇔ b = 0 ∨ b = −1.

However, by Theorem 1.1.5, b ≥ 0, which means that lim xn = 0. 11.

a) We want to show that xn+1 − xn ≥ 0 for any n ∈ N0 . Now, if n = 0, we have x1 − x0 = a > 0. If n ≥ 1, then .xn+1

− xn = xn + x2n−1 − xn = x2n−1 ≥ 0.

b) As we saw in item a) that the sequence is increasing, we have xn ≥ x1 = a > 0, ∀n ∈ N. c) Suppose there exists b ∈ R such that b = lim xn . Then, all its subsequences have limit b and .b

= lim xn+1 = lim (xn + x2n−1 ) = b + b2

from where we conclude that b = 0. d) If the sequence is bounded, it will converge because it is monotonic. We have shown in the previous item that if the sequence has a limit, it should be zero. This is not possible, however, since it is an increasing sequence of positive numbers. Therefore, we can conclude that it is unbounded. Since the sequence is increasing, it is bounded below by its first term. So, it has no upper bound. As a result, we can deduce that lim xn = +∞. 12.

a) For n = 1, x1 = 2 > 0. Induction hypothesis: xn > 0. Induction thesis: xn+1 > 0. Proof: Using the given formula xn+1 = x2n + x1 , we can see that x2n and x1 are both greater than zero n n since xn is positive by the induction hypothesis. Therefore, the sum of these two quantities, which is xn+1 , is also greater than zero. By the Principle of Induction, we have proven that xn > 0, ∀n ∈ N. √ b) For n = 1, x1 = 2 > 2. √ Induction hypothesis: xn > 2. Induction thesis: xn+1 >

√

2.

√ 2  √ √ xn − 2 xn x2n + 2 − 2 2xn 1 + Proof: By definition, xn+1 − 2 = − 2= = . As by induction xn 2xn 2xn √ √ 2 √ hypothesis xn > 2, then (xn − 2)2 > 0, and we conclude that xn+1 − 2 > 0. √ Then, using the Principle of Induction, we proved that xn > 2, ∀n ∈ N. √

1.2. Solved Exercises

57

c) Let us analyze the difference xn+1 − xn to determine whether the sequence is increasing or decreasing: .xn+1

− xn =

1 1 xn xn −x2n + 2 + + − xn = − = . 2 xn 2 xn 2xn

√ By item b), xn > 2, ∀n ∈ N; therefore, −x2n + 2 < 0, ∀n ∈ N. Then xn+1 − xn < 0, ∀n ∈ N, thus proving that the sequence is decreasing. d) If a sequence is decreasing, its √ first term is the maximum of the set of its terms; therefore, x1 = 2 ≥ xn , ∀n ∈ N, and (xn ) is bounded: 2 < xn ≤ 2, ∀n ∈ N. Thus, we can conclude that (xn ) is convergent as it is monotonic and bounded. e) Let a = lim xn . All subsequences of (xn ) have limit a and .a

Solving the equation a = √ a = 2. 13.

= lim xn+1 = lim

x

n

2

+

1

1 a2 + 2 a . = + = xn 2 a 2a

√ √ √ a2 + 2 , we get a = − 2 or a = 2. As xn > 2, ∀n ∈ N, we conclude that 2a

√ a) We will show by induction that xn − 3 > 0, ∀n ∈ N. √ If n = 1, then x1 = 3 > 3, and the proposition is satisfied. √ Induction hypothesis: xn − 3 > 0. Induction thesis: xn+1 −

√

3 > 0.

Proof: By definition, xn+1 − √ xn+1 − 3 > 0.

√

3 =

√ √ √ x2n + 3 (xn − 3)2 − 3 = . We know that xn > 3; therefore, 2 xn 2xn

Then, using the Principle of Induction, we proved that xn −

√

3 > 0, ∀n ∈ N.

b) We intend to show that the sequence is decreasing, that is, xn+1 − xn ≤ 0, ∀n ∈ N: .xn+1

− xn =

x2n + 3 −x2n + 3 − xn = . 2 xn 2xn

√ By item a), xn > 3, ∀n ∈ N; therefore, −x2n + 3 < 0, ∀n ∈ N. Then xn+1 − xn < 0, ∀n ∈ N, that is to say, the sequence is decreasing. c) If a sequence is decreasing, its first term is the maximum of the set of terms of the sequence; therefore, √ x1 = 3 ≥ xn , ∀n ∈ N. We have (xn ) bounded: 3 < xn ≤ 3, ∀n ∈ N. We can conclude that (xn ) is convergent because it is monotonic and bounded. d) Let a = lim xn . All subsequences of (xn ) have limit a and .a

Solving the equation a = √ a = 3.

= lim xn+1 = lim

x2n + 3 a2 + 3 . = 2 xn 2a

√ √ √ a2 + 3 , we get a = − 3 or a = 3. As xn > 3, ∀n ∈ N, we conclude that 2a

58 14.

Sequences of Real Numbers 25n 25 = , we must prove that the proposition 3n + 1 3 % % % 25n 25 %% .∀ε > 0 ∃ p ∈ N : n > p ⇒ %% − 0. Solving the inequality for n, we obtain % % % % % 75n − 75n − 25 % % 25n 25 25 %% 25 − 3ε % p, we get n > , that is, 9ε 9ε % % % 25n 25 %% < ε. − . %% 3n + 1 3 %

Figure 1.16: If ε = 0.2, then −ε
13

Suppose, for instance, that we want to calculate the smallest order p from which % % % 25n 25 %% < 0.2. − . %% 3n + 1 3 % 25 − 3ε , we get the value 13.5; therefore, as p ∈ N, we can choose 9ε p = Int(13.5) = 13. Thus, if n > 13, the desired result follows. This case is illustrated in Fig. 1.16. √ n = 1, we should prove that the proposition b) In order to prove by definition that lim √ n+1 Substituting ε by 0.2 in the expression

.∀ε

% % √ % % n − 1%% < ε > 0 ∃ p ∈ N : n > p ⇒ %% √ n+1

1.2. Solved Exercises

59

is true, that is, we need to show that for each positive real ε there is at least one order after which, for every natural n, % √ % % % n . %% √ − 1%% < ε. n+1 Let ε > 0. Solving the inequality for n, we obtain %√ % % % √ √ % n − n − 1% % % √ n 1−ε % 0, then the previous inequalities are still equivalent to   1−ε 2 .n > . ε     1−ε 2 1−ε 2 If we choose p ∈ N such that p ≥ , then, as n > p, it follows that n > , that is, ε ε % √ % % % n . %% √ − 1%% < ε. n+1

Figure 1.17: If ε = 0.2, then −ε
16

Suppose, for example, that we want to calculate the smallest order p from which % % √ % % n − 1%% < 0.2. . %% √ n+1   1−ε 2 Substituting ε by 0.2 in the expression , we get the value 16; therefore, we can choose p = 16. ε Thus, if n > 16, the desired result follows. This case is illustrated in Fig. 1.17. √ 2 n + 5−n = 2. For this, we must prove that the √ n+1 % % √ % % 2 n + 5−n − 2%% < ε .∀ε > 0 ∃ p ∈ N : n > p ⇒ %% √ n+1

c) We intend to prove, using the definition, that lim proposition

60

Sequences of Real Numbers is true, that is, we need to show that for each positive real ε there is at least one order after which, for every natural n, % % √ % % 2 n + 5−n − 2%% < ε. . %% √ n+1 Let ε > 0. % √ % −n % √ % % % √ −n % 2 n + 5−n − 2 n − 2 % % % 2 n + 5−n % % % < ε ⇔ % 5√ − 2 % < ε ⇔ 2√− 5 − 2%% < ε ⇔ %% < ε. . %% √ √ % n+1 % % n+1 n+1 n+1

Figure 1.18: If ε = 0.45, then −ε
11

2 − 5−n 2 < √ , ∀n ∈ N; √ n+1 n+1

therefore, if .√

then .

2 < ε, n+1

2 − 5−n < ε. √ n+1

Solving the inequality for n, we obtain .√

√ 2 2−ε

. n+1 ε

If 2 − ε ≤ 0, the last inequality is true regardless of the choice of p ∈ N. If 2 − ε > 0, then the previous inequalities are still equivalent to   2−ε 2 .n > . ε     2−ε 2 2−ε 2 If we choose p ∈ N such that p ≥ , then, as n > p, it results in n > , that is, ε ε % % √ % % 2 n + 5−n − 2%% < ε. . %% √ n+1

1.2. Solved Exercises

61

Suppose, for example, that we want to calculate an order p from which % % √ % % 2 n + 5−n − 2%% < 0.45. √ n+1

. %%



 2−ε 2 , we get the value (3, 4)2  11.86; therefore, as p ∈ N, ε we can choose p = Int(11.86) = 11. Thus, if n > 11, the desired result is obtained. This case is illustrated in Fig. 1.18.   sin(n) = 1, we must prove that the proposition d) In order to prove by definition that lim 1 + n Substituting ε by 0.45 in the expression

.∀ε

% %  % % sin(n) − 1%% < ε > 0 ∃ p ∈ N : n > p ⇒ %% 1 + n

is true, that is, we need to show that for each positive real ε there is at least one order after which, for every natural n, % %  % % sin(n) . %% 1 + − 1%% < ε. n Let ε > 0.

% % % %  % sin(n) % % % sin(n) % < ε ⇔ |sin(n)| < ε. 1+ − 1%% < ε ⇔ %% n n % n

. %%

Figure 1.19: If ε = 0.05, then −ε < 1 + But .

sin(n) n

− 1 < ε if n > 20

|sin(n)| 1 ≤ , ∀n ∈ N; n n

therefore, if .

we have .

1 < ε, n

|sin(n)| < ε. n

62

Sequences of Real Numbers Solving the inequality for n, we obtain .

If we choose p ∈ N such that p ≥

1 1 . n ε

1 1 , then, as n > p, it results in n > , that is, ε ε .

|sin(n)| < ε. n

Suppose, for example, we want to calculate an order p from which % %  % % sin(n) . %% 1 + − 1%% < 0.05. n 1 , we get the value 20; therefore, as p ∈ N, we can choose p = 20. ε Thus, if n > 20, the desired result is obtained. This case is illustrated in Fig. 1.19. Substituting ε by 0.05 in the expression

e) To prove that lim n3 = +∞ by definition, we need to show that the following proposition is true: .∀L

∈ R+ ∃ p ∈ N : n > p ⇒ n3 > L.

In other words, we must show that for each positive real L, there is at least one order after which, for every natural n, n3 > L. √ Let L > 0. Solving the inequality for n, we have n3 > L ⇔ n > 3 L. Therefore, if we choose p ∈ N such √ 3 3 that p > L, we have proven that lim n = +∞. √ In Fig. 1.20 you can see that by choosing L = 1000, then from n > p = 3 1000 = 10, all terms of the sequence are greater than L.

Figure 1.20: If L = 1000, then n3 > L if n > 10 f) In order to prove by definition that lim .∀L

n2 = +∞, we must prove that the proposition n+1

∈ R+ ∃ p ∈ N : n > p ⇒

n2 >L n+1

is true, that is, we must show that for each positive real number L, there is at least one order after which, for every natural number n, n2 > L. . n+1

1.2. Solved Exercises

63

Figure 1.21: If L = 10, then

Let L > 0. We have .

n2 > L ⇔ n2 − Ln − L > 0 ⇔ n+1

 n−

n2 n+1

L−

> L if n > 10

√

L2 + 4L 2

 n−

L+

Because L > 0 and n ∈ N, we obtain from the previous inequality that n >   √ n2 L + L2 + 4L , then for n > p, we get > L. we choose p = Int 2 n+1 In Fig. 1.21, you can see that if we choose L = 10, then from   √ 10 + 140 = 10 .n > p = Int 2

√

L+

L2 + 4L 2

√

 > 0.

L2 + 4L . Therefore, if 2

all the terms of the sequence are greater than L. 15. Let us denote Nn as the number of segments, Ln as the length of each segment, and ln as the length of the curve in the nth iteration, that is, after repeating n times the process of constructing the curve. We will prove by induction that Nn = 4n . If n = 1, we have four segments by construction, so N1 = 41 = 4. Suppose (induction hypothesis) that Nn = 4n . Then Nn+1 = Nn × 4 since each segment originates four new segments. Substituting the expression of Nn , we get Nn+1 = 4n × 4 = 4n+1 . 1 In each iteration, we divide each segment by 3. During the first iteration, the length of each segment is L1 = . 3  n  n+1 1 1 Assuming (induction hypothesis) that Ln = , we can prove that Ln+1 = as follows: 3 3  1 n  n+1 1 Ln = 3 = .Ln+1 = . 3 3 3  n  4  n The length of the curve in the nth iteration is ln = Nn × Ln . We know that Nn = 4n , so ln = 4n 13 = 3 . The length L of the Koch curve is determined by the limit of ln when n → +∞. Then, by Theorem 1.1.10,  n 4 .L = lim = +∞. 3

64

Sequences of Real Numbers

1.3

Proposed Exercises b) Prove that n! grows faster than en .

1. Prove by definition that lim un = +∞ in each item listed below: √

a) un = n

c) un =

b) un = n2

d) un = 2n

n

2. By definition, prove that the following sequences (un ) are null sequences, that is, that lim un = 0: a) un =

1 n

1 b) un = 2 n

1 c) un = √ n 1 d) un = n 2

3. Evaluate the limits of the following sequences and provide an argument for the calculations: a)

1−n 5n + 3

b)

n2 + 2 3n + 1

n2 − 1 n2 + 3n − n+2 n 3n d) 4n3 + 1 c)

e)

−n3 + 2 4n3 − 7

4. Find the limits of the following sequences and justify the calculations: √ n a) 4n + 1 √ n b) 1 √ − n 2   n2 + 1 − n2 − 1 c)  √ √  1 d) n+1− n n+ 2 1 1 1 +√ + ··· + √ e) √ 2 2 2 n +n n +1 n +2 5. We say that the sequence (un ) grows faster than the un → +∞. sequence (vn ) if vn a) Prove that nn grows faster than n!.

c) Arrange the following sequences in descending order in terms of the speed of convergence: √ .2 n, 10 n, 2n , en , n!, log(n), n3 , nn . 6. Let (un ) and (vn ) be two null sequences such that vn = 0, ∀n ∈ N. We say that (un ) is of higher un = 0. Order the following order than (vn ) if lim vn null sequences: .

1 1 1 1 1 1 1 1 , , , , . , √ , , 2n 10 n 2n en n! log(n) n3 nn

7. Calculate the limit of the sequences with general terms:    2  n + 3 2n 5n + 5 n a) c) n+1 2n + 1  n   3 n n+5 b) d) 1 − 2 2n + 1 n 8. Evaluate, if it exists: 1  n (n + 1)! 2n  1 b) lim n n(n + 1) · · · 2n n a) lim

 9. Determine p ∈ R such that lim

n

n! = 3. (p n)n

10. Find the limits of the following sequences and provide a clear explanation of the reasoning behind each calculation:  n a) cos(x) , x ∈ R   1 b) cos2 (n) sin n n (n − 1) (n − 2) (n − 3) (n + 1) (n + 2) (n + 3)    2 n d) n n! n c)

1 1 1 +√ + ··· + √ e) √ n2 + 1 n2 + 2 n2 + 2n + 1

1.3. Proposed Exercises  f) g)

1−  n

1 n

n  n

65

n+1 n

(n + 1)! − n!

1 1 1 h) √ + √ + ··· + √ n n+1 2n 1 1 1 + + ··· + n2 (n + 1)2 (2 n)2 n n n j) √ +√ + ··· + √ n4 + 1 n4 + 2 n4 + n

15. Consider the sequence of real numbers defined recursively ⎧ ⎪ ⎨ x0 = 1   . 1 2 ⎪ , ∀n ∈ N0 . xn + ⎩xn+1 = 2 xn

i)

a) Show by induction that √ .

11. For each item, provide, when possible, examples of sequences u, v, and w such that un → +∞, vn → −∞, and wn → 0. a) un + vn → 1 b) un + vn → −∞

2 ≤ xn ≤ 2, ∀n ∈ N.

b) Show that xn ≥ xn+1 , ∀n ≥ 1. c) Show that the sequence is convergent and calculate its limit. 16. The sequences (un ) and (vn ) verify the following conditions:

c) un + wn → 1 d) un × wn → 0

(i) ∀n ∈ N 0 < un < vn .

e) vn × wn → +∞ un → −1 f) wn

(ii) (vn ) is decreasing. Decide whether the following statements are true or false and give a justification for the answers provided.

12. Let (xn ) and (yn ) be two convergent sequences of real numbers such that xn → x and yn → y. Show that the sequence of general term zn = min{xn , yn } converges and that zn → min{x, y}.

a) (vn ) is convergent. b) (un ) is convergent. c) (un ) is decreasing.

13. Consider the sequence (un ) defined recursively by ⎧ ⎨u1 = 1, . ⎩un = un−1 − 1 , 2

∀n > 1.

n

Explain why (un ) is convergent and evaluate its limit. 14. Consider the sequence (un ) defined recursively ⎧ ⎨ u1 = 5 .

⎩un+1 =

5un − 4 un

17. Determine the upper and lower limits of the sequences with the following general terms and provide a reason for the answers:

∀n ≥ 1.

a) Prove by induction that ∀n ∈ N, un > 4. b) Prove that the sequence is convergent. c) Show that 4 is the infimum of the set of terms of the sequence.

a) n(−1) nπ

b) cos 3 √ √ c) n − (−1)n n − 1 nπ

d) sin 4 √ √ e) n − (−1)n n − 1 nπ

n nπ 1 f) 2 cos + cos n 10 2 (−1)n n2 + 3 g) n+1 nπ

h) sin +a , a∈R 2  n   1 1 i) + + 2 n (−1)n 3 + 3 3 2n

66

Sequences of Real Numbers  j)

 (−1)n+3 − (−1)n n3 + 2

Hint:

3n + 1

18. Show that the following are Cauchy sequences in Q:  a)

1 n2



 b)

1 2n



19. Show that the sequence of general term .1

+

1 1 + ··· + 2 n

is not a Cauchy sequence in Q. 20. Consider the sequence of general term .un

=

n+1 . n+2

Check the sequence for convergence using the definition of a Cauchy sequence.

(i) Show that vn = x2n − 2 satisfies 0 ≤ vn ≤ (ii) Use the relation xn − xm =

x2n − x2m . xn + xm

22. The Sierpinski carpet is a set constructed as follows: Consider a square, R, with side 1. Divide R into nine squares of equal area and remove the interior of the middle square. Repeat the process in the remaining eight squares: Divide each one into nine squares of equal area and remove the interior of the middle square. Let Rn be the region that remains after this process has been performed n times. The Sierpinski carpet, S, is the set of all points in R that are not removed by any of the previous operations, that is, S consists of the points in R that belong to Rn for any n ≥ 1. The figure illustrates the first three steps of this process.

21. Show that the sequence ⎧ 3 ⎪ ⎨x 1 = , 2 . 1 x ⎪ ⎩xn+1 = n + , 2 xn

∀n ∈ N

is a sequence in Q such that x2n → 2. Use this result to show that (xn ) is a Cauchy sequence in Q that does not converge in Q.

1 . 4n

a) Calculate the area An of Rn , n ≥ 1. b) Show that lim An = 0.

Numerical Series 2

The operation of adding two numbers extends without any difficulty to a finite number of terms, even if this number is very large. However, considering an infinite number of terms raises more delicate issues addressed in this chapter.

2.1

Generalization of the Addition Operation

Addition (or sum) is the operation that corresponds to each pair of real numbers with another real number, according to specific rules. This operation satisfies certain properties and can be generalized to a finite number of terms while maintaining all properties. The definition of the sum of a finite number of terms is made recursively as follows: ⎧ a , if n = 1 ⎪ ⎪ n ⎨ 1  n−1  . ai =  ⎪ ⎪ ai + an , if n > 1. i=1 ⎩ i=1

We can now consider generalizing the notion of sum to an infinite number of terms, .a1 , a2 , . . . , an , . . . If there is an order p from which all terms of the sequence are zero, the sum of all terms equals the sum of the first p terms: p   . ai = ai . n∈N

i=1

If there is a subsequence of nonzero terms, let us consider .Sn =

n 

ai , the

i=1

sum of the first n terms. It is natural to consider the sum of all terms as the limit of the sequence .(Sn ) if it exists and is finite. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 A. Alves de S´a, B. Louro, Sequences and Series, https://doi.org/10.1007/978-3-031-67202-6 2

67

68

Zeno of Elea (490–435 B.C.) Greek philosopher and disciple of Parmenides. Most of his teaching records have come to us through secondary sources, notably through Aristotle. Zeno’s great fame came from his paradoxes. These paradoxes aimed to defend his master Parmenides’ theory that reality was unique, immutable, and immobile so that change, movement, time, and plurality would be nothing more than illusions. Parmenides’ theory presented conclusions that contradicted what the senses conveyed; Zeno’s goal was not to present evidence that reinforced the theory itself but to show that opposing theories also led to contradictions. (Source of image: Jan de Bisschop, Paradigmata graphices variorum artificum, Rijksmuseum.)

Numerical Series If the terms of the sequence .(an ) are all positive, it might seem at first glance that .(Sn ) is not convergent. In fact, assuming that the sum of an infinite number of positive terms is a real number is not an intuitive concept. In this case, intuition fails precisely because we intend to generalize to infinity a concept, that of sum, which we have intuitive for a finite number of terms. It is common for intuition to deceive us in cases of “passing” from finite to infinite. It is true that we will not always be able to find the limit of the sequence .(Sn ), so, by this process, we will not be able to find the sum of an infinite number of terms if it exists. Nevertheless, we are interested in knowing how the sequence .(an ) should be so that a real number, the sum of all its terms, is associated with this sequence. One of the historical examples illustrating these difficulties in dealing with the concept of an infinite sum of terms is one of the famous Paradoxes of Zeno, which we summarize as follows: A walker wants to move from one place to another at a constant speed (Fig. 2.1). In T minutes, he covers half of the distance. He still has the other half to go. In .T /2 minutes, he will cover half of what remains (.1/4 of the total distance). Moreover, now he has .1/4 of the distance left. He will cover half of the rest (.1/8 of the total) in .T /4 minutes. This argument may be repeated indefinitely. The walker spends T+

.

T T T T + + + + ... 2 4 8 16

minutes on his journey. Given that it is the sum of infinite positive terms, we are inclined to assume that the sum is infinite. The walker will not be able to reach his goal because it will take an infinite amount of time. This reasoning contradicts intuition and practical life, which is called a paradox. This is due to the lack of understanding of the concept of an infinite sum of terms (more precisely, of series), which came to be fully understood only in the 19th century. In this case, the walker will reach his goal in time 2T , as intuition tells us.

Figure 2.1: The walker’s path in Zeno’s Paradox

2.2. Definition of Series: Convergence – General Properties

2.2

Definition of Series: Convergence – General Properties

Definition 2.2.1 Let (an ) be a sequence of real numbers. The series generated by an is the sequence (Sn ) defined as follows: S1 = a1 S2 = a1 + a 2 S3 = a1 + a 2 + a 3 . . .. . Sn = a1 + a2 + a3 + · · · + an .. . To represent a series, we can use any of the following notations: ∞  .



an ,

an ,

a1 + a2 + a3 + · · ·

n=1

The numbers a1 , a2 , . . . , are called terms of the series, an is said to be the general term of the series, and the sums S1 , S2 , . . . , are called partial sums. Definition 2.2.2 The series .



an is convergent if the limit

lim Sn = lim

n 

ai

i=1

exists and is finite. If this limit does not exist or is not finite, the series is divergent. In the case of convergence, the sum of the series is the value S of the limit of (Sn ), that is, ∞  .S = lim Sn = an . n=1

∞ Note: The identification of a series with the symbol n=1 an is a misuse of language because it is the identification of a series with its sum when it exists. However, despite being a common misuse of language, it has been found to be a valuable and harmless convention.

69

70

Numerical Series

Example 2.2.1 An important example is the geometric series. A geometric series is a series generated by a geometric sequence: If (an ) is a geometric sequence of ratio r = 1 and a1 = 0, then Sn =

n 

.

ai =

i=1

n 

a1 ri−1 = a1 ·

i=1

1 − rn . 1−r

We know that (Sn ) is convergent if and only if |r| < 1, so the geometric series is convergent if and only if the absolute value of the ratio of the geometric sequence that generated it is less than 1. In this case ∞  .

an =

n=1

a1 . 1−r

The series in Zeno’s Paradox (see Sect. 2.1) is geometric with a1 = T and r = 1/2; it is convergent with the sum 2T . If r = 1, the general term is constant; that is, ∞  .

an =

n=1

∞ 

a1 ,

n=1

and thus, Sn = na1 and, if a1 = 0, the series will diverge.

Example 2.2.2 The series ∞  .

(−1)n

n=1

3n e2n+1

=

n

∞  1 3 − 2 e e n=1

3 3 is geometric with ratio r = − 2 and first term − 3 . Since −1 < r < 1, e e the series converges and its sum is .

− e33 −3 . = e (e2 + 3) 1 − (− e32 )

Example 2.2.3 We can write the rational number 2.325 = 2.3252525 . . . in the form of a fraction using a geometric series. 2.325 = 2.3 + 0.025 + 0.00025 + . . . =

.

25 23 25 25 + + 5 + 7 + ··· 10 103 10 10

2.2. Definition of Series: Convergence – General Properties 1 After the first term, we have a geometric series with ratio r = 2 and first 10 25 term 3 . Therefore, 10 2.325 =

.

25 23 1151 3 + 10 1 = . 10 1 − 102 495

∞  1 √ ; let us study the sequence of n n=1 its partial sums and the corresponding limit:

Example 2.2.4 Consider the series

S1 = 1 1 S2 = 1 + √ 2 1 1 S3 = 1 + √ + √ 2 3 . .. . 1 1 1 Sn = 1 + √ + √ + · · · + √ n 2 3 .. . As √ 1 1 1 1 1 1 n 1 1 + √ + √ + ··· + √ ≥ √ + √ + √ + ··· + √ = √ = n n n n n n n 2 3 √ and lim n = +∞, we conclude by Theorem 1.1.2 that the sequence (Sn ) has limit +∞. The series under study is divergent.

.

∞  1 , commonly known as the harn n=1 monic series. By mathematical induction, we can prove that the subsequence (S2n ) of the partial sums of this series satisfies the inequality

Example 2.2.5 Consider the series

S 2n ≥ 1 +

.

n , ∀n ∈ N. 2

For n = 1, the proposition is true: S2 = 1 +

.

1 1 ≥1+ . 2 2

71

72

Numerical Series Assuming that the proposition is valid for n, we prove that it is also valid for n + 1. S2n+1 = S2n +

.

2n

1 1 1 n n+1 + · · · + n+1 ≥ 1 + + 2n n+1 = 1 + . +1 2 2 2 2

Thus, the desired inequality is proved using the method of induction. By Theorem 1.1.2, we can conclude that the sequence (S2n ) has limit +∞, implying that the sequence of partial sums of the harmonic series does not have a finite limit. Therefore, the harmonic series is divergent. Figure 2.2 shows the sequence of partial sums of the harmonic series and its general term. It illustrates how the sum of an infinite number of increasingly smaller positive quantities can have a value as large as desired. Figure 2.2: The sequences 1 1 an = n and Sn = 1 + · · · + n

Example 2.2.6 Consider the series

.

∞ 

1 . By observing that n(n + 1) n=1

1 1 1 = − , n(n + 1) n n+1

we can write the sequence of partial sums as follows (see Fig. 2.3): 1 2 1 1 1 S2 = 1 − + − = 1 − 2 2 3 1 1 1 + − =1− . S3 = 1 − 3 3 4 .. . 1 Sn = 1 − n+1 .. . S1 = 1 −

Figure 2.3: The sequences 1 1 and Sn= 1− n+1 an= n(n+1)

1 3 1 4

As lim Sn = 1, the series is convergent and its sum is 1: ∞ 

.

1 = 1. n(n + 1) n=1

Example 2.2.7 The sequence of partial sums of the series ∞ 



n . log n+1 n=1

 =

∞  n=1

log(n) − log(n + 1)

2.2. Definition of Series: Convergence – General Properties is the sequence S1 = log(1) − log(2) = − log(2) S2 = − log(2) + log(2) − log(3) = − log(3) S3 = − log(3) + log(3) − log(4) = − log(4) .

As lim



.. . Sn = − log(n + 1) .. .

− log(n + 1) = −∞, the series diverges.

∞  1 can be exExample 2.2.8 The general term of the series 2 n + 3n n=1

 1 1 1 pressed as − . The sequence of partial sums can be con3 n n+3 structed using this expression:

S1 =

1 3

S2 =

1 3

1 S3 = 3

.

1 = 3 1 S4 = 3 =

1 3

=

1 3

1 S5 = 3





1−

1 4

1−

1 4

1−

1 1 + − 4 2

1−

1 1 + − 4 2

1−

1 1 + − 4 2

1−

1 1 + − 4 2

1+

1 1 − + 2 5

1+

1 1 − + 2 5











 +

1 3





 1 1 1 1 1 1 − = 1− + − 2 5 3 4 2 5 

 1 1 1 1 − + 5 3 3 6  1 1 1 + − 5 3 6 

 1 1 1 1 1 1 + − − + 5 3 6 3 4 7  1 1 1 1 1 + − + − 5 3 6 4 7  1 1 1 − − 3 6 7 

 1 1 1 1 1 1 − − − + 3 6 7 3 5 8

73

74

Numerical Series

 1 1 1 1 1 1 1 1+ − + − − + − 2 5 3 6 7 5 8

 1 1 1 1 1 1 1+ + − − − = 3 2 3 6 7 8 . .. .

 1 1 1 1 1 1 − − 1+ + − Sn = 3 2 3 n+1 n+2 n+3 .. .

 1 1 1 11 , the series is convergent and As lim Sn = 1+ + = 3 2 3 18 1 = 3

∞ 

.

11 1 = . 2 + 3n n 18 n=1

The last three examples are particular cases of a type of series called telescopic series. These are series whose general term an can be expressed in the form αn − αn+k , with k ∈ N: ∞  .

(αn − αn+k ),

(2.1)

n=1

allowing us to determine the expression of the sequence of partial sums by canceling consecutive terms and, consequently, in case the sequence converges to easily calculate the sum of the series. These series are convergent if and only if lim vn , where vn = αn+1 + · · · + αn+k , exists and is finite. In the particular case of existing, finite, lim αn , we have: ∞  .

(αn − αn+k ) =

n=1

k 

αi − k a,

(2.2)

i=1

where a = lim αn . In fact, the sequence of partial sums is given by .

Sn =

n 

(αi − αi+k ) =

i=1

n  i=1

αi −

n 

αi+k

i=1

= α1 + · · · + αk + αk+1 + · · · + αn −(αk+1 + · · · + αn + αn+1 + · · · + αn+k ) .

= α1 + · · · + αk − (αn+1 + · · · + αn+k ) =

k  i=1

αi −

k  i=1

αi+n

2.2. Definition of Series: Convergence – General Properties If (αn ) is convergent, then lim αi+n exists and lim αn = lim αi+n from which it is concluded that

.

lim Sn =

k 

 αi − lim

i=1

k 

 αi+n

i=1

=

k 

αi − k a.

i=1

The following theorem establishes a necessary but not sufficient condition for the convergence of a series. Therefore, it is primarily used to determine whether a series is divergent. If the general term of a series is not a null sequence, the series will be divergent.

Theorem 2.2.1 If the series sequence.

an is convergent, then (an ) is a null

Proof: Since the series is convergent, the sequence Sn = convergent. This is true for Sn−1 as well, and we have Then .

n 

i=1 lim Sn =

ai is also lim Sn−1 .

lim an = lim (Sn − Sn−1 ) = lim Sn − lim Sn−1 = 0.

Example 2.2.9 The series

∞ 

n n is divergent because lim = 1. n + 1 n + 1 n=1

Example 2.2.10 Consider the series

.

∞  1 √ . Although n n=1

1 lim √ = 0, n

we cannot use Theorem 2.2.1 to determine if the series converges or diverges. However, in Example 2.2.4, we have already proved that this series is divergent.

75

76

Numerical Series Theorem 2.2.2 Let an and bn be convergent series with sums A and B, respectively, and λ ∈ R. Then: a) The series (an + bn ) is also convergent, and its sum is A + B: ∞ ∞ ∞    . (an + bn ) = an + bn . n=1

b) The series



n=1

n=1

λan is convergent, and its sum is λA: ∞ ∞   . λan = λ an . n=1

n=1

Proof:

an a) Let (Sn∗ ) and (Sn∗∗ ) be the sequences of partial sums of the series and bn , respectively. Because they are convergent with sums A and B, we have ∗ . lim Sn = A and lim Sn∗∗ = B. Let (Sn ) be the sequence of partial sums of the series (an + bn ), that is, Sn =

n 

.

(ai + bi ) =

i=1

n 

ai +

i=1

n 

bi = Sn∗ + Sn∗∗ .

i=1

Then .

lim Sn = lim(Sn∗ + Sn∗∗ ) = lim Sn∗ + lim Sn∗∗ = A + B,

that is, the series



(an + bn ) is convergent and has sum A + B. an . By hypothb) Let (Sn∗ ) be the sequence of partial sums of the series esis, lim Sn∗ = A. Let (Sn ) be the sequence of partial sums of the series λan . Then n n   .Sn = λai = λ ai = λSn∗ . i=1

i=1

Thus, ∗ ∗ . lim Sn = lim λSn = λ lim Sn = λA, that is, the series λan is convergent and has sum λA.

2.2. Definition of Series: Convergence – General Properties Notes: 1. From the proof of item a), it is evident that the given series may diverge, and yet the series (an + bn ) can still converge. The proof also indicates that if the sequences of partial sums have infinite limits of the same sign, that is, both series are divergent, and the sequence of partial sums will be divergent. The same thing will happen if one of the series is convergent and the other divergent. If (Sn∗ ) and (Sn∗∗ ) have infinite limits with opposite signs, the series (an + bn ) may be convergent or divergent because an indeterminate form appears in the calculation of the limit. 2. Based on the proof of part b), it can be concluded that if the value of λ is not equal to 0, then the series λan will converge if and only if the series an also converges. However, if the value of λ equals 0, then the series λan will converge since all its terms are zero. In order to better understand the previous theorem, we will examine an example that involves the application of our knowledge about telescopic series. Example 2.2.11 Consider the series ∞ 

.

1 . n(n + 3)(n + 6) n=1

We can decompose the general term as follows:



 1 1 1 1 1 1 1 = − − . − . n(n + 3)(n + 6) 18 n n + 3 18 n + 3 n + 6 The series

∞

 1 .

n=1

1 − n n+3



1 and k = 3. As lim αn = 0, the series is converis telescopic with αn = n 11 1 1 . gent, and its sum is α1 + α2 + α3 = 1 + + = 2 3 6 The series  ∞

 1 1 − . n+3 n+6 n=1

77

78

Numerical Series 1 and k = 3. In this case too, lim αn = 0; n+3 37 1 1 1 . therefore, the series is convergent and its sum is + + = 4 5 6 60 According to the previous theorem, the given series is also convergent and

is also telescopic with αn =

∞ 

1 n(n + 3)(n + 6) n=1 .

  ∞

∞

1 1 1 1  1 1  − − = − 18 n=1 n n + 3 18 n=1 n + 3 n + 6 =

1 37 73 1 11 · − · = . 18 6 18 60 1080

Example 2.2.12 The series

n+1 ∞ ∞ ∞ n+1 ∞    5 5n+1 1 1 1  5 · . = = and n+2 8 8 8 8 8 n(n + 1) n=1 n=1 n=1 n=1 52 5 are convergent. The first is geometric with ratio r = and first term 3 . 8 8 The second is telescopic, and as we saw in Example 2.2.6, it converges and has a sum of 1. We can conclude that the series  ∞ n+1  5 1 . + 8n+2 n(n + 1) n=1 is convergent and its sum is

217 25 +1= . 192 192

Theorem 2.2.3 (Cauchy’s Criterion) A series only if



an converges if and

∀δ > 0 ∃ p ∈ N : m > n > p ⇒ |an+1 + · · · + am | < δ.

.

Proof: As  m n      .|an+1 + · · · + am | =  ai − ai  = |Sm − Sn |,   i=1

i=1

2.2. Definition of Series: Convergence – General Properties our aim is to prove that the series

∞ 

an converges if and only if

n=1

∀δ > 0 ∃ p ∈ N : m > n > p ⇒ |Sm − Sn | < δ,

.

that is, we want to prove that (Sn ) is a Cauchy sequence. However, by definition, the series an converges if and only if (Sn ) is a convergent sequence, and in R a sequence is convergent if and only if it is a Cauchy sequence (see Theorem 1.1.20).

Example 2.2.13 In Example 2.2.5, we showed that the harmonic series, 1 n , is divergent. We will prove the same result using Theorem 2.2.3. If the series is convergent, then for any given δ > 0, there would exist p ∈ N such that |an+1 + · · · + am | < δ for all m > n > p. However, if we choose m = n + n, then |an+1 + · · · + am | = |an+1 + · · · + an+n | 1 1 + ··· + n+1 n+n 1 1 1 1 + ··· + =n· = . ≥ n+n n+n 2n 2

=

.

1 Therefore, the condition of the theorem is not satisfied for δ < , which 2 means that the harmonic series is divergent.

Corollary 1 The convergence or divergence of a series does not depend on the first p terms, whatever p ∈ N, that is, if an and bn are series such that there exists p ∈ N satisfying an = bn , ∀n > p, then either both series are convergent or both are divergent.

Definition 2.2.3 The remainder of order p of the series

∞  n=1

series r =

∞ 

. p

n=1

an+p =

∞  n=p+1

an .

an is the

79

80

Numerical Series According to the previous corollary, the remainder of any order will also be convergent if a series is convergent. The sum of the remainder of order p of a convergent series gives us the error made by taking the partial sum Sp as the approximate value of the sum of the series. This error is given by ∞  .

an − Sp =

n=1

∞  n=1

an −

p 

an =

n=1

∞ 

an+p = rp .

n=1

Corollary 2 The convergence or divergence of a series does not change if we suppress a finite, arbitrary number of terms.

The following theorem can be viewed as a generalization of the associative property of addition to the case of convergent series. Theorem 2.2.4 Let an be a convergent series and k1 , k2 , . . . , kn , . . . be a strictly increasing sequence of elements of N. Let bn be the sequence defined as follows: ⎧ k1  ⎪ ⎪ ⎪ ai , if n = 1 ⎪ ⎨ i=1 .bn = kn  ⎪ ⎪ ⎪ ai , if n > 1. ⎪ ⎩ i=kn−1 +1

Then, the series



bn is convergent and

∞ 

bn =

n=1

∞ 

an .

n=1

Proof: The series an is convergent, then by Definition 2.2.2 there exists and is finite the limit .

lim Sn = lim

n 

ai .

i=1

According to Theorem 1.1.15, every subsequence of (Sn ) is convergent and has the same limit.

2.2. Definition of Series: Convergence – General Properties The series

∞ 



bn is convergent if and only if Sn =

n 

n=1 

Sn =

n 

.

bi =

i=1

bi is convergent. But

i=1 k1 

ai +

i=1

k2 

kn 

ai + · · · +

i=k1 +1

ai =

i=kn−1 +1

kn 

a i = Sk n ,

i=1



that is, (Sn ) is a subsequence of (Sn ); therefore, it is convergent and has the same limit: ∞  .



bn = lim Sn = lim Sn =

n=1

Note: If the series

∞ 

∞ 

an .

n=1

an is convergent, then the theorem establishes that

n=1

the following “associative property” is valid: a +a2 +· · ·+ak1 +· · ·+ak2 +· · · = (a1 +· · ·+ak1 )+(ak1 +1 +· · ·+ak2 )+· · ·

. 1

However, this property is not valid if the series is divergent. For instance, ∞  (−1)n is divergent because its general term does not tend to the series n=1

zero. Nevertheless, (−1 + 1) + (−1 + 1) + · · · = 0.

81

82

Numerical Series

2.3

Alternating Series

Definition 2.3.1 A series is said to be alternating if its terms are alternatively positive and negative. Gottfried Wilhelm Leibniz (1646–1716) was a German philosopher and mathematician. He obtained a doctorate in Law when he was twenty years old. He pursued a career in Law and International Politics as an advisor to kings and princes. During his countless travels, Leibniz befriended the greatest intellectuals of his time. He developed differential and integral calculus simultaneously and independently of the English mathematician Isaac Newton and established the foundations of dynamics. Leibniz was one of the last intellectuals to master almost all of the knowledge of his time. (Source of image: Oil portrait by Christoph Bernhard Francke (1665– 1729), Herzog Anton Ulrich-Museum.)

Assuming that the first term of the series is positive, we can write ∞  .

(−1)n−1 an ,

an > 0, ∀n ∈ N.

n=1

Theorem 2.3.1 (Leibniz’s Test) If (an ) is a decreasing sequence of pos itive terms and lim an = 0, then the series (−1)n−1 an is convergent. Proof: Consider the sequence (Sn ) of the partial sums of the series: Sn = a1 − a2 + a3 − · · · + (−1)n−1 an .

.

We will study the subsequences of even order terms and odd order terms. Let k ∈ N be arbitrary. S2k = a1 − a2 + · · · + a2k−1 − a2k .

S2k+1 = a1 − a2 + · · · + a2k−1 − a2k + a2k+1 .

We claim that the subsequence (S2k ) is increasing. This is because, as (an ) is decreasing, we get: S2k+2 − S2k = a1 − a2 + · · · + a2k−1 − a2k + a2k+1 − a2k+2 − .

−(a1 − a2 + · · · + a2k−1 − a2k ) = a2k+1 − a2k+2 ≥ 0.

Moreover, it is a bounded sequence since: S2 ≤ S2k = a1 − [(a2 − a3 ) + (a4 − a5 ) + · · · + (a2k−2 − a2k−1 ) + a2k ] < a1 .

.

Hence, (S2k ) is monotonic and bounded, which implies that it is convergent. From S2k+1 = S2k + a2k+1 , it follows that .

lim S2k+1 = lim S2k ,

k→+∞

k→+∞

since by hypothesis (an ) is a null sequence.

2.3. Alternating Series

83

As the subsequences of even and odd terms have the same limit, (Sn ) ∞  (−1)n−1 an is convergent. converges. Therefore, the series n=1

Note: Leibniz’s test is a particular case of a more general result known as Dirichlet’s test. The Dirichlet’s test states the following: n If the sequence ( i=1 bi ) is bounded, and if (an ) is a decreasing sequence ∞ of positive terms satisfying lim an = 0, then the series n=1 bn an is convergent. If we let bn = (−1)n in Dirichlet’s test, we obtain Leibniz’s test. ∞ 

∞ 

log(n) = is alter(−1)n n n n=1 n=2 nating. Let us check the conditions for applying Leibniz’s test: Example 2.3.1 The series

(−1)

n log(n)

log(x) log(n) = 0 (it is enough to note that lim is an indeterx→+∞ n x ∞ minate form of type ; we can calculate the limit using L’Hˆ opital’s ∞ Rule:

 log(x) 1 = 0). . lim = lim  x→+∞ x→+∞ (x) x

(i) lim

(ii)

log(n) > 0, ∀n > 1. n

log(x) is a decreasing function on ]e, +∞[ (iii) The function f (x) = x because

 log(x) 1 − log(x)  .f (x) = = < 0, ∀x > e. x x2

 log(n) This implies that the sequence is decreasing for n ≥ 3. n Thus, we can conclude that the alternating series ∞  .

n=3

(−1)n

log(n) n

is convergent. As this series differs from the initial series in one term, we can affirm that the series under study is convergent.

Peter Gustave Lejeune Dirichlet (1805–1859) was a German mathematician who made valuable contributions to number theory, analysis, and mechanics. He was a professor at the universities of Breslau (1827) and Berlin (1828–1855). In 1855, he succeeded Carl Friedrich Gauss at the University of G¨ ottingen. (Source of image: Portrait Collection of the German Museum, Munich.)

84

Numerical Series

Example 2.3.2 The series

∞ 

cos(nπ) sin

π

n=2 .

cos(nπ) = (−1)n and

an = sin

n π n

is alternating because > 0, ∀n > 1

π π ≤ , ∀n > 1). We can write the series in the form (note that 0 < n 2 ∞ π  n . Let us check the conditions of Leibniz’s test: (−1) sin n n=2 π = 0. (i) lim sin n π (ii) We have already seen that sin > 0, ∀n > 1. n π π (iii) As we also justified earlier, 0 < ≤ , ∀n > 1; the sequence n 2   π π is decreasing and the sine function is increasing on 0, , so 2  n  π  is decreasing for n ≥ 2. sin n We can conclude that the alternating series is convergent. ∞ 

1 , α ∈ R. α n n=1 If α ≤ 0, the series diverges because its general term does not tend to zero. 1 If α > 0, the series is convergent because an = α is a decreasing sequence n with positive terms and lim an = 0. Example 2.3.3 Consider the series

(−1)n

Not all alternating series meet the conditions of Leibniz’s test. In this situation, the test is not applicable, and it is necessary to look for another approach. Example 2.3.4 Consider the alternating series



  ∞ ∞  1 (−1)n (−1)n (−1)n n √ √ + . (−1) . 1+ √ = n n n n n=1 n=1 (−1)n 1 is a null sequence and an > 0, ∀n > 1, we Although an = √ + n n cannot apply Leibniz’s test because (an ) is not decreasing. However, we can

2.3. Alternating Series

85

show that the given series is divergent because it is the sum of a convergent  1 1 series, the series (−1)n √ , with a divergent series, the series n n (see Note 1 on page 77).

Theorem 2.3.2 Let (an ) be a decreasing sequence of positive terms such that lim an = 0, and S be the sum of the series (−1)n−1 an . Then 0 ≤ (−1)n (S − Sn ) ≤ an+1 , ∀n ∈ N.

.

Proof: We know from the proof of the previous theorem that (S2k ) is an increasing subsequence of (Sn ), and it has the same limit, S, as the subsequence (S2k+1 ). Similarly, (S2k+1 ) is decreasing. This leads to the following inequalities: S2k ≤ S

.

and

S ≤ S2k+1 , ∀k ∈ N.

From these inequalities, we can deduce that: 0 ≤ S2k−1 − S ≤ S2k−1 − S2k = a2k , .

0 ≤ S − S2k ≤ S2k+1 − S2k = a2k+1 .

Therefore:

0 ≤ S2k−1 − S ≤ a2k , .

0 ≤ S − S2k ≤ a2k+1 .

This implies that 0 ≤ (−1)2k−1 (S − S2k−1 ) ≤ a2k , .

0 ≤ (−1)2k (S − S2k ) ≤ a2k+1 .

From these last two inequalities, it follows that 0 ≤ (−1)n (S − Sn ) ≤ an+1 .

.

Corollary 1 Let (an ) be a decreasing sequence of positive terms such that lim an = 0 and S be the sum of the series (−1)n−1 an . Then |S − Sn | ≤ an+1 , ∀n ∈ N.

.

86

Numerical Series Note: According to the previous corollary, it is evident that under the conditions of Leibniz’s test, the absolute value of the error made in using a partial sum as an estimation for the sum of an alternating series is, in absolute value, always less than the absolute value of the first of the disregarded terms.

Example 2.3.5 Consider the series ∞ 

.

1 (−1)n , n n=1

which is known as the alternating harmonic series. According to Leibniz’s test, this series is convergent, as an = n1 is a null decreasing sequence of positive terms. If we use the partial sum S9 as an estimation of the sum of the series, we will make an error that in absolute value is less than or equal 1 to , which is the value of a10 . 10

2.4. Absolute Convergence

2.4

87

Absolute Convergence

Theorem 2.4.1 If the series is also convergent.



|an | is convergent, then the series

Proof: By Theorem 2.2.3, the series

∞ 



an

|an | is convergent if and only if

n=1

∀δ > 0 ∃ p ∈ N : m > n > p ⇒ | |an+1 | + · · · + |am | | < δ.

.

Since |an+1 + · · · + am | ≤ |an+1 | + · · · + |am |

.

and | |an+1 | + · · · + |am | | = |an+1 | + · · · + |am |,

.

we have ∀δ > 0 ∃ p ∈ N : m > n > p ⇒ |an+1 + · · · + am | < δ,

.

that is, the series

∞ 

an is convergent.

n=1

Note: It is worth noting that the converse of this theorem is not true. This means that the series an can be convergent even if the series of modules, |an |, is not. For instance, consider the harmonic series, which diverges, and the alternating harmonic series, which converges. The harmonic series is the series of modules of the alternating harmonic series. Definition 2.4.1 A series an is absolutely convergent if the series |an | is convergent. A series an is conditionally convergent if it is convergent and the series |an | is divergent. Definition 2.4.2 We say that the series bn is a rearrangement of the series an , or that it is obtained by reordering its terms, if there exists a bijection φ from N into N such that bn = aφ(n) .

88

Numerical Series Theorem 2.4.2 If the series an is absolutely convergent, then any series obtained by rearranging its terms is absolutely convergent and has the same sum. This theorem generalizes the commutative property of the usual addition to absolutely convergent series. However, it should be noted that this property is not valid for conditionally convergent series. By rearranging the terms of a conditionally convergent series, we can obtain a series with a predetermined sum and even a divergent series. Theorem 2.4.3 Let



an be a conditionally convergent series. Then: a) There are bijections φ : N → N such that the series aφ(n) is divergent.

b) For every real number k, there exists a bijection φ : N → N such that the series aφ(n) is convergent and its sum is equal to k. ∞ 

1 , n n=1 which we know to be conditionally convergent. Let us rearrange its terms such that each positive term is followed by two negative terms. We will obtain the following series: Example 2.4.1 Consider the alternating harmonic series,

1−

.

1 1 1 1 1 1 1 1 − + − − + − − + ··· 2 4 3 6 8 5 10 12

For this series, we have the following partial sums: 

S1 = 1 

S2 = 1 −

1 2

1 1 1 1 1 − = − = . 2 4 2 4 2 1 1 1  S4 = 1 − − + 2 4 3 1 1 1 1  S5 = 1 − − + − 2 4 3 6 

S3 = 1 −

1−

1 2

 =

1 S2 2

(−1)n−1

2.4. Absolute Convergence

89



  1 1 1 1 1 1 1 1 1 1 − S6 = 1 − − + − − = 1 − − + − 2 4 3 6 8 2 4 3 6 8

 1 1 1 1 1 1 1 1 1 = − + − = 1− + − = S4 2 4 6 8 2 2 3 4 2 .. .



 1 1 1 1 1 1 1 1 1 1   − − = − 1− + − + − . S9 = S6 + 5 10 12 2 2 3 4 5 10 12



 1 1 1 1 1 1 1 1 1 1 1 1 − = = 1− + − + 1− + − + − 2 2 3 4 10 12 2 2 3 4 5 6 1 = S6 2 .. . 

where Sn =

n 

1 (−1)i−1 . i i=1

1 S2n , which implies that 2 if we designate by S the limit of the sequence (Sn ), then we have 

Using induction, we can demonstrate that S3n =



.





As S3n+1 = S3n +

lim S3n =

1 1 1   and S3n+2 = S3n + − , we have 2n + 1 2n + 1 4n + 2 

.



lim S3n+1 = lim S3n + lim

and 

.

1 S. 2



lim S3n+2 = lim S3n + lim

1 2n + 1

1 1 − lim , 2n + 1 4n + 2

that is, 

.





lim S3n+1 = lim S3n+2 = lim S3n = 

1 S. 2

Therefore, we conclude that lim Sn = 12 S. In other words, the series obtained by this rearrangement of the terms of the alternating harmonic series is convergent, and its sum equals half the original series’ sum.

90

Numerical Series

2.5

Series of Nonnegative Terms

In this section, we will establish convergence tests for a series of nonnegative terms. These tests apply to investigating the absolute convergence of series in general. We should note that convergence and absolute convergence are equivalent for a series of nonnegative terms. Theorem 2.5.1 Let . an be a series of nonnegative terms. The series an is convergent if and only if the sequence of its partial sums is . bounded. Proof: Let .(Sn ) be the sequence of partial sums of the series . an . If the series is convergent, then by definition, the sequence .(Sn ) converges. Consequently, it is a bounded sequence (see Theorem 1.1.7). Suppose that .(Sn ) is bounded. As .an ≥ 0, we have Sn+1 ≥ Sn , ∀n ∈ N,

.

that is, .(Sn ) is an increasing sequence. According to Theorem 1.1.12, the two previous statements imply the convergence of .(Sn ), which is equivalent to saying that the series . an is convergent. Theorem 2.5.2 (Cauchy’s Condensation Test) Let .(an ) be a decreas ing sequence of positive numbers. The series . an converges if and only n if the series . 2 a2n converges. Proof: Consider the partial sums of the series . an and . 2n a2n : Sn =

n 

.

ai and Tn =

i=1

n 

2i a2i .

i=1

Note that the sequences .(Sn ) and .(Tn ) are increasing, with positive terms. Suppose that the series . 2n a2n is convergent. By Theorem 2.5.1, the sequence .(Tn ) is bounded. Then S2n −1 = a1 + (a2 + a3 ) + (a4 + a5 + a6 + a7 ) + · · · +(a2n−1 + a2n−1 +1 + · · · + a2n −1 ) .

≤ a1 + 2 a2 + 4 a4 + · · · + 2n−1 a2n−1 = a1 + Tn−1 .

2.5. Series of Nonnegative Terms As the sequence .(Tn ) is bounded, the sequence .(S2n −1 ) is also bounded. Given that n .Sm ≤ S2n −1 , ∀m ≤ 2 − 1, the sequence .(Sn ) is bounded; therefore, according to the previous theorem, the series . an is convergent. Conversely, suppose that the series . an is convergent. Then, using again the previous theorem, the sequence .(Sn ) is bounded. We can observe that S2n = a1 + a2 + (a3 + a4 ) + (a5 + a6 + a7 + a8 ) + · · · + +(a2n−1 +1 + a2n−1 +2 + · · · + a2n ) .

≥ a1 + a2 + 2 a4 + 4 a8 + · · · + 2n−1 a2n 1 = a1 + (2 a2 + 4 a4 + 8 a8 + · · · + 2n a2n ) 2 1 = a1 + Tn . 2

This implies that the sequence .(Tn ) is also bounded. Hence, the series n 2 a2n is convergent. .

∞  1 , .p ∈ R, usually referred to as np n=1

Example 2.5.1 Consider the series . the p-series.

If .p ≤ 0, the series is divergent because its general term does not tend to zero. 1 If .p > 0, we can apply Theorem 2.5.2 because the sequence .an = p is n decreasing with positive terms. The series n ∞ ∞

  1 2n . 2 p = n = 2p−1 (2n ) (2p ) n=1 n=1 n=1 ∞ 

n

1

1 1 is geometric, with ratio . p−1 , and is convergent if and only if . p−1 < 1, 2 2 that is, if and only if .p > 1. ∞  1 converges if and only if .p > 1. Conclusion: The p-series . p n n=1

91

92

Numerical Series

Example 2.5.2 Consider the series function

∞ 

1

α , where .α ∈ R. The n log(n) n=2

.

1

α f (x) = x log(x)

.

is positive and continuous on .[2, +∞[. We can find the derivative of f to determine whether it is increasing or decreasing. We have



f (x) = 0 .

α



α−1 + α log(x) ⇔− =0

2α x2 log(x)

α−1

log(x) log(x) + α ⇔− =0

2α x2 log(x) log(x)

⇔ log(x) + α = 0. If .α ≥ 0, this equation has no root in .[2, +∞[, which implies that .f  (x) < 0, .∀x ∈ [2, +∞[ . If .α < 0, then .f  (x) < 0, .∀x > e−α , and this implies that .f  (x) < 0, −α .∀x ∈ ]e , +∞[ . 1

α is decreasing from the order q if n log(n) −α .q ∈ N is such that .q ≥ max{2, e }. The series

Then, the sequence .an =

∞  .

n=q

2n



∞ ∞   2n 1 1 1

α =

α

α = α α log(2) n log(2n ) log(2) n n=q n=q

is convergent if and only if .α > 1 as we saw in the previous example. Remember that the convergence or divergence of the series does not change if we omit a finite number of terms. Hence, we can conclude that the series ∞ 

.

1

α n log(n) n=2

is convergent if and only if .α > 1.

Note: In Theorem 2.5.2, the hypothesis that .(an ) is a decreasing sequence cannot be removed. In fact, consider the series . an with

2.5. Series of Nonnegative Terms ⎧ 1 ⎪ ⎨ , k 2 .an = ⎪ ⎩0, The series

∞  .

2n a2n =

n=1

93

if n = 2k if n = 2k .

∞ ∞   2n = 1 2n n=1 n=1

is divergent. However, we can write the original series as ∞ ∞ ∞ k    1 1 . an = = , k 2 2 n=1 k=1

k=1

which is a geometric series with ratio .1/2 and is therefore convergent.

Theorem 2.5.3 (Integral Test) Let .f : [1, +∞[ → R be a continuous, positive, and decreasing function. For each .n ∈ N, let .an = f (n).  ∞ ∞  an and the improper integral . f (x) dx are both Then, the series . 1

n=1

convergent or both divergent. Proof: Let us partition the interval .[1, n] into .n − 1 intervals of length 1, as shown in Fig. 2.4. The total area of the rectangles inscribed in the figure is .a2 + a3 + a4 + · · · + an . We have the following inequality:  n .Sn − a1 = a2 + a3 + a4 + · · · + an ≤ f (x) dx. 1

If we consider the area of the circumscribed rectangles (see Fig. 2.5), we have the inequality  n .Sn−1 = a1 + a2 + a3 + · · · + an−1 ≥ f (x) dx.

Figure 2.4: Inscribed rectangles in the graph of f

1

From the two previous inequalities, we can conclude that  n .Sn − a1 ≤ f (x) dx ≤ Sn−1 . 1



If the integral is divergent,

.

n

lim

n→+∞

f (x) dx = +∞ since f is positive. 1

Then, by the inequality on the right, the limit of the sequence of the partial

Figure 2.5: Circumscribed rectangles in the graph of f

94

Numerical Series sums of the series is also .+∞; that is, the series diverges. If the integral  n f (x) dx. Conseconverges, then there exists and it is finite . lim n→+∞

quently, the sequence .Sn =

n 

1

ai is bounded. As the terms of the series

i=1

are positive, it is convergent by Theorem 2.5.1.



If the series is convergent, by the inequality on the right,

.

n

f (x) dx is 1

bounded, so the improper integral is convergent. If the series is divergent,  n f (x) dx = +∞. based on the inequality on the left, . lim n→+∞

1

Note: When we use the Integral Test, it is not necessary for the series or the integral to start at .n = 1. Additionally, the function f does not need to decrease on the interval .[1, +∞[, but only on some interval of the form .[N, +∞[, .N ∈ N. ∞ 

en . The real-valued function 1 + e2n n=1 ex f of a real variable defined by .f (x) = is continuous and positive 1 + e2x on .[1, +∞[. Let us study the monotonicity of f : Example 2.5.3 Consider the series .

f  (x) =

.

ex (1 + e2x ) − ex 2e2x (1 +

2 e2x )

=

ex (1 − e2x ) (1 + e2x )

2

< 0, ∀x ∈ [1, +∞[.

Thus, the function f decreases on .[1, +∞[. Furthermore,  .

t

lim

t→+∞

1



π ex dx = lim arctan(et ) − arctan(e) = − arctan(e). 2x t→+∞ 1+e 2

 Since the improper integral . ∞ 

+∞ 1

ex dx converges, then, by the Inte1 + e2x

en gral Test, the series . also converges. 1 + e2n n=1 ∞  log(n) . The real-valued function n n=2 log(x) f of a real variable defined by .f (x) = is continuous and positive on x

Example 2.5.4 Consider the series .

2.5. Series of Nonnegative Terms

95

[2, +∞[. We can analyze the monotonicity of f by taking the derivative:

.

1 · x − log(x) 1 − log(x) x .f (x) = = < 0, ∀x ∈ ]e, +∞[. 2 x x2 Thus, the function f decreases on .]e, +∞[. Now, we compute the integral:  t

log(x) 1 dx = lim (log(t))2 − (log(2))2 = +∞. . lim t→+∞ 1 t→+∞ 2 x 

According to the Integral Test, the series  improper integral .

+∞ 1

∞  log(n) diverges since the n n=2

.

log(x) dx is also divergent. x

Theorem 2.5.4 (General Comparison Test) Let . an and . bn be two series of nonnegative terms such that .an ≤ bn , .∀n ∈ N: a) If the series . bn is convergent, then the series . an is convergent. b) If the series . an is divergent, then the series . bn is divergent. Proof: Let .(Sn ) and .(Sn ) be the sequences of partial sums of series . an and . bn , that is, Sn =

n 

.



ai and Sn =

i=1

n 

bi .

i=1

Since .0 ≤ an ≤ bn , .∀n ∈ N, we have 

0 ≤ Sn ≤ Sn , ∀n ∈ N.

.



(2.3)

a) If the series . bn is convergent, then, by Theorem 2.5.1, the sequence of  its partial sums, .(Sn ), is bounded. By inequality (2.3), the sequence .(Sn ) is also bounded, that is, the series . an is convergent. b) If the series . an is divergent, then by Theorem 2.5.1, the sequence .(Sn )  is not bounded. Inequality (2.3) implies that the sequence .(Sn ) is also not bounded; thus, again by Theorem 2.5.1, the series . bn is divergent. Note: Omitting a finite number of terms does not change the convergence or divergence of the series, as we have seen. Therefore, the previous theorem remains valid if there exists .p ∈ N such that .an ≤ bn , .∀n ≥ p.

96

Numerical Series ∞  1 Example 2.5.5 Consider the series . . We know that n! n=1

1 1 1 = ≤ n−1 , ∀n ∈ N. n! n(n − 1)(n − 2) . . . 2 2

0
1 and is, therefore, convergent. n2 n=1 By applying the General Comparison Test and taking into  account inequality ∞   cos(n)    (2.4), we can conclude that the series .  n2  is convergent. This

The series .

n=1

∞  cos(n) is absolutely convergent. means that the series . n2 n=1

Example 2.5.7 Consider the series .

∞ 

n=1

(−1)n

sin(n)

n . Let us analyze log(10)

the series of modules:   ∞  ∞  sin(n)   | sin(n)|  n

n  =

n . . (−1)  log(10)  n=1 log(10) n=1 We notice that the following inequality holds for every natural number n: 0<

.

| sin(n)| 1

n ≤

n . log(10) log(10)

2.5. Series of Nonnegative Terms The series .

∞ 



n=1

97

1 1

n is geometric with ratio .r = . Since log(10) log(10)

10 > e ⇒ log(10) > 1 ⇒ 0
0, the series . cn and . bn are either both convergent or both divergent. From this fact, we can conclude the desired result. Corollary 2 Let . an and . bn be two series such that .an ≥ 0 and an .bn > 0, .∀n ∈ N. If .lim = k ∈ R+ , then both series are either bn convergent or divergent. an = k. By Definition 1.1.7, bn    an   < δ. .∀δ > 0 ∃ p ∈ N : n > p ⇒  − k  bn 

Proof: Suppose .lim

Let .δ =

k . From a certain order, we have 2    an  k k an k an k 3   .  bn − k  < 2 ⇔ − 2 < bn − k < 2 ⇔ 2 < bn < 2 k,

98

Numerical Series which implies a
0, .∀n ∈ N. If .lim = 0, then: bn a) If the series . bn is convergent, then the series . an is also convergent. b) If the series . an is divergent, then the series . bn is also divergent. Proof: Let .lim

an = 0. By Definition 1.1.7, bn    an   .∀δ > 0 ∃ p ∈ N : n > p ⇒   bn  < δ.

From a certain order, we have    an  an   .  bn  = bn < δ, since .an ≥ 0 and .bn > 0. Consequently, .0 ≤ an < δbn , and from Corollary 1 of Theorem 2.5.4, the result follows. Corollary 4 Let . an and . bn be two series such that .an ≥ 0 and an .bn > 0, .∀n ∈ N. If .lim = +∞, then: bn a) If the series . bn is divergent, then the series . an is also divergent. b) If the series . an is convergent, then the series . bn is also convergent. Proof: Let .lim

an = +∞. By Definition 1.1.6, bn ∀δ > 0 ∃ p ∈ N : n > p ⇒

.

an > δ. bn

2.5. Series of Nonnegative Terms

99

From order p, we have a > δbn > 0,

. n

since .bn > 0. The result is a consequence of Corollary 1 of Theorem 2.5.4. Corollary 5 Let . an and . bn be two series such that .an > 0 and .bn > 0, .∀n ∈ N. If there exists .p ∈ N such that .

an+1 bn+1 ≤ , ∀n ≥ p, an bn

then: a) If the series . bn is convergent, then the series . an is convergent. b) If the series . an is divergent, then the series . bn is divergent. Proof: As .an > 0 and .bn > 0, we can deduce that: .

an+1 bn+1 an+1 an ≤ ⇔ ≤ , ∀n ≥ p. an bn bn+1 bn

 an This means that the sequence . is decreasing after order p. Therefore, bn an ap ) such that . ≤ k. In other there exists a constant k (we can take .k = bp bn words: a ≤ k bn , ∀n ≥ p.

. n

We can use Corollary 1 of Theorem 2.5.4 to arrive at the final result.

Example 2.5.8 The terms of the series

∞  1 + (−1)n are nonnegative. n2 n=1

.

As 0≤

.

1 + (−1)n 2 ≤ 2 , ∀n ∈ N, n2 n

∞  1 is convergent, Corollary 1 of Theorem 2.5.4 allows us 2 n n=1 to conclude that the given series is convergent.

and the series .

100

Numerical Series ∞  2n2 + n . Since 3n5 + 3 n=1

Example 2.5.9 Consider the series of positive terms .

2n2 + n 2 2n5 + n4 3n5 + 3 = , . lim = lim 5 3n + 3 3 1 n3 ∞  2n2 + n by Corollary 2 of Theorem 2.5.4, the series . is convergent be3n5 + 3 n=1 ∞  1 converges. cause . 3 n n=1 Example 2.5.10 The series ∞  1 . is convergent and 2 n n=1

∞  log(n) is of nonnegative terms. Since n3 n=1

.

log(n) log(n) n3 = 0, . lim = lim n 1 n2 by Corollary 3 of Theorem 2.5.4, the original series is also convergent.

Example 2.5.11 Consider the series ∞  .

n=1

bn =

∞  1 × 3 × · · · × (2n − 1) 2 × 4 × · · · 2n n=1

and

∞  1 . n n=1

Both series are of positive terms, and the second is divergent. We have bn+1 . = bn

1 × 3 × · · · × (2n − 1)(2n + 1) 2n + 1 2 × 4 × · · · 2n(2n + 2) = 2n + 2 1 × 3 × · · · × (2n − 1) 2 × 4 × · · · 2n

and

an+1 n . = an n+1

n 2n + 1 ≤ , which allows us to conclude, by n+1 2n + 2 ∞  Corollary 5 of Theorem 2.5.4, that the series . bn is divergent.

It is easy to verify that

.

n=1

2.5. Series of Nonnegative Terms

101 ∞  arctan(n) + n √ are positive. n n+1 n=1

Example 2.5.12 The terms of the series . Let us consider the series

∞  .

n=1

series with .p =

1 n1/2

, which is divergent because it is a p-

1 . The limit 2

arctan(n) arctan(n) + n √ +1 1/2 (arctan(n) + n) n n n n+1 √ = lim . lim = lim =1 n n+1 1 1 1 + 3/2 n1/2 n is finite and different from zero. Therefore, by Corollary 2 of the General ∞  arctan(n) + n √ diverges. Comparison Test, the series . n n+1 n=1

Theorem 2.5.5 (Ratio Test) Let . an be a series of positive terms: a) If there exist .r < 1 and .p ∈ N such that then the series . an is convergent.

.

an+1 ≤ r < 1 .∀n ≥ p, an

an+1 b) If there exists .p ∈ N such that . ≥ 1, .∀n ≥ p, then the series an an is divergent. . Proof: n a) The series . bn = r is geometric with ratio r and is convergent since .0 < r < 1. Moreover, we have .

an+1 rn+1 ≤ n = r, ∀n ≥ p, an r

which implies by Corollary 5 of the General Comparison Test that the series an is convergent. b) The series . bn = 1 is divergent. Additionally, we have

.

.

an+1 bn+1 ≥1= , ∀n ≥ p. an bn

By Corollary 5 of the General Comparison Test, we can conclude that the series . an is divergent.

102

Jean Le Rond D’Alembert (1717–1783) French mathematician and philosopher who was elected to the Academy of Sciences of Paris at the age of 24, and later to the Berlin Academy. With Diderot, he began publishing the “Encyclopedia” in 1747, writing articles on mathematics and literature. In 1772, he was appointed perpetual secretary of the French Academy. As a mathematician and scientist, D’Alembert continued the work of Newton and Leibniz and researched differential equations with partial derivatives. He was also responsible for stating the fundamental theorem of algebra, which Gauss later proved. Additionally, D’Alembert studied hydrodynamics and mechanics and formulated what is now known as “D’Alembert’s principle.” This principle, when applied to the Earth’s movement, explains the variations of the rotation axis of the globe. (Source of image: Oil portrait by Maurice-Quentin de La Tour (1704–1788), Louvre Museum.)

Numerical Series Corollary 1 (D’Alembert’s Test) Let . an be a series of positive terms. an+1 If there exists .lim = a . (a ∈ R+ 0 or .a = +∞), then: an a) If .a < 1, the series . an is convergent. b) If .a > 1, the series . an is divergent. Proof: We know that if .a ∈ R,

   an+1  an+1  . lim = a ⇔ ∀δ > 0 ∃p ∈ N : n > p ⇒  − a < δ. an an

a) If .a < 1, then .1 − a > 0. Let .δ be such that .0 < δ < 1 − a. There exists p ∈ N such that for all .n > p,    an+1  an+1 an+1   .  an − a < δ ⇔ −δ < an − a < δ ⇔ a − δ < an < a + δ.

.

Since .δ < 1 − a, then .a + δ < 1. Item a) of the Ratio Test allows us to conclude that the series . an converges. b) If .a > 1, let .δ = a − 1. There exists .p ∈ N such that, for all .n > p,    an+1  an+1   .  an − a < a − 1 ⇔ 1 < an < 2a − 1. By item b) of the Ratio Test, the series . an diverges. If .a = +∞, there exists .p ∈ N such that an+1 . > 1, ∀n > p. an Again, by the Ratio Test, the series . an diverges. an+1 = 1, D’Alembert’s test is inconclusive as there are an divergent and convergent series in this situation. For example, the harmonic ∞  1 is divergent and series . n n=1

Note: If .lim

.

lim

an+1 n = 1, = lim an n+1

∞  1 is convergent and n2 n=1

and the series .

an+1 . lim = lim an



n n+1

2 = 1.

2.5. Series of Nonnegative Terms an+1 = 1 and the convergence is for values greater than 1, an an+1 that is, there is an order .p ∈ N from which . ≥ 1, then, by the Ratio an ∞  an diverges. Test, the series .

However, if .lim

n=1 ∞  k n n! Example 2.5.13 Let .k > 0. The series . is of positive terms. As nn n=1

k n+1 (n + 1)!

n n k n+1 (n + 1)! nn 1 (n + 1)n+1 . lim = lim n = lim k · =k· , n+1 k n n! k n! (n + 1) n+1 e nn k k we can apply D’Alembert’s test: If . < 1, the series is convergent; if . > 1, e e the series is divergent.

n n+1 k If . = 1, D’Alembert’s test is inconclusive. However, as . is an e n

n n increasing sequence with limit e, . is a decreasing sequence with n +

n1 n 1 limit . , which implies that .e · is decreasing with limit 1, that is, e n+1 an+1 . tends to 1 from above. Then, if .k = e, the series is divergent. an

2 (n + 1)! is alternating. Let us (2n)! 5n n=1

2 ∞  (n + 1)! start by studying the series of modules, . , using D’Alembert’s (2n)! 5n n=1 test. Let .an be the general term of the series of modules: Example 2.5.14 The series



lim .

an+1 an

∞ 

.

(−1)n

2 (n + 2)!

2

(n + 2)! (2n)! 5n 2(n + 1) ! 5n+1 = lim = lim

2



2 (n + 1)! 2(n + 1) ! 5n+1 (n + 1)! (2n)! 5n

2 (n + 2)! (2n)! = lim

2 (2n + 2)(2n + 1)(2n)! 5 (n + 1)!

103

104

Numerical Series

= lim

(n + 2)2 ((n + 1)!)2 (2n + 2)(2n + 1) 5 ((n + 1)!)2

= lim

1 (n + 2)2 = . 5 (2n + 2)(2n + 1) 20

.

Since this value is less than 1, the series of modules is convergent; that is, the original series is absolutely convergent. ∞  (n!)2 + n! is of positive terms and (4n)! + n4 n=1

Example 2.5.15 The series .

(n!)2 + n! 2(n!)2 , ∀n ∈ N. < 4 (4n)! + n (4n)!

0
p, then the series . an is convergent. b) If there exist .p ∈ N and a subsequence, .(akn ), of .(an ) such that √ an is divergent. . kn akn ≥ 1, ∀kn > p, then the series . Proof: √ a) If . n an ≤ r, ∀n > p, then .an ≤ rn < 1, .∀n ≥ p. The series . rn is convergent because it is a geometric series with ratio r, where .0 < r < 1. Therefore, the series . an is convergent. √ b) If . kn akn ≥ 1, ∀kn > p, then .akn ≥ 1, .∀kn > p, so it does not tend to zero. As a result, .(an ) is not a null sequence, which implies that the series an is divergent. . Corollary 1 Let . an be a series of nonnegative terms: √ a) If .lim n an < 1, the series . an is convergent. √ b) If .lim n an > 1, the series . an is divergent. √ Proof: Let .a = lim n an . a) Let r be such that .a < r < 1. We can affirm that √ n .∃ p ∈ N : n > p ⇒ an < r, which implies by item a) of the theorem that the series . an converges. √ b) By definition of upper limit, there is a subsequence of . n an with limit .a > 1, so this sequence has an infinity of values greater than 1. By item b) of the theorem, the series diverges. Corollary 2 (Cauchy’s Root Test) Let . an be a series of nonnegative √ terms. If there exists .lim n an = a . (a ∈ R+ 0 or .a = +∞), then: a) If .a < 1, the series . an is convergent. b) If .a > 1, the series . an is divergent.

2.5. Series of Nonnegative Terms Proof: If there exists .lim over, Corollary 1 applies.

107

√ √ √ n an = a, then .lim n an = lim n an = a. More-

√ Note: If .lim n an = 1, the test is inconclusive, as there exist both convergent and divergent series in this situation. For instance, the harmonic series ∞  1 . is divergent and n n=1  1 n 1 = lim √ = 1, . lim n n n ∞  1 is convergent and 2 n n=1  1 1 n . lim = lim √ = 1. n 2 n n2

and the series .

n2 ∞

 n+1 Example 2.5.18 The series . is of positive terms. As n n=1 



n2 n n+1 n+1 n . lim = lim = e > 1, n n by Cauchy’s Root Test, the series is divergent. ∞  1 Example 2.5.19 Let us consider the series . . (3 + (−1)n )n n=1 ⎧ ⎪ ⎪ 1 ⎪  ⎪ ⎨ , if n is even 1 4 n . = ⎪ (3 + (−1)n )n 1 ⎪ ⎪ ⎪ ⎩ 2 , if n is odd.

√ 1 Then . n an ≤ < 1, .∀n ∈ N, and by the Root Test, the series is convergent. 2

2n

2n ∞  log(n) log(n) Example 2.5.20 The series . = is of nn nn n=1 n=2 positive terms. We will study this series using the Cauchy’s Root Test: ∞ 

 .

lim

n



log(n) nn

2n





log(n) = lim n

2 =0

108

Numerical Series because using L’Hˆ opital’s Rule, we have  .

log(x)

lim

2 

(x)

x→+∞

= lim

x→+∞

2 log(x) = 0. x

Since the limit is less than 1, the series is convergent. Example 2.5.21 Let us study the convergence of the series . an where

a =

. n

⎧ n ⎨(1 − √ n)n ,

if n is odd

⎩n2 e−n ,

if n is even.

It is an alternating series. Let us study the series of modules.  n .

⎧ √ ⎨ n |(1 − n n)n |, |an | = √ ⎩ n n2 e−n , ⎧√ ⎨ n n − 1, = n ⎩e−1 √ n2 ,

if n is odd if n is even

if n is odd if n is even.

√  √ n Since .lim( n n−1) = 0 and .lim e−1 n2 = e−1 , we obtain .lim n |an | = e−1 . ∞   |an | is convergent. Therefore, the series As .lim n |an | < 1, the series . ∞  .

n=1

an is absolutely convergent.

n=1

Note: The Cauchy’s Root Test is more general than D’Alembert’s test. This means that if Cauchy’s Root Test is inconclusive about a series, then D’Alembert’s test will also be inconclusive. In fact, by Theorem 1.1.11 √ an+1 n a .lim n = a. This theorem is significant because if .a = 1, an .= a ⇒ lim then D’Alembert’s test is inconclusive, and the same happens for Cauchy’s Root Test. However, it is essential to note that the reciprocal is not true. It may be possible to draw conclusions through Cauchy’s Root Test, even if D’Alembert’s test fails to do so.

2.5. Series of Nonnegative Terms

109

Example 2.5.22 Let us consider the series .

∞ 

2−n−(−1) . Using Cauchy’s n

n=1

Root Test,  .

lim

n

2−n−(−1) = lim 2−1 2− n

(−1)n n

=

1 < 1, 2

we conclude that the series is convergent, whereas D’Alembert’s test fails to be applied. In fact, 2−(n+1)−(−1) .



n+1

−n−(−1)n

= 2−n−1−(−1)

n+1

+n+(−1)

2

n

=

2, −3

2

if n is even ,

if n is odd.

Theorem 2.5.7 (Kummer’s Test) Let . an and . bn be two series of positive terms, with . bn divergent. If there exists

 1 an 1 . lim · − = k, k ∈ R, bn an+1 bn+1 then: a) If .k > 0, the series . an is convergent. b) If .k < 0, the series . an is divergent.

Proof: If . k ∈ R, .lim

1 an 1 · − bn an+1 bn+1

 = k is equivalent to

  1  an 1 ∀δ > 0 ∃p ∈ N : ∀n > p,  · − − k  < δ. bn an+1 bn+1

.

But   1  1 an 1 an 1   .  bn · an+1 − bn+1 − k  < δ ⇔ k − δ < bn · an+1 − bn+1 < k + δ.

Ernst Eduard Kummer (1810–1893) was a German mathematician. In 1855, he succeeded Dirichlet at the University of Berlin, where he worked with Weierstrass and Kronecker. In 1857, the Academy of Sciences of Paris awarded him a prize for his work on Fermat’s Theorem. (Source of image: Photo archive of the Mathematical Research Institute Oberwolfach.)

110

Numerical Series k a) Let .k ∈ R+ and .δ = . There exists an order .n0 ∈ N from which we 2 have 1 1 k an 1 k an 1 < · − ⇔ < · − 2 bn an+1 bn+1 2 bn an+1 bn+1 



1 2 1 an 1 2 an 1 · − · − ⇔ an+1 < an+1 .⇔ 1 < k bn an+1 bn+1 k bn an+1 bn+1 

2 an an+1 ⇔ an+1 < − . k bn bn+1 k−

Adding the two members of the inequality from .n0 + 1 to .n + 1, we obtain n+1 

.

⇔

n+1  i=n0 +1

ai−1 ai − bi−1 bi



i=n0

ai
1, the series . an is convergent.

Proof: In Kummer’s test, consider .bn =

∞  1 1 . The series . is divergent, n n n=1

and we have

 

1 an 1 an lim · − −n−1 = lim n · bn an+1 bn+1 an+1 .

 an = lim n − 1 − 1 = a − 1. an+1 This demonstrates the corollary. Note: Often, cases that are inconclusive by D’Alembert’s test can be solved by Raabe’s test.

Example 2.5.23 Consider the series ∞ ∞   1 × 3 × · · · × (2n − 1) 1 · = . an . 2 × 4 × · · · × 2n n n=1 n=1

We have .

lim

an+1 n(2n + 1) = 1, = lim an (n + 1)(2n + 2)

Joseph L. Raabe (1801– 1859), was a Swiss mathematician and physicist. He was a professor at the Zurich Polytechnic Institute. His name is mainly associated with the convergence test for a series of positive terms that extends D’Alembert’s Test. He also studied various aspects of planetary movements. (Source of image: Lithographie by Carl Friedrich Irminger (1813– 1863).)

112

Numerical Series and therefore, D’Alembert’s test is inconclusive. As 



an (n + 1)(2n + 2) −1 lim n − 1 = lim n an+1 n(2n + 1) .

= lim

(n + 1)(2n + 2) − n(2n + 1) 3n + 2 3 = lim = > 1, 2n + 1 2n + 1 2

Raabe’s test shows that the series is convergent.

2.6. Products of Series

2.6 Let

113

Products of Series ∞ 

an and

∞ 

bn be two convergent series with sums A and B, re∞  ∞    spectively. When thinking about the product . an × bn , it will .

n=1

.

n=1

n=1

n=1

be natural to define it in such a way that the resulting series, if convergent, has sum .A × B. We can define, for example, ∞  ∞    ∞ ∞     . an × bn = bk an n=1

n=1

n=1

k=1

obtaining ∞  .

n=1

 an

∞ 

 bk

=

∞ 

an · B = B ·

n=1

k=1

∞ 

an = B × A.

n=1

However, one may ask if it is possible to form a series whose terms are products of the form .an bk , in some order, so that the sum of this series is .A × B. The answer to this question is given in the following theorem: Theorem 2.6.1 Let . an and . bn be two convergent series, with sums A and B, respectively. Let .φ be a bijective function, .φ : N2 → N, cn , with .φ(i, j) = n. To each .φ, it corresponds to a series . c = cφ(i,j) = ai × bj .

. n



The series . cn converges, regardless of the function .φ considered if and only if . an and . bn are both absolutely convergent, and, in this case, we have . cn = A×B, and the series . cn is also absolutely convergent. Note: Saying that . cn converges for any function .φ considered is equivalent to state that the product series converges regardless of the order in which its terms are taken. ∞  Definition 2.6.1 The Cauchy product of two convergent series, . an n=1   ∞ ∞ n    and . bn , is the series . ak bn−k+1 . n=1

n=1

k=1

114

Numerical Series Note: If .n ∈ N0 , then the Cauchy product is written as follows:  n  ∞   . ak bn−k . n=0

k=0

Corollary 1 If . an and . bn are absolutely convergent series with sums A and B, respectively, their Cauchy product is absolutely convergent and has sum .A × B.

∞  xn , .x ∈ R. As n! n=0

Example 2.6.1 Consider the series .

 n+1    x    (n + 1)!  |x|  n = lim = 0, ∀x ∈ R, . lim x  n +1    n!  the series is absolutely convergent for every .x ∈ R. Then, the Cauchy product of two series of this type is absolutely convergent. Let us form the product and verify that the obtained series is absolutely convergent.  ∞   ∞  n ∞  xn  yn   xk y n−k × = · n! n! k! (n − k)! n=0 n=0 n=0 k=0   ∞ n  n! 1  k n−k .= ·x y n! k!(n − k)! n=0 k=0

=

∞ 

(x + y)n . n! n=0

Note: The Cauchy product of two series that are not absolutely convergent may lead to a divergent series. ∞  (−1)n √ is conditionally convergent. Caln+1 n=0 culating the Cauchy product of the series with itself, we get

Example 2.6.2 The series

.

2.6. Products of Series

115

 (−1)k (−1)n−k  · (k + 1) (n − k + 1) n=0 k=0  n  ∞   (−1)n √ √ .= k+1 n−k+1 n=0 k=0  n  ∞ ∞    1 n √ √ = (−1) (−1)n an , = k + 1 n − k + 1 n=0 n=0 k=0  n ∞  

which is an alternating series. However, .(an ) is not a null sequence as n 

 1 1 √ √ √ √ .an = ≥ = 1; n + 1 n+1 k + 1 n − k + 1 k=0 k=0 n

thus, the product series is divergent.

Theorem 2.6.2 (Mertens) If at least one of the convergent series, . an and . bn , is absolutely convergent, then their Cauchy product is convergent, and its sum is the product of the sums of the given series. Theorem 2.6.3 If the series . an and . bn are convergent, with sums A and B, respectively, then if their Cauchy product is convergent, its sum is .A × B. ∞  (−1)n is conditionally convergent, and the n n=1

Example 2.6.3 The series .

∞  (−1)n series . is absolutely convergent. By Mertens’s Theorem, the n2 n=1 Cauchy product of the series, which is alternating, is convergent:

 ∞   n  ∞ ∞   (−1)n   (−1)k (−1)n−k+1 (−1)n × · = n n2 k (n − k + 1)2 n=1 n=1 n=1 k=1 .   ∞ n   1 n+1 = (−1) . k(n − k + 1)2 n=1 

k=1

116

2.7

Numerical Series

Solved Exercises

1. Show that the following series are convergent and find their sums: a)

∞  n=1

b)

1 n2 + 3n + 2

∞  

arctan(n + 3) − arctan(n)



d)

 ∞   1 1 − n+1 n e e n=0 ∞   n=1

e)

∞  n=0

g)

h)

∞  log(n7 + 1) n2 n=1   ∞  √ 1 d) n+ − n n n=1 √ ∞  n + 1 − √n √ e) √ n+1+ n n=1



f)

g)

∞  cos(n) n2 + 4 n=1 ∞  n=1

23n−2 h)

1 n3 + 10 cos(n)

∞  1 − (−1)n √ n n=1

5. Investigate the convergence of the following series: 1 4n−1

∞  n=2

−

1



3n + 7n 3n · 7n

∞  n=0

1 22n−1

n=1

f)

1 1 − arctan(n) arctan(n + 1)

∞  

 2 ∞   1 sin n n=1

c)

n=1

c)

b)

2 3 · n · n+1 5 6

a)

∞ 

(−1)n

n=1

log( n+2 ) n log(n) log(n + 2)

b)

∞ 

log(n) n

(−1)n+1

n=1

2. Show that the series ∞  . n=2



1 1 + n (n + 1)2 π



c)

∞ 

∞ 

1 log(3n) 1 sin n n

(−1)n

n=1

d)

(−1)

n

n=1

is convergent and find its sum, knowing that ∞  1 π2 . = 2 n 6 n=1

e)

3. Write the following repeating decimals as fractions:

f)

a) 0, 5

c) 1, 345

b) 0, 34

d) 0, 324101

4. Test the convergence of the following series using a comparison test:

a)

∞  n=1

2 n2 − 1 3 n5 + 2 n + 1

∞  n=1

2 en + e−n

(−1)n

3n e2n+1

  ∞  sin nπ √ 2 n+1 n=1

6. Consider the series

∞  (−1)n+1 : n n=1

a) Study it for convergence. b) Indicate a partial sum Sn that approximates the sum of the series with an error less than 1 . 1000

2.7. Solved Exercises c) Indicate an upper bound of the error committed when S5 is taken as the sum of the series. 7. Use the Ratio Test or D’Alembert’s Test to determine the convergence or divergence of the following series: a)

b)

∞  nn π n n! n=1 ∞  n=0

n 2n 4 n3 + 1

∞  (n + 1)! − n! c) n! (n + 1)! n=1 ∞  2 · 4 · 6 · · · · · (2n + 2) d) 1 · 4 · 7 · · · · · (3n + 1) n=1

e)

∞  n=1

f)

g)

∞  3n + n! n! + nn n=1 ∞  n=1

h)

(n + 1)! √ nn 3n + 2

∞  n=0

n! (2 n)! + 2n n! + 3n ((n + 1)!)2

8. Use the Root Test or Cauchy’s Root Test to determine the convergence or divergence of the following series: a)

∞  

1−

n=3

b)

∞ 

e−n

2 n

n2

c)

n=1

∞  (−1)n n) log(n n=2     ∞  n2 2 n √ b) + 1+ n n5 + 1 n=1   ∞  1 c) log 1 − 2 n n=2

a)

d)

e)

∞  sin(n) + 2n n + 5n n=1 ∞ 

en+1 nn

(−1)n

n=1

f)

g)

h)

∞ (−1)n +  n=1 ∞  n=1 ∞ 

1 n

n   1 log 1 + n 2 

1−

n √ n n

n=1 ∞ 

(−1)n 1 − (−1)n n2 n=1  ∞  (−1)n + 4  3 2n j) + n + 4n n+2 n=1   ∞  1 n 1− k) n n=1 i)

l)

m) n 2−1

n  ∞   1 π + cos 6 n n=1 n  ∞  n 1 + (−1)n e) 2 4n + 1 n=0 n ∞  2  n −2 f) 3n2 n=1

d)

9. Test for convergence or divergence the following series. If convergence occurs, indicate whether it is conditional or absolute:

2

n=0 ∞ √  n

117

n)

∞  n=2 ∞  n=1 ∞ 



1

3 n log(n) n+2 1 dx 2 x n √

n=1 ∞ 

(−1)n √ n+1+ 3n+1

1 √  n + 1+ n n=1    ∞  1 p) (−1)n 1 − cos n n=1   ∞  1 (−1)n 2n sin n q) 3 n=1 o)

n

√

118

Numerical Series √ n! n √ nn n + 1 n=1   ∞  √ 1 ( n + 1) sin s) n2 n=1 r)

t)

∞ 

∞  n=2

(−1)n

1 log(n!)

Hint: Note that n! < nn 10. Assess the convergence or divergence of the given series. If the series converges, specify whether the convergence is absolute or conditional: a)

b)

∞  n cos(n π) √ n3 + 1 n=0 ∞  n=1

c)

∞  n=1

n)

n=1

∞  3n (n + 1)! (n + 1)n n=0   ∞  (−1)n e) sin n n=1 √ ∞  n f) 2 + cos(n) n n=1

c)

d)

g)

n=1

h)

3n 2n + nn

∞  1 + n(−1)n 1 + 2n3 n=1 ∞ 

2n i) n! + 1 n=1   ∞  1 n! j) − n (2n)! 2 n=0 ∞  n (2n)! k) 4n (n!)2 n=1 ∞ 

sin(n) l) 3 log(n) n n=2   ∞  1 n m) (−1)n 1 − 2n n=1

  1 n

∞  cos(πn) √ n2 − 1 n=2 ∞  n=2

(−1)n √ n log(n)

e)

∞  4 + sin(n) √ 3 n+1 n=1

f)

∞  n2n sin(n3 ) (3n2 + 5)n n=1

d)

∞ 

1 arcsin 2n + 5

11. Determine whether the following series are convergent or divergent. If convergence occurs, indicate if it is conditional or absolute:   ∞  1 1 a) sin n n 2 n=1 √ ∞ 3  n2 b) √ n n + 2n2 n=1

n! 3 × 5 × · · · × (2 n + 1) 1 + cos2 (n) √ n

∞ 

g)

∞ 



(−1)n arcsin

n=3

h)

n−1 n2 + 1



∞  4 × 7 × · · · × (3n + 1) 8 × 11 × · · · × (3n + 5) n=1

 ∞   arctan(n3 ) 2n (2n)! + √ n + n2 3n (2n + 1)! n=1

 ∞ sin √1  n j) √ n+ n n=1 i)

k)

l)

m)

n)

    ∞   1 1 n sin − (n + 2) sin n n+2 n=1 ∞  2n + 3 (n + 1)! n=1 ∞  arctan(n + 1) − arctan(n) n2 n=1 ∞  n=1

o)

log(n)   1 n sin n

∞   n=1

log(2)

n

log



n+1 n



2.7. Solved Exercises p)

∞  n=3

12.

 n log(n)

119 1   log log(n)

p , p ∈ R

a) Test for convergence or divergence the series ∞  (n + 1)n . 3n n! n=1

b) Based on the previous item, find the limit of (n + 1)n the sequence with general term . 3n n!

13. Test for convergence or divergence the series an where ⎧  n2 ⎪ n ⎪ ⎪ , if n = 3k, k ∈ N ⎪ ⎪ ⎪ n+1 ⎪ ⎪ ⎪ ⎨1 , if n = 3k + 1, k ∈ N0 .an = n! ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (−1)n ⎪ ⎪ , if n = 3k + 2, k ∈ N0 . ⎪ ⎩ (n + 1)n If convergence occurs, indicate whether it is conditional or absolute. 14. Investigate, by two distinct processes, the conver gence or divergence of the series an where ⎧ n 2 ⎪ , if n is even ⎪ ⎪ 2 ⎨ .an = (n + 1)(n − 1) ⎪ ⎪ , if n is odd. ⎪ ⎩ 22 If convergence occurs, indicate whether it is conditional or absolute. 15. Let (an )n∈N be a sequence of positive terms such

that the series nan is convergent. Prove that the

2 series (an ) is convergent.

a) Show that Leibniz’s test cannot be applied. b) Show that the series is divergent. 18. Let (an ) be a sequence of real numbers. Consider the sequence (bn )⎧defined by ⎨a n+1 , if n is odd 2 .bn = ⎩−a n , if n is even. 2

Show that:

a) The series bn is convergent if and only if lim an = 0.

b) The series bn is absolutely convergent if

and only if the series an is absolutely convergent.

19. Show that if (an ) is a decreasing sequence and an converges, then lim(n an ) = 0. Hint: Relate the convergence of the series to the fact that the sequence of partial sums is a Cauchy sequence (choose m and n conveniently.)

20. Let an be a convergent series of positive terms.  √a n 1 Show that the series converges if p > . np 2 Hint:Consider the inequality ab ≤

21. Let an be a sequence of positive terms such that

1 an+1 ≥ 1 − , ∀n ∈ N. Show that the series an an n is divergent. 22. Let g : R → R be a function of class C 2 . Suppose there exists p ∈ N such that g, g  , (g  )2 −g g  are all positive functions on [p, +∞[. Show that the series

16. Show that, if an > 0, ∀n ∈ N, and the series an

converges, then the series log(1 + an ) converges. 17. Consider the series

∞ 

.

∞  g  (n) g(n) n=p

is convergent if and only if (−1)

n−1

an where

n=1

⎧ ⎪ ⎪ ⎪1, ⎪ ⎨n .an = ⎪ 1 ⎪ ⎪ ⎪ ⎩ n2 ,

a2 + b 2 , ∀a, b ∈ R. 2

lim g(x) exists and is

x→+∞

finite. 23. Test for convergence or divergence the series ∞ 

if n is odd .

if n is even.

arctan(vn )

n=2

where v2 = k > 0 and vn+1 = vn sin

  π , n ≥ 2. n

120

Numerical Series

24. The Sierpinski triangle is a geometric figure constructed as follows: A square is divided into four equal squares, and the upper right square is removed. Each of the remaining three squares is divided in the same way, and the upper right square of each is removed.

to the sides of the triangle and to the circles that precede and follow it in the sequence. Find the sum of the areas of the circles.

Continuing this process to infinity, we obtain a figure called the Sierpinski triangle. 26. Consider a square inscribed in a circle of radius 1. In this square, inscribe a circle and in it a new square, and so on, as suggested by the figure. Find the area of the colored region.

Suppose the length of the side of the initial square is equal to 1. What is the total area of the squares removed from the initial square? What can we conclude about the area of the Sierpinski triangle? 25. Consider an equilateral triangle with side length 1. Construct a sequence of circles approaching the vertices of the triangle such that each circle is tangent

2.7. Solved Exercises

121 SOLUTIONS

1.

a) Let an be the general term of the series

∞  n=1

an = .

from which



1 . Since n2 + 3n + 2 = (n + 1)(n + 2), we get n2 + 3n + 2

A B 1 = + n2 + 3n + 2 n+1 n+2

=

A(n + 2) + B(n + 1) (n + 1)(n + 2)

=

(A + B)n + 2A + B , (n + 1)(n + 2)

A+B =0

.

2A + B = 1

 ⇔

A=1 B = −1.

Therefore, .an

=

1 1 − . n+1 n+2

1 and k = 1, using the notation of formula (2.1) on page 74. In n+1 this case, lim αn = 0, so we can conclude that the series converges and, using formula (2.2) on the page mentioned above, its sum is  ∞   1 1 1 − . = α1 = . n + 1 n + 2 2 n=1

The series is telescopic with αn =

Alternatively, we can write the sequence of partial sums and determine its limit as follows: 1 1 − 2 3 1 1 1 1 1 1 = − + − = − 2 3 3 4 2 4 1 1 1 1 1 1 = − + − = − 2 4 4 5 2 5 .. . 1 1 = − 2 n+2 . . .

S1 = S2 .

S3

Sn

As lim Sn =

1 1 , the series is convergent and its sum is : 2 2 ∞  . n=1

1 1 = . n2 + 3n + 2 2

122

Numerical Series ∞  

b) Let us consider the series

+∞     arctan(n + 3) − arctan(n) = − arctan(n) − arctan(n + 3) . It is

n=1

n=1

telescopic with αn = arctan(n) and k = 3, using the notation of formula (2.1) on page 74. In this case, π lim αn = , therefore, we can conclude that the series converges, and, using the formula (2.2), its sum is 2 .

−

5π π = − arctan(2) − arctan(3). (arctan(n) − arctan(n + 3)) = − α1 + α2 + α3 − 3 · 2 4 n=1 ∞ 

An alternative path to solve this problem is to write the sequence of partial sums and find its limit: π 4 π arctan(4) − + arctan(5) − arctan(2) 4 π arctan(4) − + arctan(5) − arctan(2) + arctan(6) − arctan(3) 4 π arctan(4) − + arctan(5) − arctan(2) + arctan(6) − arctan(3) + arctan(7) − arctan(4) 4 π − + arctan(5) − arctan(2) + arctan(6) − arctan(3) + arctan(7) 4 π − + arctan(5) − arctan(2) + arctan(6) − arctan(3) + arctan(7) + arctan(8) − arctan(5) 4 π − − arctan(2) + arctan(6) − arctan(3) + arctan(7) + arctan(8) 4 π − − arctan(2) − arctan(3) + arctan(6) + arctan(7) + arctan(8) 4

S1 = arctan(4) − arctan(1) = arctan(4) − S2 = S3 = S4 = = .

S5 = = =

Sn

. . . π = − − arctan(2) − arctan(3) + arctan(n + 1) + arctan(n + 2) + arctan(n + 3) 4 . ..

As lim Sn = − and

3π 5π π − arctan(2) − arctan(3) + = − arctan(2) − arctan(3), the series is convergent 4 2 4 +∞  . n=1



 5π arctan(n + 3) − arctan(n) = − arctan(2) − arctan(3). 4

 ∞   1 1 1 c) The series − is telescopic, with αn = n and k = 1, using the notation of formula (2.1) n n+1 e e e n=0 1 on page 74. As lim n = 0, the series is convergent. The sum of the series is e  ∞   1 1 1 1 . − = 0 − lim n = 1. en en+1 e e n=0 Another process to find the sum is to write the general term in the form

1 en



1−

 1 , highlighting the fact e

1 1 . As 0 < < 1, the series converges and: e e      ∞   ∞  ∞   1 1 1 1  1 1 1 1 = 1 − = 1. = 1 − . − = 1 − en en+1 en e e n=0 en e 1− 1 n=0 n=0 e

that this series is geometric with ratio r =

2.7. Solved Exercises d) The series

∞   n=1

123 

1 1 − arctan(n) arctan(n + 1)

is telescopic, with αn =

notation of formula (2.1) on page 74. In this case, lim αn = converges and, using formula (2.2), has sum ∞   . n=1

1 1 − arctan(n) arctan(n + 1)

 =

1 and k = 1, using the arctan(n)

2 . Therefore, we can conclude that the series π

2 4 2 2 1 − = − = . arctan(1) π π π π

Alternatively, we can write the sequence of partial sums and find its limit: 1 4 1 1 − = − arctan(1) arctan(2) π arctan(2) 1 1 1 4 1 4 − + − = − = π arctan(2) arctan(2) arctan(3) π arctan(3) 1 1 1 4 1 4 − + − = − = π arctan(3) arctan(3) arctan(4) π arctan(4) . .. 4 1 = − π arctan(n + 1) . . .

S1 = S2

.

S3

Sn



As . lim Sn

= lim

4 1 − π arctan(n + 1)

 =

4 2 2 − = , π π π

the series is convergent, and its sum is ∞   . n=1

e) The following two series

∞ 

1 22n−1

n=1

∞  . n=1

and

∞  . n=1

1 22n−1

1 23n−2

1 1 − arctan(n) arctan(n + 1) ∞ 

and

n=1 ∞ 

=

n=1

=

∞  n=1

 =

2 . π

1 are geometric series. To see this, observe that: 23n−2

∞ ∞   1 1 1 = 2 = 2 2n n 22n · 2−1 2 4 n=1 n=1 ∞ ∞   1 1 1 = 4 = 4 . 3n n 23n · 2−2 2 8 n=1 n=1

1 1 The first series has ratio r = , and the second has ratio r = . The series are convergent as in both cases 4 8 |r| < 1. The sum of the series can be calculated as follows: ∞  . n=1

1 22n−1

By Theorem 2.2.2 the series

=2·

∞   n=1

1 4

1−

1 4

1 22n−1

=

−

2 3

and

∞  n=1

1 23n−2

1 23n−2

=4·

1 8

1−

1 8

=

4 . 7

 is convergent, and its sum is equal to

4 2 2 − = . 3 7 21

124

Numerical Series  ∞ ∞  ∞ ∞     1 3n + 7n 1 1 1 . Both can be written in the form + and are n · 7n n n n n 3 7 3 7 3 n=0 n=0 n=0 n=0 1 1 geometric series with ratios r = and r = , respectively. 7 3 The series are convergent as in both cases |r| < 1. Using the formula for the sum of a convergent geometric series, we get ∞ ∞   1 7 1 3 1 1 . = = = = . and 1 1 n n 7 6 3 2 1 − 1 − 7 3 n=0 n=0

f) The series

By applying Theorem 2.2.2, we can conclude that the given series

∞  3n + 7n is convergent, and its sum 3n · 7n n=0

7 3 8 + = . 6 2 3 ∞  1 2 3 1 because · n · n+1 is geometric with ratio r = g) The series n−1 4 5 6 120 n=0 is equal to

.

1 4 2 3 6·4 = · · = n n n . 4n−1 5n 6n+1 4 ·5 ·6 ·6 120n

Thus, we can calculate the sum as follows: ∞  . n=0

1 4n−1

·

∞  1 2 3 1 480 . · = 4 =4· = 1 n 5n 6n+1 120 119 1 − 120 n=0

h) Using the properties of logarithms, we can obtain the following equality: ∞  . n=2

log( n+2 ) n log(n) log(n + 2)

=

∞   n=2

1 1 − log(n) log(n + 2)

 .

1 and k = 2, using the notation of formula (2.1) on page 74. log(n) 1 = 0, and The series is convergent because lim log(n)

Thus the series is telescopic, with αn =

∞   . n=2

2. By reindexing the series, we can write

1 1 − log(n) log(n + 2)

∞  n=2

.

Knowing that

 =

1 1 + . log(2) log(3)

∞  1 1 = . Additionally, (n + 1)2 n2 n=3 ∞ ∞   1 1 1 =1+ + . 2 2 n 4 n n=1 n=3

∞ ∞   1 1 1 π2 5 π2 π2 , then −1− = − , that is, = = 2 2 n 6 n 6 4 6 4 n=1 n=3 ∞  . n=2

1 π2 5 = − . 2 (n + 1) 6 4

2.7. Solved Exercises

125

∞  1 1 is convergent because it is geometric with ratio . By Theorem 2.2.2, the sum of these two n π π n=2 series is convergent and

The series

∞   . n=2

3.

1 1 + n (n + 1)2 π

 =

∞  n=2

1 ∞  1 1 5 5 1 π2 π2 2 − + π 1 = − + : + = 2 n (n + 1) π 6 4 6 4 π(π − 1) 1 − π n=2

a) The rational number 0, 5 = 0, 55555 . . . can be written in the form of a fraction using geometric series. By writing 5 5 5 5 + 2 + 3 + 4 + ··· .0, 5 = 0, 5 + 0, 05 + . . . = 10 10 10 10 5 1 and first term . Therefore, it is clear that we have a geometric series with ratio r = 10 10 .0, 5

=

5 10

1−

1 10

=

5 . 9

b) As in the previous item, we can use geometric series to write the rational number 0, 34 = 0, 343434 . . . in the form of a fraction. By writing .0, 34

= 0, 34 + 0, 0034 + . . . =

34 34 34 34 + 4 + 6 + 8 + ··· 102 10 10 10

it is evident that we have a geometric series with ratio r =

.0, 34

=

34 102

1−

1 102

1 34 and first term . Therefore, 102 102

=

34 . 99

c) We can write the rational number 1, 345 in the following form: .1, 345

= 1 + 0, 345 + 0, 000345 + . . . = 1 +

345 345 345 345 + 6 + 9 + 12 + · · · 103 10 10 10

After the first term, we have a geometric series with ratio r =

.1, 345

=1+

1

345 103 − 1013

=1+

1 345 and first term . Therefore, 103 103

345 448 = . 999 333

d) By writing the rational number 0, 324101 in the form .0, 324101

= 0, 324 + 0, 000101 + 0, 000000101 + . . . =

324 101 101 101 + 6 + 9 + 12 + · · · 103 10 10 10

we see that after the first term, we have a geometric series with ratio r = 101

.0, 324101

=

1 101 and first term 6 . Therefore, 103 10

324 324 101 323777 6 . + 10 1 = + = 103 103 999 × 103 999000 1 − 103

126 4.

Numerical Series a) The series

∞  n=1

2 n2 − 1 is of positive terms. To study this series, we will compare it with the series 3 n5 + 2 n + 1

∞  1 , which is convergent because it is a p-series with p = 3: n3 n=1

2 n2 − 1 (2 n2 − 1) n3 2 n5 − n3 2 3 n5 + 2 n + 1 . lim = lim = lim = . 5 1 3n + 2n + 1 3 n5 + 2 n + 1 3 n3 2 ∈ R+ , according to Corollary 2 of the General Comparison Test, we can conclude that the given 3 ∞  2 n2 − 1 is convergent. series 5 + 2n + 1 3 n n=1

Since

 2  π 1 1 is of positive terms because ∈ ]0, 1] ⊂ 0, , ∀n ∈ N. To determine its n n 2 n=1 ∞  1 , which we know is convergent as it is a p-series with convergence, we compare it with the series 2 n n=1 p = 2. We calculate the limit

b) The series

∞  

sin

  2 ⎡   ⎤2 1 1 sin sin ⎢ n n ⎥ ⎥ = 1, ⎢ . lim = lim ⎣ ⎦ 1 1 n2 n which belongs to R+ . Therefore, by Corollary 2 of the General Comparison Test, the given series is convergent. ∞  log(n7 + 1) is a series of positive terms. We can analyze this series by comparing it with n2 n=1 ∞  1 3 , which is convergent as it is a p-series with p = : the series 3/2 2 n n=1

c) The series

log(n7 + 1) n3/2 log(n7 + 1) log(n7 + 1) n2 . lim = lim = lim = 0. 2 1 n n1/2 3/2 n ∞ 

1 is convergent, the fact that this limit is zero allows us to conclude, by Corollary 3 3/2 n n=1 ∞  log(n7 + 1) of the General Comparison Test, that the series is also convergent. n2 n=1 Since the series

Note: For calculating the limit lim lim

x→+∞

log(n7 + 1) , we used L’Hˆ opital’s Rule applied to the calculation of n1/2

log(x7 + 1) , where the indeterminate form of type x1/2

∞ ∞

arises:

2.7. Solved Exercises

127

 .

lim

log(x7 + 1)

 =

(x1/2 )

x→+∞

lim

7x6 √ 14x6 x +1 = lim = 0. 1 x→+∞ x7 + 1 √ 2 x

x7

x→+∞

d) Let us begin by rewriting the general term of the series:  √ √ 1 1 1 n+ + n − n n n √ 1 n  =  = n+ − n= n √ √ 1 1 n+ + n n+ + n n n 1 1 1 ⎞ = = ⎛ =  √ . 3 + n + n3 √ 1 2 n √ n + 1 n n+ + n + n⎠ n⎝ n n 



an .

n+

As an > 0, ∀n ∈ N, the series is of positive terms. Let us compare it with the series

∞  n=1

convergent because it is a p-series with p = √ . lim

1 n3 + n + 1 n3/2

1 , which is n3/2

3 . The limit 2 √

n3

√ = lim

n3

n3

+n+

√

n3

=

1 2

is finite and different from zero; therefore, by Corollary 2 of the General Comparison Test, the series ∞  √ 1 n + − n is convergent. n n=1 e) The series

∞  n=1

√ √ n+1− n √ √ is of positive terms. Considering that n+1+ n √ .√

√ √ √ √  √ n+1− n n+1+ n n+1− n 1 = √ = √ √ √ 2 √ 2 , n+1+ n n+1+ n n+1+ n

we can write the series in the form

∞ 

√

n=1

the harmonic series: √ . lim

1 n+1+ 1 n

1 n+1+

√ 2 n

√ 2 . We will analyze this series by comparing it with n

= lim √

n n+1+

1 √ 2 = . 4 n

1 ∈ R+ , we can conclude, by Corollary 2 of the General Comparison Test, that the given series Since 4 √ √ ∞  n+1− n √ √ is divergent. n+1+ n n=1

128

Numerical Series ∞  cos(n) is not of positive terms. However, by bounding the modulus of the general term, we 2+4 n n=1 can obtain the following inequality:

f) The series

.0

! ! ! cos(n) ! 1 1 !≤ ≤ 2 , ∀n ∈ N. < !! 2 n + 4! n2 + 4 n

(2.5)

∞  1 is convergent since it is a p-series with p = 2. Using the inequality 2 n n=1 ! ∞ !  ! cos(n) ! ! ! (2.5), we can apply the General Comparison Test to show that the series ! n2 + 4 ! converges. Therefore,

It is worth noting that the series

n=1

∞  cos(n) is absolutely convergent. the original series 2+4 n n=1

g) Recall that −1 ≤ cos(n) ≤ 1, ∀n ∈ N. Using this fact, we obtain n3 + 10 cos(n) ≥ n3 − 10, ∀n ∈ N. Thus, we have that .n

3

+ 10 cos(n) ≥ n3 − 10 > 0, ∀n ≥ 3.

This inequality implies that .0

The series

∞  n=3

series


0, ∀n > 1. n

log(x) 1 − log(x) (iii) The function f (x) = is decreasing on ]e, +∞[. Indeed, f  (x) = < 0 ⇔ x > e, x  x2  log(n) which implies that is decreasing for n ≥ 3. n Therefore, we can state that the alternating series is convergent. As the series of modules is divergent, the given series is conditionally convergent, taking into account Definition 2.4.1. Note: The series terms. ∞ 

∞ ∞   log(n) log(n) and are both divergent since they differ only in a finite number of n n n=2 n=3

2 2 is alternating, given that n > 0, ∀n ∈ N. To begin with, let us n + e−n e e + e−n n=1 study the series of modules:

b) The series

(−1)n+1

.

! ∞ ! ∞   ! ! 2 2 !(−1)n+1 != . ! ! n −n n + e−n e + e e n=1 n=1

130

Numerical Series We have .0


0, ∀n ∈ N. To study the convergence (−1)n c) The series log(3n) log(3n) n=1 ∞  1 of the series, let us first consider the series of modules, . We can compare this series with the log(3n) n=1 harmonic series: 1

The series

. lim

n log(3n) = lim = +∞. 1 log(3n) n ∞ 

1 is divergent. Since the log(3n) n=1 initial series is alternating, we can check the conditions of Leibniz’s test to see if it converges: 1 = 0. (i) lim log(3n) 1 (ii) > 0, ∀n ≥ 1. log(3n)     1 (iii) The sequence log(3n) is increasing; therefore, is a decreasing sequence. log(3n) According to Leibniz’s test, the alternating series is convergent. However, as the series of its modules is divergent, the given series is conditionally convergent, considering Definition 2.4.1. It follows by Corollary 4 of the General Comparison Test that the series

d) The series

∞ 

(−1)n

n=1

sin

1

n

n

is alternating, because an =

∀n ≥ 1). The series of modules is

∞  sin n=1

1 n

sin

1

n

n

> 0, ∀n ≥ 1 (recall that 0
0, ∀n ∈ N. To analyze the absolute convergence 2n+1 e e n=1 of the series, we consider the series of modules:

e) The series

(−1)n

.

!   n ∞ ! ∞ ∞ ∞  n !    ! 3 3n 1 1  3 n !(−1)n 3 != = = . ! ! 2n+1 2n+1 2 2 e e e e e n=1 e n=1 n=1 n=1

 ∞   3 n 3 is geometric with ratio r = 2 . Since |r| < 1, the series is convergent. We know by 2 e e n=1 Theorem 2.2.2 that the product of a constant by a convergent series is convergent; therefore, the series of modules is convergent. Then, the given series is absolutely convergent, again considering Definition 2.4.1. The series

f) Considering that



nπ . sin 2

we can write the series

∞  n=1

 =

⎧ ⎨0,

if n is even

⎩(−1) n−1 2 ,

if n is odd,

 nπ 

sin 2 √ as n+1

.

∞ ∞  1  (−1)n−1 (−1)n−1 √ = √ . √ n 2n 2 n=1 n=1

∞  1 √ , which is divergent. We will now apply n n=1 Leibniz’s test to determine the convergence of our given series:

This series is alternating, and its series of modules is

1 (i) lim √ = 0. n 1 (ii) √ > 0, ∀n ∈ N. n   1 1 1 is decreasing. > 0, ∀n ∈ N, and therefore, the sequence √ (iii) √ − √ n n n+1 Thus, the alternating series is convergent. However, since the series of absolute values is divergent, we know the given series is conditionally convergent.

6.

a) The series

∞ 

(−1)n+1

n=1

∞  1 1 is alternating. The series of modules is the harmonic series, which is n n n=1

divergent. To determine the convergence of the alternating series, let us apply Leibniz’s test: (i) lim

1 = 0. n

1 > 0, ∀n ∈ N. n 1 1 1 − = > 0, ∀n ∈ N, and therefore, the sequence is decreasing. (iii) n n+1 n(n + 1) (ii)

Thus, the alternating series is convergent. As its series of modules is divergent, the given series is conditionally convergent.

132

Numerical Series b) From the Corollary of Theorem 2.3.2, we know that |S − Sn | ≤ an+1 . It is sufficient to require that 1 , that is, n > 999. Therefore, the intended sum is S1000 . an+1 < 1000 c) By the corollary referred to in the previous item, |S − S5 | ≤ a6 = the error.

7.

1 , 6

and this value is an upper bound of

∞  nn is of positive terms. Let an be the general term of the series. We can find the limit π n n! n=1 of the ratio of successive terms as follows:

a) The series

lim .

an+1 an

(n + 1)n+1 (n + 1)n+1 π n n! π n+1 (n + 1)! = lim = lim n n+1 n n n π (n + 1)! π n n!   (n + 1)(n + 1)n n! 1 n+1 n e = lim = lim = . n n π (n + 1) n! π n π

Since this value is less than 1, D’Alembert’s test guarantees that the series is convergent. ∞ 

n 2n is of positive terms. Let an be the general term of the series. We will take the 3+1 4 n n=0 limit of the ratio of successive terms of the series:

b) The series

(n + 1)2n+1 (n + 1) 2n+1 (4n3 + 1) an+1 4(n + 1)3 + 1   = lim lim = lim n n2 an n 2n 4(n + 1)3 + 1 . 4n3 + 1 = lim

2 (n + 1) (4n3 + 1) 2(n + 1)(4n3 + 1)   = lim = 2. 3 n (4n3 + 12n2 + 12n + 5) n 4(n + 1) + 1

As this limit has a value greater than 1, the series is divergent by D’Alembert’s test. c) Noting that .

∞ ∞ ∞    (n + 1)! − n! (n + 1)n! − n! n = = , n! (n + 1)! n! (n + 1)! (n + 1)! n=1 n=1 n=1

let an be the general term of the series. As n+1 (n + 1)(n + 1)! n+1 an+1 (n + 2)! = lim = lim = 0 < 1, . lim = lim n an n(n + 2)! n(n + 2) (n + 1)! it follows from D’Alembert’s test that the series is convergent. d) The series have

∞  2 · 4 · 6 · · · · · (2n + 2) is of positive terms. Let an be the general term of the series. We 1 · 4 · 7 · · · · · (3n + 1) n=1

2 · 4 · 6 · · · · · (2n + 2)(2n + 4) 2 an+1 2n + 4 1 · 4 · 7 · · · · · (3n + 1)(3n + 4) = . . lim = lim = lim 2 · 4 · 6 · · · · · (2n + 2) an 3n + 4 3 1 · 4 · 7 · · · · · (3n + 1)

2.7. Solved Exercises

133

As this value is less than 1, we can conclude by D’Alembert’s test that the series is convergent. e) The series

∞  n=1

lim

n→∞

(n + 1)! √ is of positive terms. Let an be the general term of the series. Evaluating nn 3n + 2

an+1 , we obtain an (n + 2)! √ √ an+1 (n + 2)! nn 3n + 2 (n + 1)n+1 3n + 5 √ lim = lim = lim (n + 1)! an (n + 1)! (n + 1)n+1 3n + 5 √ . n n 3n + 2  √  n 1 n n + 2 3n + 2 (n + 2) nn 3n + 2 √ = . = lim = lim n+1 n + 1 3n + 5 n + 1 e (n + 1) 3n + 5

Since this value is less than 1, the series is convergent according to D’Alembert’s test. ∞  3n + n! is of positive terms. It is easy to prove by induction that 3n < n!, ∀n ≥ 7, which n! + nn n=1 implies that

f) The series

3n + n! 2 n! ≤ n , ∀n ≥ 7. n! + nn n ∞  2 n! be the general term of the series . D’Alembert’s test confirms that the series converges nn n=1 .

Let an since

2 (n + 1)! n  an+1 (n + 1)! nn n 1 (n + 1)n+1 . lim = lim = lim = lim = < 1. 2 n! an n! (n + 1)n+1 n+1 e nn ∞  3n + n! By the General Comparison Test, we can conclude that the series also converges. n! + nn n=1 Note: Two series that differ only in a finite number of terms are both convergent or both divergent. g) The series

∞  n=1

n! is of positive terms. We can establish the inequality (2 n)! + 2n .

n! n! , ∀n ∈ N. ≤ (2 n)! + 2n (2 n)!

Let us investigate the convergence of the series

∞  n=1

then

n! . If an denotes the general term of this series, (2 n)!

(n + 1)!   2 (n + 1) ! (n + 1)! (2 n)! n+1 an+1   = lim = 0. = lim . lim = lim n! an (2n + 2)(2n + 1) n! 2 (n + 1) ! (2 n)!

By D’Alembert’s test, the series is convergent since this limit is less than 1. Applying the General Comparison ∞  n! Test, the original series is convergent. (2 n)! + 2n n=1

134

Numerical Series h) The series ∞ 

∞ 



n=0

∞ 

∞ ∞   n! + 3n n! an = 2 is of positive terms. Let us consider the series  2 and (n + 1)! n=0 n=0 (n + 1)!

3n 2 . If both series are convergent, it follows from Theorem 2.2.2 that the given series (n + 1)! a : is convergent. Let us begin by finding the limit of n+1 a

n=0

bn =



n=0

n

(n + 1)!  2  2 (n + 2)! (n + 1)! (n + 1)! an+1 n+1 . lim = lim = lim = lim = 0 < 1.   2 n! an (n + 2)2 n! (n + 2)!  2 (n + 1)! As the limit is less than 1, we can use D’Alembert’s test to conclude that the series Similarly, the series

∞ 

an is convergent.

n=0

∞ 

bn is convergent:

n=0

. lim

bn+1 bn

3(n + 1) 2  2 (n + 2)! (n + 1) (n + 1)! n+1 = lim = lim = lim = 0 < 1.   2 3n n(n + 2)2 n (n + 2)!  2 (n + 1)! 

Since both series are convergent, we can conclude that the given series is also convergent as it is the sum of two convergent series.

8.

∞  

a) Consider the series of positive terms

1−

n=3

. lim

√ n

 an = lim

n

2 n

n2 and let an represent its general term. Since

   2  2 n 1 2 n = lim 1 − = e−2 = 2 < 1, 1− n n e

the series converges by Cauchy’s Root Test.

b) The series

∞ 

2

e−n =

n=0

∞  n  1 n=0

2

is of positive terms. We have

e

 . lim

n

 n   n2 1 1 = lim = 0. e e

As this value is less than 1, Cauchy’s Root Test ensures the convergence of the series. c) The series

∞ √  n

n 2−1 is of positive terms. We find that:

n=1 . lim



√ n

n

n √  n 2−1 = lim 2 − 1 = 0,

√ since lim n 2 = 1. As this value is less than 1, Cauchy’s Root Test allows us to conclude that the series is convergent.

2.7. Solved Exercises

135

∞  



n π 1 1 π π + + < , ∀n ∈ N. Applying is of positive terms because 0 < 6 n 6 n 2 n=1 Cauchy’s Root Test, we conclude that the series is convergent. In fact:

d) The series

cos

 . lim

e) The series

n

 cos

1 π + 6 n



n = lim cos

1 π + 6 n

 = cos

π 6

√ =

3 < 1. 2

n ∞   1 n + (−1)n is of positive terms. The sequence of general term 2 4n + 1 n=0  .

n

1 n + (−1)n 2 4n + 1

n =

n 1 + (−1)n 2 4n + 1

1 3 and . Since the upper limit is less than 1, we can apply Corollary 1 of the Root Test 4 4 to conclude that the series is convergent.

has two sublimits:

n n ∞  2 ∞  2   n −2 n −2 1 + . The series is of positive terms 2 2 3 n=2 3n 3n n=1 n=2 and differs from the given series in only one term. We can state that the two series are either both convergent or both divergent. Using Cauchy’s Root Test, we can show that the series converges because

f) We can write

n ∞  2  n −2 3n2

=−

 . lim

9.

n

n2 − 2 3n2

n = lim

n2 − 2 1 = < 1. 3n2 3

∞  (−1)n 1 is alternating because an = > 0, ∀n ≥ 2; its general term can be written n) n) log(n log(n n=2 ! ∞ ! ∞   ! (−1)n ! 1 (−1)n ! = ! . Let us start by studying the series of modules, , in the form ! ! n log(n) n log(n) n log(n) n=2 n=2   1 applying the Cauchy’s Condensation Test. The sequence is obviously decreasing. The n log(n) n∈N\{1} series ∞ ∞ ∞    2n 1 1 1 . = = n n 2 log(2 ) n log(2) log(2) n=2 n n=2 n=2

a) The series

is divergent as it is the product of a constant by the harmonic series. Then, the series of modules is divergent by Cauchy’s Condensation Test. As a result, if convergent, the series under study is conditionally convergent. Since the series is alternating and satisfies the following conditions: 1 >0 (i) n log(n) 1 (ii) lim =0 n log(n)   1 (iii) As mentioned earlier, the sequence is decreasing n log(n) n∈N\{1} it follows from the Leibniz’s test that the alternating series is convergent. From the study of the series of modules, we can state that the series is conditionally convergent, considering Definition 2.4.1.

136

Numerical Series b) Let us evaluate the limit of the general term of the series:  . lim

√



n2 n5

+1

+

2 1+ n

n 

n2



2 = lim √ + lim 1 + n n5 + 1



n = lim

n4 + e2 = e2 . n5 + 1

As the general term of the series does not converge to zero, the series is divergent (refer to Theorem 2.2.1).   1 1 < 1, ∀n ≥ 2, which implies that log 1 − 2 < 0, ∀n ≥ 2. Thus, the 2 n n series is of negative terms. Next, we consider the series of modules

c) First, we note that 0 < 1 −

.

 !   ∞ ! ∞    ! ! 1 !log 1 − 1 ! = − log 1 − 2 . ! ! 2 n n n=2 n=2

We can compare this series with the series

∞  1 , which is convergent because it is a p-series with p = 2. 2 n n=2

The limit   1     2 − log 1 − 2 1 1 n n . lim = − lim n2 log 1 − 2 = − lim log 1 − 2 = − log(e−1 ) = 1 1 n n n2 is finite and different from zero. Therefore, the series of modules is convergent by Corollary 2 of the General Comparison Test. Then, by Theorem 2.4.1, the given series is absolutely convergent, considering Definition 2.4.1. ∞  sin(n) + 2n is of positive terms because −1 ≤ sin(n) ≤ 1 and 2n > 1, ∀n ∈ N. The n + 5n n=1 previous inequalities allow us to obtain the following estimate:

d) The series

.0

The series


0, ∀n ∈ N. To determine the convergence n n nn n=1 ∞  en+1 , using D’Alembert’s test: of the series, we first examine the series of modules, nn n=1

e) The series

(−1)n

2.7. Solved Exercises

137

en+2  n an+1 en+2 nn e 1 n (n + 1)n+1 . lim = lim = lim = lim = 0 × = 0. en+1 an en+1 (n + 1)n+1 n+1 n+1 e nn As this value is less than 1, the series of modules is convergent. Therefore, the given series is absolutely convergent by Theorem 2.4.1, considering Definition 2.4.1.

f) To study the convergence of the series

∞ (−1)n +  n=1

n

1 1 ∞ (−1)n +  n = n , we will start by rewriting its n n=2

general term: (−1)n + .

n

1 n n = (−1)n · n + (−1) . n2

n + (−1)n Let an = . It is evident that an > 0, ∀n ∈ N \ {1}, and therefore the series under study is n2 ∞  n + (−1)n alternating. Let us study the series of modules, , by comparison with the harmonic series. n2 n=1 The limit n + (−1)n n(n + (−1)n ) n + (−1)n n2 = lim . lim = lim =1 2 1 n n n belongs to R+ . As the harmonic series is divergent, the series of modules is also divergent, by Corollary 2 of the General Comparison Test. 1 ∞ ∞ (−1)n +   (−1)n 1 (−1)n 1 n and = + 2 Let us consider the series . Taking into account that 2 n n n n n n=1 n=1 and that the two previous series are convergent (the first one is the alternating harmonic series, and the 1 ∞ (−1)n +  n second one is a p-series with p = 2), the series is convergent by Theorem 2.2.2. From n n=1 the study of the series of modules, we can affirm that the series is conditionally convergent, according to Definition 2.4.1.   n + (−1)n is not monotonic (see Note: The Leibniz’s Test is not applicable because the sequence n2 Fig. 2.6).   1 1 g) Bearing in mind that 1 + n > 1, ∀n ≥ 1, we conclude that log 1 + n > 0, ∀n ≥ 1. Thus, the 2 2   ∞ ∞   1 1 series log 1 + n is of positive terms. Let us compare this series with the series , which is n 2 2 n=1 n=1 1 convergent as it is a geometric series with ratio r = . The limit 2   1     n log 1 + n 1 1 2 2 . lim = lim 2n log 1 + n = lim log 1 + n = log(e) = 1 1 2 2 2n

138

Numerical Series

Figure 2.6: The sequences (−1)n

n + (−1)n n + (−1)n and n2 n2

is finite and different from zero; therefore, the series is convergent by Corollary 2 of the General Comparison Test. It is absolutely convergent since it is a series of positive terms. h) The series

∞  

1−

∞  √ n n √ n n = (−1)n n n − 1 is an alternating series. Let us begin by studying the

n=1

n=1

series of modules using Cauchy’s Root Test: . lim

" √ n

n

n √  n − 1 = lim n n − 1 = 0

√ since lim n n = 1. As this value is less than 1, the series of modules is convergent. It follows by Theorem 2.4.1 that the given series is convergent, and by Definition 2.4.1, it is absolutely convergent. i) To determine the convergence of the series

∞  n=1

.

(−1)n , we first rewrite its general term as 1 − (−1)n n2

(−1)n 1 −1 = = 2 . 1 − (−1)n n2 (−1)n − n2 n + (−1)n+1

We can simplify the series as follows: ∞  . n=1

∞ ∞   (−1)n −1 1 = =− . n 2 2 n+1 2 + (−1)n+1 1 − (−1) n n + (−1) n n=1 n=1

Note that the general term of this last series is positive, whatever n ∈ N. We can compare the series with ∞  1 the series , which we know is convergent because it is a p-series with p = 2. The limit n2 n=1 1 n2 1 n2 + (−1)n+1 . lim = lim 2 = lim =1 1 (−1)n+1 n + (−1)n+1 1 + n2 n2 belongs to R+ . Therefore, by Corollary 2 of the General Comparison Test, the series is convergent, and the convergence is absolute.

2.7. Solved Exercises

139

2n ∞ ∞ ∞  ∞     3 (−1)n + 4 and b = . The series an is of n n + 4n n+2 n=1 n=1 n=1 n=1 n=1 positive terms; in fact, the expression for an is as follows: ⎧ 5 ⎪ ⎪ if n is even ⎨ n + 4n .an = 3 ⎪ ⎪ ⎩ if n is odd. n + 4n

j) Consider the series

∞ 

an =

∞  5 5 , ∀n ∈ N. The series is a convergent geometric series because the ratio is n 4n 4 n=1 ∞  5 1 an is convergent. r = . Since an ≤ n , ∀n ∈ N, the General Comparison Test implies that 4 4 n=1

Consequently, an ≤

We can use the Cauchy’s Root Test to study the series of positive terms,

∞ 

bn :

n=1

 . lim

This implies that the series Since both

∞ 

an and

n=1

∞ 

3 n+2

n

2n

 = lim

3 n+2

2 = 0 < 1.

bn is convergent.

n=1 ∞ 

bn are convergent, we can conclude that the given series is convergent by

n=1

Theorem 2.2.2. Since it is a series of positive terms, it is absolutely convergent. k) The series

 ∞   1 n 1− has a general term that is not a null sequence. Specifically, n n=1  . lim

1−

1 n

n

= e−1 .

Theorem 2.2.1 can be applied to establish that this series is divergent. ∞  1 l) We will use the Cauchy Condensation Test to determine the convergence of the series  3 . n log(n) n=2  1 It is clear that the sequence  3 is decreasing with positive terms. We can apply the Cauchy n log(n) Condensation Test by considering the series ∞  . n=2



2n

2n log(2n )

3 =

∞  n=2



1

n3 log(2)

3 = 

1 log(2)

3

∞  1 , 3 n n=2

which is convergent as it is the product of a constant by a convergent p-series, Cauchy Condensation Test, the series

∞  n=2

positive terms, it is absolutely convergent.

∞  1 . Therefore, by the n3 n=2

1  3 is convergent. Furthermore, since it is a series of n log(n)

140

Numerical Series 1 > 0 for all x = 0, we can say that the integral is positive, which implies that the series consists of x2 positive terms. As $ # n+2 1 1 1 n+2 1 , . dx = − = − x2 x n n n+2 n

m) If

we can write the series as

∞ 

n+2

. n=1

n

 ∞   1 1 1 − . dx = x2 n n+2 n=1

1 and k = 2, using the notation of formula (2.1) on page 74. The This series is telescopic with αn = n  ∞   1 1 telescopic series converges if (αn ) is a convergent sequence. As αn → 0, the series − is n n+2 n=1 convergent. Furthermore, it is absolutely convergent since it consists of positive terms. ∞  (−1)n 1 √ √ √ is alternating. The series of modules is the series , n+1+ 3n+1 n+1+ 3n+1 n=1 n=1 ∞  1 1 , which is divergent because it is a p-series with p = . As which we can compare with the series 1/2 2 n n=1 the value of the limit 1 √ √ √ n n+1+ 3n+1 √ =1 . lim = lim √ 1 n+1+ 3n+1 n1/2

n) The series

∞ 

√

belongs to R+ , Corollary 2 of the General Comparison Test guarantees that the series of modules is divergent. Let us study the alternating series by applying the Leibniz’s test: 1 √ (i) lim √ = 0. n+1+ 3n+1 1 √ > 0, ∀n ∈ N. (ii) √ n+1+ 3n+1   √  √ 1 √ (iii) The sequence √ n + 1 + 3 n + 1 is obviously increasis decreasing, because n+1+ 3n+1 ing. The test conditions are satisfied; therefore, the series is convergent. As the series of modules is divergent, we conclude that the alternating series is conditionally convergent. o) The series series

∞  n=1

∞  n=1

n

√

1 √  is of positive terms. Let us study this series by comparing it with the n+1+ n

1 3 , which we know to be convergent, as it is a p-series with p = . The limit 2 n3/2

. lim

n

√

1 √  n+1+ n 1 n3/2

n3/2 1 = lim √ √  = 2 n n+1+ n

belongs to R+ . It follows from Corollary 2 of the General Comparison Test that the given series is convergent. It is absolutely convergent since it is a series of positive terms.

2.7. Solved Exercises

141

    1 1 is alternating, because 1 − cos > 0, ∀n ≥ 1. The series of n n n=1    ∞  1 modules is the series 1 − cos . As the value of the limit n n=1

p) The series

∞ 



(−1)n

1 − cos

1 − cos . lim

  1 n

1 n2

=

1 2

∞  1 is convergent because it is a p-series with p = 2, then by Corollary 2 of 2 n n=1 the General Comparison Test, the series of modules is convergent. Therefore, by Theorem 2.4.1, the given series is convergent, and it is absolutely convergent (see Definition 2.4.1).

belongs to R+ . As the series

q) The series

∞ 

 (−1)n 2n sin

n=1

1 3n



 is alternating, because an = 2n sin 



1 3n

 > 0, ∀n ≥ 1 (note that

∞  π 1 1 < , ∀n ≥ 1). The series of modules is 2n sin n . Considering that if x > 0, then n 3 2 3 n=1 sin(x) ≤ x, we get    n  n 1 2 1 n .2 sin = , ∀n ∈ N. ≤ 2n 3n 3 3

0
0, ∀n ∈ N. Let us study the series nn n + 1 nn n + 1 n=1 √ √ ∞  n! n n! n √ √ of modules, = , using D’Alembert’s test. Taking a : n nn n + 1 nn n + 1 n=1

r) The series

∞ 

(−1)n

lim .

an+1 an

√ (n + 1)! n + 1 √ (n + 1)! nn (n + 1) (n + 1)n+1 n + 2 √ = lim = lim n! n n! (n + 1)n+1 n(n + 2) √ nn n + 1 n  (n + 1)2 nn n+1 1 n = lim = lim = . e (n + 1)n+1 n(n + 2) n(n + 2) n + 1

Since this value is less than 1, the series is convergent. As it is the series of modules of the initial series, we √ ∞  n! n (−1)n n √ is absolutely convergent. can affirm that the series n+1 n n=1 s) The series

  ∞  √ 1 π 1 is of positive terms because 0 < 2 < . Let us consider the series ( n + 1) sin 2 n n 2 n=1

142

Numerical Series ∞  n=1

1 3 , which we know to be convergent because it is a p-series with p = . The limit 2 n3/2 √ ( n + 1) sin . lim



1 n2

1 n3/2



 sin = lim

1 n2 1 n2

 √ ·

n+1 =1 √ n

belongs to R+ , and therefore, by Corollary 2 of the General Comparison Test, the series is convergent. As it is of positive terms, it is absolutely convergent. ∞ 

1 is of positive terms. Knowing that n! < nn and that the logarithmic function is log(n!) n=2 increasing, we have 1 1 1 = < , ∀n ≥ 2. .0 < log(nn ) n log(n) log(n!)

t) The series

We proved in exercise 9a) that the series series

∞  n=2

10.

∞  n=2

1 is divergent. By the General Comparison Test, the n log(n)

1 is divergent. log(n!)

∞ ∞   n cos(nπ) n cos(nπ) n √ √ = is alternating because cos(nπ) = (−1)n and √ > 0, 3+1 3+1 3+1 n n n n=0 n=1 ∞  n √ ∀n ∈ N. First, we will examine the series of modules, , by comparing it with the series n3 + 1 n=1 ∞  1 1 , which is divergent because it is a p-series with p = . The limit 1/2 2 n n=1

a) The series

√ . lim

n √ n3 n3 + 1 = lim √ =1 3 1 n +1 n1/2

is finite and different from zero. Therefore, by Corollary 2 of the General Comparison Test, the series ∞  n √ is divergent. 3+1 n n=1 Next, we will apply Leibniz’s test to the alternating series: n = 0. (i) lim √ n3 + 1 n > 0, ∀n ∈ N. (ii) √ n3 + 1 (iii) To verify that it is a decreasing sequence, consider the real function of real variable −x3 + 2 x √ f (x) = √ . The derivative of f is f  (x) = , and therefore, f  (x) < 0, x3 + 1 2(x3 + 1) x3 + 1 √ ∀x > 3 2. The sequence then decreases for n ≥ 2. Since the test conditions are satisfied, we can conclude that the alternating series is convergent. As the series of modules is divergent, the given series is said to be conditionally convergent.

2.7. Solved Exercises

143

∞ 

n! is of positive terms. Let us study it using D’Alembert’s test. Let 3 × 5 × · · · × (2 n + 1) n=1 an be the general term of the series:

b) The series

(n + 1)! n+1 1 an+1 3 × 5 × · · · × (2 n + 1)(2 n + 3) = lim = . . lim = lim n! an 2n + 3 2 3 × 5 × · · · × (2 n + 1) Since this limit is less than 1, it follows from D’Alembert’s test that the series is convergent. The series is absolutely convergent, as it is of positive terms. c) The series

∞  1 + cos2 (n) is of positive terms. We have the following inequality √ n n=1

1 1 + cos2 (n) ≤ , ∀n ∈ N. √ n n

.√ ∞ 

1 1 √ is divergent as it is a p-series with p = , the General Comparison Test allows us n 2 n=1 to conclude that the given series is divergent. Since the series

∞  3n (n + 1)! is of positive terms. Let us study it using D’Alembert’s test. Let an be the (n + 1)n n=0 general term of the series:

d) The series

3n+1 (n + 2)!   an+1 n+1 n 3 (n + 2)n+1 . lim = lim = 3 lim = . n 3 (n + 1)! an n+2 e (n + 1)n This limit value is greater than 1; therefore, by D’Alembert’s test, the series is divergent. e) We know that the function sin(x) is odd; therefore, the series

∞  n=1

 sin

(−1)n n

 is alternating and can be

    ∞ ∞   1 1 (−1)n sin sin . Let us begin by studying the series of modules, by written in the form n n n=1 n=1 comparing it with the harmonic series. The limit sin . lim

  1 n =1 1 n

belongs to R+ . It follows from Corollary 2 of the General Comparison Test that the series of modules is divergent. Let us study the alternating series by applying Leibniz’s test:   1 = 0. (i) lim sin n   1 (ii) sin > 0, ∀n ∈ N. n

144

Numerical Series   1 (iii) The sequence is decreasing, the sine function is increasing on [0, n    1 sin is decreasing. n

π ], 2

and therefore, the sequence

We conclude that the alternating series is convergent. Since the series of modules is divergent, the alternating series is conditionally convergent. f) The series

∞  n=1

√

n is of positive terms. The limit n2 + cos(n) √ n n2 n2 + cos(n) =1 . lim = lim 2 1 n + cos(n) n3/2 ∞ 

1 3 is convergent, since it is a p-series with p = . 3/2 2 n n=1 The General Comparison Test ensures that the original series is convergent. As it is of positive terms, it converges absolutely.

is finite and different from zero. The series

g) The series

∞  n=1

3n 3n 3n is of positive terms, and the inequality ≤ n holds in N. Let us study 2n + nn 2n + nn n

∞  3n using the Cauchy’s Root Test. Let an be the general term of the series: the series n n n=1

√

. lim n

an = lim

3 = 0. n

Since this limit is less than 1, the series is convergent. Using the General Comparison Test, we can affirm that the series under study is convergent and it is absolutely convergent since it is of positive terms. ∞  1 + n(−1)n n + (−1)n is alternating, since we can write its general term as (−1)n . Let 3 1 + 2n 1 + 2n3 n=1 ∞  n + (−1)n . The limit us start by studying the series of modules, 1 + 2n3 n=1

h) The series

n + (−1)n 1 n3 + (−1n )n2 1 + 2n3 = . lim = lim 1 2n3 + 1 2 n2 ∞ 

1 is a p-series with p = 2, it converges. Then, 2 n n=1 ∞  n + (−1)n Corollary 2 of the General Comparison Test allows us to conclude that the series is con1 + 2n3 n=1 vergent. Therefore, by Theorem 2.4.1, the alternating series is convergent, and it converges absolutely (see Definition 2.4.1). is finite and different from zero. Since the series

2.7. Solved Exercises i) The series

∞  n=1

145 2n 2n 2n is of positive terms, and the inequality ≤ holds in N. Let us study the n! + 1 n! + 1 n!

∞  2n series using D’Alembert’s test. Let an be the general term of the series: n! n=1

. lim

an+1 an

2n+1 2 (n + 1)! = lim = 0. = lim 2n n+1 n!

This limit has a value less than 1; therefore, according to D’Alembert’s test, the series is convergent. By the General Comparison Test, the initial series is convergent and absolutely convergent as it is of positive terms. ∞  n! 1 and . The first one, which is of positive terms, can be studied n (2n)! 2 n=0 n=0 using D’Alembert’s test. Let an be the general term of the series:

j) Let us consider the series

∞ 

(n + 1)! (n + 1)!(2n)! an+1 n+1 (2n + 2)! = lim . lim = lim = lim = 0. n! an n!(2n + 2)! (2n + 2)(2n + 1) (2n)! This limit is less than 1; therefore, by D’Alembert’s test, the series is convergent. As it is of positive terms, it is absolutely convergent. 1 The second series is geometric with ratio r = ; therefore, it is convergent. Again, as it is of positive terms, 2 it converges absolutely. ∞ ∞   n! 1 and − n , so it is convergent and absolutely The original series is the sum of the series (2n)! 2 n=0 n=0 convergent (consider the inequality: |a − b| ≤ |a| + |b|, ∀a, b ∈ R). k) The series

∞  n (2n)! is of positive terms. Let an be the general term of the series, n (n!)2 4 n=1

lim .

an+1 an

(n + 1)(2n + 2)!  2 4n+1 (n + 1)! (n + 1)(2n + 2)! 4n (n!)2 = lim = lim  2 n(2n)! n(2n)! 4n+1 (n + 1)! 4n (n!)2 (n + 1)(2n + 2)(2n + 1) 2(n + 1)2 (2n + 1) = lim = lim 2 4 n(n + 1) 4 n(n + 1)2 = lim

2n + 1 = 1, 2n

 2n + 1 tends to 1 by values greater than 2n 1, from which we conclude, by the Ratio Test, that the series is divergent. 

and therefore, D’Alembert’s test is inconclusive. The sequence

146

Numerical Series Note: We can also, alternatively, study the series using Raabe’s test (a common procedure when D’Alembert’s test is inconclusive). In this case,  . lim

n

an an+1

  − 1 = lim n

 2n −n 1 − 1 = lim =− . 2n + 1 2n + 1 2

As this value is less than 1, the series is divergent, as we already saw. l) The modulus of the general term of the given series is bounded by the general term of the series

∞  n=2

since

1 , n3 log(n)

! ! ! sin(n) ! 1 !≤ , ∀n ∈ N \ {1}. . !! n3 log(n) ! n3 log(n)

As the series

∞  n=1

1 is convergent as it is a p-series with p = 3 and n3 1 1 n3 log(n) = 0, . lim = lim 1 log(n) n3 ∞ 

1 is convergent. n3 log(n) Consequently, the series of modules is convergent, making the given series absolutely convergent.

Corollary 3 of the General Comparison Test allows us to conclude that the series

n=2

∞ 

m) The general term of the series

 (−1)n

1−

n=1

 . lim

1−

1 2n

1 2n

n is not a null sequence. In fact,



n = lim

1−

1 2n

2n 12

 which implies that the sequence of general term (−1)n

1−

1 2n

= e−1/2 ,

n does not have a limit. From Theo-

rem 2.2.1, it follows that the series is divergent.   ∞  1 1 1 · arcsin is of positive terms. Consider the series , which we know to be 2 2n + 5 n n n=1 n=1 convergent as it is a p-series with p = 2. The limit

n) The series

∞ 

    1 1 1 1 arcsin · arcsin 1 2n + 5 n 2n + 5 n · = . lim = lim 1 1 1 2 n2 n n belongs to R+ . By Corollary 2 of the General Comparison Test, the series

∞  n=1

convergent. As it is of positive terms, it is absolutely convergent.

  1 1 · arcsin is 2n + 5 n

2.7. Solved Exercises 11.

a) The series

147

  ∞  1 1 1 π sin n consists of positive terms because 0 < n < , ∀n ∈ N. Consider the series n 2 2 2 n=1

∞  1 1 . This series is convergent since it is a geometric series with r = . To show that the first series is 2n 2 n=1 also convergent, we can use Corollary 3 of the General Comparison Test. Specifically, we can take the limit

    1 1 1 sin n sin n 1 n 2 2 . lim = lim · = 0 × 1 = 0. 1 1 n 2n 2n Since the limit value equals zero, we can conclude that the series under consideration is convergent. Moreover, as it is of positive terms, it is absolutely convergent. √ ∞ ∞ 3   n2 1 b) The series is of positive terms. Consider the series , which is convergent as it is √ 2 4/3 n n + 2n n n=1 n=1 4 a p-series with p = . The limit 3 √ 3 n2 √ n2 1 n n + 2n2 . lim = lim √ = 1 2 n n + 2n2 n4/3 is finite √ and different from zero; therefore, by Corollary 2 of the General Comparison Test, the series ∞ 3  n2 is convergent. As it is a series of positive terms, it is absolutely convergent. √ n n + 2n2 n=1 ∞  cos(πn) 1 √ is alternating because cos(nπ) = (−1)n and √ > 0. We will study the n2 − 1 n2 − 1 n=2 ∞  1 √ series of modules, , by comparing it with the harmonic series. The limit 2−1 n n=2

c) The series

√ . lim

1 √ n2 n2 − 1 =1 = lim √ 2 1 n −1 n

is finite and different from zero; therefore, by Corollary 2 of the General Comparison Test, the series +∞  1 √ is divergent. 2−1 n n=2 Let us apply Leibniz’s test to study the alternating series: 1 = 0. (i) lim √ n2 − 1 1 (ii) √ > 0, ∀n ∈ N \ {1}. n2 − 1   √ 1 is decreasing for n ≥ 2. (iii) The sequence ( n2 − 1) is increasing so √ n2 − 1

148

Numerical Series The conditions for applying the test are satisfied; therefore, the alternating series is convergent. As the series of modules is divergent, the given series is conditionally convergent. ∞ 

(−1)n 1 is alternating because an = √ > 0, ∀n ≥ 2. Let us start by studying √ n log(n) n log(n) n=2 the series of modules ! ∞ !   ! (−1)n ! +∞ 1 !√ ! . . √ ! n log(n) ! = n log(n)

d) The series

n=2

Given the limit √ . lim

n=2

1 n1/4 n3/4 n log(n) = lim = +∞, = lim √ 1 n log(n) log(n) 3/4 n

∞ 

1 3 is divergent because it is a p-series with p = , by Corollary 4 of the 3/4 4 n n=2 General Comparison Test, the series of modules is divergent. Consequently, if convergent, the original series will be conditionally convergent. Since the series is alternating, let us check if we are in the conditions of Leibniz’s test: 1 > 0. (i) √ n log(n) 1 (ii) lim √ = 0. n log(n)   √ 1 (iii) Since ( n log(n)) is clearly increasing, the sequence √ is decreasing. n log(n) n∈N\{1}

and the fact that the series

As we are in the conditions of Leibniz’s test, we can conclude that the alternating series is convergent. From the study on the series of modules, we can affirm that the series is conditionally convergent. ∞  4 + sin(n) 3 4 + sin(n) is of positive terms. We can observe that 0 < √ ≤ √ , ∀n ∈ N. √ 3 3 3 n + 1 n + 1 n+1 n=1 ∞  1 1 As the series is a p-series with p = , it is divergent. Since the limit √ 3 n 3 n=1

e) The series

√ 3 . lim

3 √ 33n n+1 =3 = lim √ 3 1 n+1 √ 3 n

is finite and different from zero, Corollary 2 of the General Comparison Test allows us to conclude that the ∞ ∞   3 4 + sin(n) is divergent. Hence, the General Comparison Test ensures that the series series √ √ 3 3 n + 1 n+1 n=1 n=1 is divergent. f) We begin by analyzing the series of modules, as the series inequality .0

∞  n2n sin(n3 ) is not of positive terms. The (3n2 + 5)n n=1

! ! 2n ! n sin(n3 ) ! n2n !≤ ≤ !! ! 2 n 2 (3n + 5) (3n + 5)n

2.7. Solved Exercises

149

holds true in N. We evaluate the limit

 . lim

n

n2n 1 n2 = , = lim (3n2 + 5)n 3n2 + 5 3 ∞ 

n2n is convergent. By 2 + 5)n (3n n=1 ! ! ∞  ! n2n sin(n3 ) ! ! ! the General Comparison Test, the same happens to the series ! (3n2 + 5)n !. The original series is n=1 absolutely convergent since the series of modules is convergent, considering Definition 2.4.1.

which is less than 1. Consequently, by Cauchy’s Root Test, the series

 n−1 n−1 . Since 0 < 2 < 1, ∀n > 2, we have that an > 0, and therefore, the series 2 n +1 n +1 +∞ ∞   (−1)n an is alternating. Let us start by studying the series of modules an . The limit 

g) Let an = arcsin

n=3

n=3



n−1 n2 + 1 n−1 n2 + 1



arcsin . lim

=1

∞  n−1 is divergent (compare it with the harmonic series). We can then 2+1 n n=3 conclude that the series of modules is divergent. Consequently, if convergent, the series under study is conditionally convergent. Since the series is alternating, let us check if we are in the conditions of Leibniz’s test:   n−1 (i) arcsin > 0, ∀n > 2. n2 + 1   n−1 (ii) lim arcsin = 0. 2 n +1   x−1 . This function is (iii) Let f be the real function of a real variable defined by f (x) = arcsin x2 + 1 decreasing on [3, +∞[, because

belongs to R+ , and the series

.f



(x) =

−x2 + 2x + 1 √ < 0, (x2 + 1) x4 + x2 + 2x

Therefore, the sequence

 .

 arcsin

n−1 n2 + 1

∀x ∈ [3, +∞[.

 n∈N\{1,2}

is decreasing. As we are in the conditions of Leibniz’s test, we can conclude that the alternating series is convergent. The study done on the series of modules shows that the series is conditionally convergent. ∞  4 × 7 × · · · × (3n + 1) is of positive terms. Let us use D’Alembert’s test to study it. If an 8 × 11 × · · · × (3n + 5) n=1 is the general term of the series, then:

h) The series

150

Numerical Series

4 × 7 × · · · × (3n + 1)(3n + 4) an+1 3n + 4 8 × 11 × · · · × (3n + 5)(3n + 8) . lim = lim = 1. = lim 4 × 7 × · · · × (3n + 1) an 3n + 8 8 × 11 × · · · × (3n + 5) Since this limit is 1, we cannot draw any conclusions using D’Alembert’s test. Therefore, let us use Raabe’s test, which is a common procedure when D’Alembert’s test is inconclusive. In this case,  . lim

n

an an+1

   4n 4 3n + 8 − 1 = lim = . − 1 = lim n 3n + 4 3n + 4 3

As this value is greater than 1, we can conclude that the series is convergent. Moreover, since it is a series of positive terms, it converges absolutely. ∞ ∞   arctan(n3 ) 2n (2n)! . The first one is of positive terms, and we and √ 2 n (2n + 1)! n + n 3 n=1 n=0 ∞  1 compare it with the p-series , which we know to be convergent. Indeed, the limit 2 n n=1

i) Consider the two series

arctan(n3 ) √ n2 π n + n2 . lim = lim √ arctan(n3 ) = 1 n + n2 2 n2

is finite and different from zero, so, by Corollary 2 of the General Comparison Test, the series

∞  arctan(n3 ) √ n + n2 n=1

is convergent. The second series is also of positive terms. We can study it using D’Alembert’s test. Let an be the general term of the series: 2n+1 (2n + 2)! 2(2n + 1) 2 an+1 2n+1 (2n + 2)! 3n (2n + 1)! 3n+1 (2n + 3)! = lim = . . lim = lim = lim n 2 (2n)! an 2n (2n)! 3n+1 (2n + 3)! 3(2n + 3) 3 3n (2n + 1)! This limit is less than 1; therefore, by D’Alembert’s test, the second series is convergent. The original series is the sum of the two series studied, so it is convergent. Since it is of positive terms, the series is absolutely convergent. ∞ sin 



√1 n



π 1 consists of positive terms, since 0 < √ < , ∀n ∈ N. To check its convergence, n 2 n=1 ∞  1 , which we know to be convergent because it is a p-series with let us compare it with the series n3/2 n=1

j) The series

n+

√

n

2.7. Solved Exercises p=

151

3 . Taking the limit, we have 2



  sin √1n sin √1n

 √ √ sin √1n n n+ n n+ n . lim = lim = lim · √ = 1. 1 1 1 n+ n √ √ n n n n3/2

Since the

limit  is finite and different from zero, by Corollary 2 of the General Comparison Test, the series ∞ sin √1  n is convergent. Moreover, as a series of positive terms, it converges absolutely. √ n + n n=1 k) The series sums:

    ∞   1 1 n sin − (n + 2) sin is telescopic. We can write the sequence of partial n n+2 n=1   1 3   1 sin +2 3   1 sin +2 3   1 sin −4 2   1 −4 sin 2   1 −5 sin 2

S1 = sin(1) − 3 sin S2 = sin(1) − 3 S3 = sin(1) − 3 = sin(1) + 2 .

S4 = sin(1) + 2 = sin(1) + 2 . ..

Sn = sin(1) + 2 sin

  1 −4 2   1 sin −4 2   1 sin −5 4   1 sin −5 4   1 sin −6 5 sin

  1 4       1 1 1 sin + 3 sin − 5 sin 4 3 5   1 sin 5       1 1 1 sin + 4 sin − 6 sin 5 4 6   1 sin 6 sin

      1 1 1 − (n + 1) sin − (n + 2) sin 2 n+1 n+2

. ..   1   sin 1 n = lim =1 . lim n sin 1 n n

As

we have . lim Sn

= sin(1) + 2 sin

  1 − 2. 2

The series is convergent and .

      ∞   1 1 1 − (n + 2) sin = sin(1) + 2 sin − 2. n sin n n + 2 2 n=1

152

Numerical Series To determine whether it is absolutely convergent,  we will term.  study the sign of the general   For this, we 1 1 will prove that the sequence of general term n sin is increasing. Let f (x) = x sin , x ≥ 1. We n x have     1 1 1  − cos .f (x) = sin x x x     1 1 1 1 and, therefore, f  (x) = 0 ⇔ tan = . However, this equation has no zeros. In fact, tan > , x x x x   1 ∀x ≥ 1. Then f  (x) > 0, ∀x ≥ 1, which implies that f is increasing. As f (n) = n sin , we can conclude n ∞ ∞   |an | = − an , that the sequence is increasing and, therefore, the series is of negative terms. Since n=1

n=1

the given series is absolutely convergent. l) The series

∞  2n + 3 is of positive terms. We can establish the following inequality: (n + 1)! n=1

.

2n + 3 2n ≤2 , ∀n ≥ 2. (n + 1)! (n + 1)!

To determine the convergence of the series

∞  n=1

2n , we can apply D’Alembert’s test. Let an be the (n + 1)!

general term of the series: . lim

an+1 an

2n+1 2 (n + 2)! = lim = 0. = lim 2n n+2 (n + 1)!

This limit is less than 1, indicating that the series is convergent according to D’Alembert’s test. Using the General Comparison Test, we can affirm that the series under study is convergent and, hence, absolutely convergent. ∞  arctan(n + 1) − arctan(n) is of positive terms because arctan(x) is an increasing function n2 n=1 ∞  1 , which we know is convergent as it is a p-series on R. Let us compare this series with the series n2 n=1 with p = 2. By Corollary 3 of the General Comparison Test, given the value of the limit

m) The series

arctan(n + 1) − arctan(n) n2 . lim = lim (arctan(n + 1) − arctan(n)) = 0, 1 n2 we can conclude, as we are comparing with a convergent series, that the series under study is convergent. As it is a series of positive terms, it converges absolutely. n) The general term of the series

∞  n=1

log(n)  1  is not a null sequence. In fact, n sin n

2.7. Solved Exercises

153

1 log(n) n   = lim log(n) ·   = +∞. . lim 1 1 n sin sin n n Therefore, by Theorem 2.2.1, the series diverges. o) Consider the series

∞   n=1

log(2)

∞  

Let us analyze the series

n

 log

log(2)

n

 n+1 n+1 > 1, ∀n ∈ N. . It consists of positive terms because n n

. It converges because it is a geometric series with 0 < r = log(2) < 1.

n=1

  n n+1   log(2) log n+1 n  n =0 . lim = lim log n log(2)

The limit



allows us to use Corollary 3 of the General Comparison Test to conclude that the series under study is convergent because we are comparing it with a convergent series. As the series is of positive terms, it is absolutely convergent.     p p) Given that n ≥ 3 implies log(n) > 1, then log log(n) > 0 so n log(n) log log(n) > 0, ∀p ∈ R. Therefore, the series is of positive terms.   Let us assume that p ≤ 0. If n > ee (for example, if n ≥ 33 = 27), then log log(n) > 1. Thus,   −p ≥ 1, with equality holding for p = 0. Consequently, we can write: log log(n) .

Since the series

+∞  n=27

  −p log log(n) 1 1  p = ≥ . n log(n) n log(n) n log(n) log log(n) 

1 is divergent as seen in Example 2.5.2, we can apply the General Comn log(n)

parison Test to deduce that the series

+∞ 

1   p is divergent. Therefore, the series n log(n) log log(n)

n=27 +∞  n=3

1   p is also divergent. n log(n) log log(n)

Now, suppose that p > 0. With the application of the Integral Test in mind, let us consider the function .f (x)

=

1   p x log(x) log log(x)

  p and let g(x) = x log(x) log log(x) . We know that g(x) > 0, ∀x ∈ [3, +∞[. Additionally, since the logarithm function is continuous and increasing, g is continuous and increasing on [3, +∞[, so f is positive, +∞ continuous, and decreasing on [3, +∞[. We can now study the improper integral f (x) dx. If p = 1,

.

3



y 3

#

$   −p+1 y 1 log log(x) −p + 1 3

  −p+1   −p+1  1 log log(y) . = − log log(3) −p + 1

1  p dx = x log(x) log log(x) 

154

Numerical Series If p = 1, y . 3

    y       1   dx = log log log(x) = log log log(y) − log log log(3) . 3 x log(x) log log(x)

We can then determine that .

y

lim

y→+∞

3

⎧ ⎪ ⎪ ⎨+∞,

1  p = ⎪ x log(x) log log(x) ⎪ ⎩ 



+∞

This means that the improper integral

if 0 < p ≤ 1

−p+1 1   log log(3) p−1

if p > 1.

f (x) dx is convergent if p > 1 and divergent if 0 < p ≤ 1. By

3

the Integral Test, the series under consideration is convergent if p > 1 and divergent if p ≤ 1. +∞  1   p is convergent if and only if p > 1. Conclusion: The series n log(n) log log(n) n=3

12.

∞  (n + 1)n is of positive terms. To analyze this series, we will use D’Alembert’s test. Let an 3n n! n=1 be the general term of the series:

a) The series

(n + 2)n+1   an+1 e n+2 n+2 n 1 3n+1 (n + 1)! . lim = lim = . = lim n (n + 1) an 3 n+1 n+1 3 3n n! Since this limit is less than 1, according to D’Alembert’s test, the series converges. ∞  (n + 1)n b) In the previous item, we proved that the series is convergent, which implies, by Theorem 2.2.1, 3n n! n=1 (n + 1)n that its general term is a null sequence. Therefore, lim = 0. 3n n! 13. We will study the convergence of the series of modules,

∞ 

|an |. We will use the Root Test or one of its corollaries,

n=1

to do this. We have

.

n

⎧   n2 n ⎪ ⎪ n n n ⎪ ⎪ = , ⎪ ⎪ ⎪ n+1 n+1 ⎪ ⎪ ⎪ ⎨ 1 1 n |an | = = √ , n ⎪ n! n! ⎪ ⎪ ⎪  ⎪ ! ⎪ !! ⎪ (−1)n !! 1 ⎪ n ! ⎪ ⎪ ⎩ ! (n + 1)n ! = n + 1 ,

if n = 3k, k ∈ N if n = 3k + 1, k ∈ N0 if n = 3k + 2, k ∈ N0 .

1 1 and 0; therefore, lim n |an | = . Since this value is less than 1, we can affirm, by e e ∞ ∞   |an | is convergent. Consequently, the series an is absolutely Corollary 1 of the Root Test, that the series

The sublimits of

convergent.

n

|an | are

n=1

n=1

2.7. Solved Exercises

155

14. One initial approach is to analyze the general term of the series

∞ 

an . Since lim a2n = lim a2n+1 = +∞, then

n=1

lim an = +∞, and by Theorem 2.2.1, the series diverges. Another approach:

We find that

⎧ n+2 ⎪ ⎪ , ⎨ an+1 n . = n+1 ⎪ an ⎪ , ⎩ n−1

if n is even if n is odd, n > 1.

an+1 > 1, ∀n ∈ N \ {1}, and therefore by the Ratio Test, the series diverges. an

15. If (an ) is a sequence of positive terms such that the series

∞ 

nan is convergent, then the sequence (an ) is

n=1 (an ) was

bounded. We can prove this by reductio ad absurdum: If not bounded, then lim (nan ) would not be ∞  zero, and in this case, the series nan would not be convergent. It follows from the fact that (an ) is bounded n=1

that . lim

a2n an = 0. = lim nan n

By Corollary 3 of the General Comparison Test, the series

∞ 

a2n is convergent.

n=1

16. As the series

∞ 

an converges, lim an = 0. By hypothesis, an > 0, ∀n ∈ N, so log(1 + an ) > 0. Thus, we have

n=1

two series of positive terms. The limit . lim

log(1 + an ) =1 an

is finite and different from zero; therefore, by Corollary 2 of the General Comparison Test, the series

∞ 

log(1+an )

n=1

also converges. Note: The limit lim

x→0

log(1 + x) 0 is an indeterminate form of type . Applying L’Hˆ opital’s Rule, we get x 0  .

therefore, lim

x→0

17.

lim

x→0

log(1 + x) (x)

 = lim

x→0

1 =1 1+x

log(1 + x) log(1 + an ) = 1, which implies that lim = 1. x an

a) The series

∞ 

(−1)n−1 an is alternating. Leibniz’s test cannot be applied because the sequence (an ) does

n=1

not decrease. In fact, if n is even, .an+1

− an =

1 n2 − n − 1 1 − 2 = 2 > 0. n+1 n n (n + 1)

156

Numerical Series b) Let us prove that the series

∞ 

(−1)n−1 an is divergent. If it was convergent, then, by Theorem 2.2.4, any

n=1

series obtained from it by association of its terms would be convergent. Consider the series

∞ 

bn , where

n=1

(bn ) is the sequence defined by .bn

= a2n−1 − a2n =

The limit

1 4n2 − 2n + 1 1 (2n)2 − 2n + 1 − = . = 2n − 1 (2n)2 (2n)2 (2n − 1) 4n2 (2n − 1)

4n2 − 2n + 1 1 4n2 − 2n + 1 4n2 (2n − 1) = . lim = lim 1 4n(2n − 1) 2 n

is finite and different from zero; therefore, based on Corollary 2 of the General Comparison Test, the series ∞ ∞   bn diverges because the harmonic series is divergent. However, we obtained the series bn by n=1

n=1

association of the terms of the original series. Hence, the original series must be divergent. 18.

a) Suppose that the series

∞ 

bn is convergent. Then, its general term is a null sequence, which implies

n=1

that all its subsequences have limit zero. Therefore, we have lim b2n−1 = 0, or equivalently, lim an = 0. ∞  Conversely, let lim an = 0 and (Sn ) be the sequence of partial sums of the series bn . We can see that n=1

S2n = 0 and S2n−1 = an for all n ∈ N. Since lim an = 0, it follows that lim Sn = 0. By definition, if the sequence of partial sums of a series has a finite limit, the series converges. As a result, it can be inferred ∞  that the series bn converges. n=1 ∗ denote the general terms of the sequences of partial sums of the series b) Let Sn and Sn

∞ 

|bn | and

n=1

∞ 

|an |,

n=1

respectively. We can observe that S1 = |b1 | = |a1 | = S1∗ S2 = |b1 | + |b2 | = |a1 | + |a1 | = 2S1∗ S3 = |b1 | + |b2 | + |b3 | = |a1 | + |a1 | + |a2 | = S1∗ + S2∗ .

S4 = |b1 | + |b2 | + |b3 | + |b4 | = |a1 | + |a1 | + |a2 | + |a2 | = 2S2∗

.

S5 = |b1 | + |b2 | + |b3 | + |b4 | + |b5 | = |a1 | + |a1 | + |a2 | + |a2 | + |a3 | = S2∗ + S3∗ .. . So, we can infer that .S2n

If the series

∞ 

∗ = 2Sn

and

∗ ∗ S2n+1 = Sn + Sn+1 .

bn converges absolutely, then the limits lim S2n and lim S2n+1 exist, are finite, and equal.

n=1 ∗ = Since lim Sn

1 2

lim S2n , we can conclude that the series

∞  n=1

an converges absolutely.

2.7. Solved Exercises

157 ∞ 

Reciprocally, if the series

∗ exists and is finite. This implies that an converges absolutely, then lim Sn

n=1 ∗ + S∗ ∗ lim S2n+1 = lim(Sn n+1 ) = lim 2Sn = lim S2n , and we can state that the series

∞ 

bn converges

n=1

absolutely.

19. If the series

∞ 

an is convergent, then the sequence of its partial sums, (Sn ), is also convergent. This implies

n=1

that all its subsequences are convergent and have the same limit. Since the sequence (an ) is decreasing, we have .S2n

− Sn = a2n + · · · + an+1 ≥ a2n + · · · + a2n = n a2n ,

and as lim (S2n − Sn ) = 0 and an ≥ 0, we can conclude that lim n a2n = 0, which implies that lim 2 n a2n = 0. Similarly, we have .S2n+1

− Sn+1 = a2n+1 + · · · + an+2 ≥ a2n+1 + · · · + a2n+1 = n a2n+1 .

As lim (S2n+1 − Sn+1 ) = 0 and (an ) is a sequence of nonnegative terms, we conclude that lim n a2n+1 = 0. This also implies that lim (2n + 1) a2n+1 = 0 because lim an = 0 since, by hypothesis, the series is convergent. We have shown that the subsequence of even order terms of (n an ) has the same limit as its subsequence of odd order terms, which is zero. Hence lim n an = 0. a2 + b 2 , ∀ a, b ∈ R, along with the fact that the sequence (an ) is of positive 2 terms, to establish the following expression:

20. We can use the inequality ab ≤

.0

1 √ an ≤ np

1 n2p

+ an 2

=

1 2



1 + an n2p

 .

∞  1 1 , then 2p > 1, which implies that the p-series 2p 2 n n=1 n=1 is convergent. The sum of these two series is convergent, and given the previous inequality, we can state by the ∞ √  an General Comparison Test the convergence of the series . np n=1

By hypothesis, the series

21. Consider the series

∞  n=2

∞ 

≤

an is convergent, and if p >

1 1 , which is divergent as it is the harmonic series. Let bn = . Then n−1 n−1

bn+1 . = bn

1 n−1 1 an+1 n = =1− ≤ , ∀n ≥ 2. 1 n n an n−1

By Corollary 5 of the General Comparison Test, we can determine that the series

∞  n=1

an is divergent.

158

Numerical Series

g 22. The functions g and g  are, by hypothesis, continuous and positive on [p, +∞[, so is continuous and positive. g      2 g  (x)g(x) − g  (x) g g (x) = < 0, so is decreasing on [p, +∞[. We are in the conditions of In addition,  2 g(x) g g(x) the Integral’s Test. But y      g (x) . dx = log g(y) − log g(p) ; g(x) p +∞    g (x) therefore, the improper integral dx is convergent if and only if there exists and is finite lim log g(x) . x→+∞ g(x) p   By the continuity of the logarithm function, lim log g(x) exists finite if and only if there exists finite and difx→+∞

ferent from zero the

lim g(x). But, by hypothesis, g is positive and increasing (g  is positive), so it cannot

x→+∞

have a limit 0 when x tends to +∞. Conclusion: The series

+∞  n=p

g  (n) is convergent if and only if g(n)

lim g(x) exists and is finite.

x→+∞

23. We can prove by induction that (vn ) is a sequence of positive terms. Applying D’Alembert’s test, we find that ∞  the series vn is convergent because n=2

. lim

  vn+1 π = 0 < 1. = lim sin vn n

Furthermore, . lim

arctan(vn ) = 1, vn

which implies by Corollary 2 of the General Comparison Test that the series

∞ 

arctan(vn ) is convergent.

n=2

24. Let us look at the numbers of the construction of the Sierpinski triangle in a table. Number of squares

Number of squares removed

Length of the edges

Area of each square

Area of the squares removed

1

0

1

1

0

3

1

1 2

 2 1 2

32

3

1 22

33

32

1 23

. . .

. . .

. . .





1 22 1 23 . . .

1×

2

 2 1 2

 3×

1 22



2 32 ×

. . .

2

1 23

2

2.7. Solved Exercises

159

So the total area of the squares removed is the sum of the geometric series .A

=

∞ 

 3n

n=0

2

1

=

2n+1

∞   1  3 n = 1. 4 n=0 4

Since the initial square has area 1, we conclude that the area of the Sierpinski triangle is zero. 25. The sum of the areas of the circles is given by the series πr12 +

∞ 

2 3πrn , where r1 is the radius of the largest

n=2

circle, r2 is the radius of the circles tangent to it, etc.

π We need to find the expression of rn as a function of r1 . In the figure, the angle ∠BAC measures radians and 3

π r1 1 r1 1 r2 cos = , and therefore, = . Similarly, = . Then |AB| = 2r1 and |DB| = 2r2 . As a 3 |AB| 2 |AB| 2 |DB| 1 result, we have the equality 2r1 = r1 + r2 + 2r2 , from which we conclude that r2 = r1 . Similarly, we can show 3  n−1 1 1 1 that r3 = r2 = r1 and, more generally, rn = r1 , n ≥ 2. 3 9 3 The series now takes the form  2  ∞ ∞ ∞    11 2 1 1 2 2 2 2 1 + 3 = πr1 . .πr1 + 3πrn = πr12 + 3π r = πr 1 1 n−1 n−1 3 9 8 n=2 n=2 n=2 What is the value of r1 ? Considering that |BC| =

π |BC| 1 1 , |AB| = 2r1 , and sin = , we have r1 = √ . 2 3 |AB| 2 3

Finally, the sum of the areas of the circles is 2 .πr1

+

∞ 

2 3πrn =

n=2

26. The area of the colored region can be written as A = is the area of the square inscribed in that circle. Let side of the inscribed square.

∞ 

n=1 rn be

11π . 96

(An − Bn ), where An is the area of the circle and Bn the radius of the nth circle and ln be the length of the

160

Numerical Series

2 + l2 = (2r )2 , that is, l2 = 2r 2 . We can write the series in the From Pythagoras’s Theorem, it follows that ln n n n n form ∞ ∞   2 2 2 .A = (πrn − ln )= (π − 2) rn . n=1

n=1

What is the relationship between r1 and rn ? We know that 2r2 = l1 , so 4r22 = l12 . As l12 = 2r12 , we have that 1 1 2 = n−1 r12 . Thus, r22 = r12 . By induction, it is easy to show that rn 2 2 .A

=

∞ 

2 (π − 2) rn =

n=1

As r1 = 1, we conclude that A = 2(π − 2).

∞  n=1

(π − 2)

1 r 2 = 2(π − 2) r12 . 2n−1 1

2.8. Proposed Exercises

2.8

161

Proposed Exercises

1. Find the general term and the sum of each of the following series: 1 1 1 1 + + + + ··· 3 8 15 24 1 1 1 b) + + + ··· 1×2×3 2×3×4 3×4×5 1 1 1 + + +· · · c) 1×2×3×4 2×3×4×5 3×4×5×6 a)

∞ 

(−1)n

n=1

2n + 1 n(n + 1)

∞

a   1 tan n , a ∈ R \ {kπ, k ∈ Z}. n 2 2 n=1

x

x = cot − 2 cot(x) Hint: tan 2 2 √ √ ∞  n+1− n √ c) 2 n +n n=1

∞  (−1)n − 8 3n n=1    n+1 ∞  1 n 1 1− e) − 1− n n+1 n=1

d)

f)

2−n − 2−3n



n=1 ∞  arctan(n) + arctan(−n + 2) g) arctan(n) arctan(n − 2) n=3

3. Let

∞ 

an be a convergent series. n=1 ∞  a3 + 5n √n is divergent. series n2 + 1 n=1

Show that the

4. Find the values of x for which the following series converge and, when possible, evaluate the sum of the series: a)

∞  n=0

b)

8n (x + 1)3n

∞   n=1

n |x| − 1

(−1)n x2n+1

n=0

5. Show that if

∞ 

an = A ∈ R, then

n=0 ∞  .

(an−1 + an + an+1 ) = 3A − a1 − 2a0 .

n=1

a)

∞  (−1)n+1 n n=1

b)

∞  (−1)n−1 n2 n=1

b)

∞  

∞ 

6. Test for convergence or divergence the following series. If convergence occurs, indicate whether it is conditional or absolute:

2. Find the sum of the series: a)

c)

c)

∞ 

(−1)n

n=1

n2 1 + n2

7. Consider the series

∞  (−1)n : (n + 1)2 n=1

a) Verify that it is convergent. b) Calculate its sum with an error less than 1 . 10000

8. Let a and bn be two convergent series, cn

n and dn be two divergent series, and α = 0 be a real number. What can be said about the convergence of the following series?   a) (an + bn ) e) (an cn ) b) c) d)

  

(an bn )

f)

(αan )

g)

(an + cn )

h)

  

(αcn ) (cn + dn ) (cn dn )

9. Test the convergence of the following series using a comparison test: a)

∞  n=1

1 n3 + 3

162

Numerical Series b)

∞ 



n=1

c)

n+1 n2 + 1

∞ 

1

n(n2 + 1) √ ∞  n n √ d) 3 n3 + 1 n=1 (n + 1)

b)

∞  n! n 3 n=1

c)

∞  n3 n! n=1

d)

∞  nn (2n)! n=1

e)

∞  n! 2 n n=1

n=1

e)

∞  

n2 − 1

n−



n=1

f)

∞



sin

π 2 n

n=1

h)

n=1

g)

a)

n2 + 1 n3 + 1

h)

n=2

b)

c)

1

n log(n)

n e−n

e)

∞  arctan(n) n2 + 1 n=2 ∞  n=1

f)

√

1 n+1−1 √

∞    1 n=1

b)

∞ 

c)

11. Use the Ratio Test or By the continuity of the logarithm function to determine the convergence or divergence of the following series: ∞  n n 3 n=1

2 × 4 × 6 × · · · × (2n) 2 × 5 × 8 × · · · × (3n − 1)

1 (n + 1)n 

d)

 ∞   n+1 n ∞ 

2

n

n=1

e)

nn/2 n log(n)

∞  kn ,k∈R n! n=1

2−2n+(−1)

n

n

n=1

f)

g)

 n ∞  1 1 − 1 nn n n=1 ∞

 n=1

a)

n+1

n! 3 × 5 × 7 × · · · × (2n − 1)

n=2

n

2

∞  n=1

2

n=2

d)

n

2−n+(−1) 3−n+(−1)

12. Use the Root Test or Cauchy’s Root Test to determine the convergence or divergence of the following series:

2

∞  log(3n) n2 n=2 ∞ 

∞  n=1

a) 

∞  n=1

10. Using the Cauchy Condensation Test or the Integral Test, determine the convergence of the following series: ∞ 

∞  n=1

√ ∞  n+ n g) n2 − n n=2 ∞ 

f)

sin

π n n

13. Test the convergence of the following series using Raabe’s test: a)

∞  n=1

n (2n + 1)!

2.8. Proposed Exercises b)

∞  n=1

c)

∞ 

1 n(n + 1)(n + 2) √

n=1

d)

∞  n=1

n−1 n

∞ 

(2n + 1)! n × n!

√

n=0

b)

p)

1 n2 + 1

∞ 

∞  en n n c) n! n=1

d)

n=1

e)

2 ,p∈R n2 + p 2

∞  cos(n π2 )

n2

n=1

f)

∞  n=0

g)

∞ 

i)

(n + 1)!

tan

π n n

∞

nπ   1 cos n 2 n=1 ∞  n=1

1 n

1) (1+ n

∞  n2 + 2n + 1 j) 3n2 + 2 n=1   n ∞  sin 3π 2 k) n2 + 1 n=1

l)

∞  n=2

m)

1 nn log(n)

∞  n=2

n=2 ∞  n=0 ∞  n=1 ∞  n=0 ∞ 

ne log(n) 2n (2n + 1)! 2 cos(nθ) ,θ∈R n5/2 3n n2n (2n)!

log(n!) + n! nn + 2n n=1 √ ∞ 3  n2 + 1 s) (−1)n n+3 n=0 r)



n

15. Examine the convergence or divergence of the following series. If convergence occurs, indicate whether it is conditional or absolute:

1 1 1+ log(n)



∞ 

sin(n + 1) 2 log(n + 1) n n=1  ∞   1 (n!)2 b) + n 3 (2n)! n=3 a)

c)

52n

n=3

h)

q)

∞ 

1 n(n2 − 1)

n=2

∞ 

n)

o)

14. Investigate the convergence or divergence of the following series. If convergence occurs, indicate whether it is conditional or absolute: a)

163

∞ 

1

(n + 1)(n + 2) n=1 ∞

e  d) cos(nπ) tan n n=3 ∞  1 + (−1)n n n2 + 5 n=1 3n−1  ∞  2n f) 4n + 1 n=1  ∞   n! n 1 g) 2 + nn n2 + n n=1

e)

h)

∞  (n + p)! , p, q ∈ N n!(n + q)! n=1 ∞ 

n! (π + 1)(π + 2) . . . (π + n) n=1   ∞  (−1)n 1 +√ j) n(n + 1) 3n n=1 i)

k)

∞  n=0

(−1)n

2n + 3 (n + 1)(n + 2)

164

Numerical Series l)

∞   n=0 ∞ 

a a+3

n

19. Knowing that an is convergent and of positive be said about terms and bn > 0, ∀n ∈ N, what  can an ? the convergence of the series 1 + bn

, a ∈ R \ {−3}

1 , a ∈ R+ 0 n+a 2 n=0 !  ! ∞  ! 1 !! n) log !!log ! n! n=2

m)

o)

∞ 

bn are convergent se20. Knowing that an and ries of positive terms, study the convergence of the following series:

(3n)! 27n (n!)2

n=0

1 1 1×3 1 1×3×5 1 · + · + · + ··· 2 3 2×4 5 2×4×6 7 √ ∞  2n − 1 log(4n + 1) q) n(n + 1) n=1

p) 1 +

∞  3 × 5 × 7 × · · · × (2n + 1) r) 2n (2n − 1) n! n=1

s)

t)

u)

∞ 

a)

b)

n=1 ∞ 

1 + sin(α)

(−1)n

n=0

17.

a) Test ∞  n=1

e

∞ 

a)

n=1

n

convergence  n2 .

sn



an ,a>0 (1 + a)(1 + a2 ) . . . (1 + an )

∞  (α + 1)(α + 2) . . . (α + n) , β ∈ R \ Z− : (β + 1)(β + 2) . . . (β + n) n=1

b)

1 (n + 1)α

for  2n

√

23. Study the conditional and absolute convergence of the series:

16. Indicate for which values of α the following series are conditionally or absolutely convergent: ∞  

sn+1 −

a0 np +···+ap in which a0 , 22. Prove that the series b0 nq +···+bq . . . , ap , b0 , . . . , bq are real numbers and a0 > 0, b0 > 0, is convergent if and only if q − p > 1.

1 2n − 1 + sin2 (n3 )

n=1

 √

is divergent.

2 + sin3 (n + 1) 2n + n2

n=1 ∞ 

21. Let an be a divergent series, an ≥ 0, and sn be the sum of its first n terms. Show that the series .

(−3)n 3 × 5 × 7 × · · · × (2n + 1)

n=1 ∞ 

  1 1 + an bn  n+1 b) an n a)

i. If α ∈ R \ Z− ii. If α ∈ Z− the

series

n n+3

b) Based on the previous item, indicate the limit    n2 n 2n . of the sequence e n+3

18. Let an and bn be two convergent series of pos √ an bn also itive terms. Show that the series converges. an + b n ≥ an b n . Hint: Prove that 2

un+1 2 1 24. Let un > 0 and ≤ 1 − + 2 ∀n ∈ N \ {1}. un n n

Show that un is convergent. 25. Consider the series .

∞ ∞   (−1)n 1 √ and : n! (n + 1) n+1 n=0 n=0

a) Calculate the partial sum of order three of Cauchy’s product of the two series. b) Study the convergence of the product series.

2.8. Proposed Exercises 26. The figure shows a sequence of cubes constructed as follows: The edge of the larger cube has length 1, and the length of the edge of each of the following cubes is half the length of the edge of the previous cube. Find the sum of the volumes of the cubes.

165 connect the midpoints of the sides of this square to form another square inside it. Repeat this process with the new square, connecting its midpoints to form another square. Keep repeating this process infinitely, and we will obtain infinite squares within the first one. Determine the sum of the areas of all the squares.

27. The figure shows a sequence of squares constructed as follows: Start with a square of side length 4. Now,

Series of Functions 3

It is possible to express various functions as a combination of “simpler” functions, such as a series of monomials (known as Taylor series) or trigonometric functions (known as Fourier series). These function series are useful not only in approximating other functions but also in other areas of mathematics, such as differential equations.

3.1

Introduction: Sequences of Functions

In many Analysis issues, it is interesting to consider sequences of functions of the form .f1 (x), .f2 (x), . . . , .fn (x), . . ., and naturally the question of passing to the limit arises.

Definition 3.1.1 Let .(fn ) be a sequence of functions, .fn : D ⊂ R → R. The sequence .(fn ) converges at a point .a ∈ D if the numerical sequence   . fn (a) is convergent. If the sequence .(fn ) converges at all points of D, we can define a function .f : D → R by .f (x) = lim fn (x), n→+∞

which is called the limit of .(fn ) on D. We also say that .(fn ) converges pointwise to f on D.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 A. Alves de S´a, B. Louro, Sequences and Series, https://doi.org/10.1007/978-3-031-67202-6 3

167

168

Series of Functions  x n Example 3.1.1 The sequence of functions . 1 + , defined on .R, conn verges whatever .x ∈ R (see Fig. 3.1). The limit function is .f (x) = ex :  x n . lim 1 + = ex ∀x ∈ R. n

Figure 3.1: The function .f (x) = ex and the first terms of the sequence

Example 3.1.2 Consider the functions .fn (x) = xn , .n ∈ N, on the interval .[0, 1]. They are continuous, and the limit function exists, but it is not continuous: ⎧ ⎨0, if 0 ≤ x < 1 n .f (x) = lim x = ⎩1, if x = 1 (see Fig. 3.2).

Figure 3.2: Sequence of continuous functions converging to a discontinuous function

3.1. Introduction: Sequences of Functions Based on Example 3.1.2, we can observe that a sequence of continuous functions may converge pointwise to a discontinuous function. We will see later that even if all the integrals in question exist, the integral of the limit function may differ from the limit of the integrals of the sequence’s functions. This means that the limit does not always commute with the integral. However, there is a stronger form of convergence known as uniform convergence, which guarantees both the continuity of the limit of a sequence of continuous functions and the ability to interchange limit and integration symbols. Definition 3.1.2 The sequence of functions .(fn ) converges uniformly to f on .D ⊂ R (.D = ∅) if ∀δ > 0 ∃ p ∈ N : n > p ⇒ |fn (x) − f (x)| < δ, ∀x ∈ D.

.

This condition is equivalent to ∀δ > 0 ∃ p ∈ N : ∀n > p, sup |fn (x) − f (x)| < δ,

.

x∈D

that is, .

lim

sup |fn (x) − f (x)| = 0.

n→+∞ x∈D

It is evident that if a sequence of functions .(fn ) converges uniformly to f on D, then it also converges pointwise to f on D. The converse is not true. If we take the sequence .(fn ) from Example 3.1.2, which converges pointwise to f on .[0, 1], we have .

sup |fn (x) − f (x)| = 1, ∀n ∈ N, x∈[0,1]

so .(fn ) does not converge uniformly to f on .[0, 1]. Note that if .(fn ) converges uniformly to f on D, then .(fn ) converges uniformly to f on any subset of D. The definition of uniform convergence means that, for every fixed .δ > 0, there is an order from which the images of all functions are in the .δ neighborhood of .f (x) for any .x ∈ D, that is, there is an order from which the images of all functions are in the region of the plane defined by .f (x) − δ and .f (x) + δ, in the set D. This is illustrated in Fig. 3.3.

169

170

Series of Functions

Figure 3.3: An illustration of uniform convergence

Obviously, there are sequences of continuous functions that converge pointwise to continuous functions, but not uniformly, as shown in the following example. Example 3.1.3 Let .α ∈ R. Consider the sequence of functions of general term α −nx .fn (x) = x n e , x ∈ R+ 0. This sequence converges pointwise to the function (Fig. 3.4) f (x) = 0, ∀x ∈ R+ 0.

.

However, this convergence is not uniform if .α ≥ 1: Note that

1 nα−1 , . sup |fn (x) − f (x)| = max fn (x) = fn = n e x∈R+ x∈R+ 0 0

which implies that nα−1 . lim sup |fn (x) − f (x)| = lim e x∈R+ 0

⎧ ⎪ ⎪+∞, ⎪ ⎨ 1 = , ⎪e ⎪ ⎪ ⎩ 0,

if α > 1 if α = 1 if α < 1.

Figure 3.5 illustrates the uniform convergence of the sequence of functions f (x) = x n1/2 e−nx .

. n

Remark the variation of .p ∈ N in relation to .δ > 0.

3.1. Introduction: Sequences of Functions

  Figure 3.4: The sequence of functions . x nα e−nx , .α > 1, .α = 1, and .α < 1

  Figure 3.5: The sequence of functions . x n1/2 e−nx

171

172

Series of Functions

3.2

Pointwise and Uniform Convergence of Series of Functions

The concepts of pointwise and uniform convergence extend to series of functions. Definition 3.2.1 Let .(fn ) be a sequence of functions, .fn : X ⊂ R → R. The series of general term .fn is the sequence of functions .(Sn ) defined by .Sn (x) = f1 (x) + f2 (x) + · · · + fn (x), ∀x ∈ X; the series is also represented by .

∞

fn . fn or by .

n=1

Definition 3.2.2 The series . fn converges at a point .a ∈ X if the

numerical series . fn (a)is convergent. If the series is convergent at all points of .D ⊂ X, we can define a function

fn (x) at each .f : D → R that corresponds to the sum of the series . point .x ∈ D; the function f is called the sum function of the series.

We also say that . fn (x) converges pointwise to f on D. ∞

x2 . (1 + x2 )n n=0 If .x = 0, all terms are zero; therefore, the series is convergent. If .x = 0, we can write

n ∞ ∞ ∞ 1 x2 1 2 2 . = x = x , (1 + x2 )n (1 + x2 )n 1 + x2 n=0 n=0 n=0 Example 3.2.1 Consider the series .

and this is a geometric series with ratio .r = is convergent and ∞

.

x2 = x2 · 2 )n (1 + x n=0

The sum function is f (x) =

.

1 ; as .|r| < 1, the series 1 + x2

1 1 1− 1 + x2

= 1 + x2 .

⎧ ⎨1 + x2 ,

if x = 0

⎩0,

if x = 0.

3.2. Pointwise and Uniform Convergence of Series of Functions

∞ xn . We can use a numerical series n! n=0 test to study the pointwise convergence of series of functions. In this case, we will apply D’Alembert’s test to study the series

Example 3.2.2 Consider the series .

.

∞  n x   .  n! 

n=0

 n+1    x    (n + 1)!  |x|  n = lim = 0, . lim x  n +1    n! 

∀x ∈ R.

Thus, we conclude that the original series is absolutely convergent .∀x ∈ R, defining a function f on .R. We will see later that .f (x) = ex , that is,

.

∞ xn = ex , n! n=0

Example 3.2.3 Consider the series .

∀x ∈ R.

∞

x (1 − x)n , x ∈ [0, 1].

n=0

If .x = 0, all terms are zero. Therefore, the series is convergent. ∞ If .x = 0, as the series . (1 − x)n is geometric with ratio .r = 1 − x and n=0

|r| < 1 since .x ∈ ]0, 1], the series converges. In this case,

.

∞ .

x (1 − x)n = x ·

n=0

Then, the series

∞ .

1 = 1. 1 − (1 − x)

x (1 − x)n , .x ∈ [0, 1], converges pointwise to the

n=0

function f (x) =

.

⎧ ⎨1,

if 0 < x ≤ 1

⎩0,

if x = 0.

173

174

Series of Functions

Definition 3.2.3 We say that a series . fn converges uniformly to the function f on .D ⊂ R, .D = ∅, if   n     .∀δ > 0 ∃ p ∈ N : n > p ⇒ f (x) − fi (x) < δ, ∀x ∈ D.   i=1

This condition is equivalent to ∀δ > 0 ∃ p ∈ N : n > p ⇒ sup |f (x) − Sn (x)| < δ,

.

x∈D

that is, .

lim sup |f (x) − Sn (x)| = 0.

n→+∞ x∈D

Note: Uniform convergence implies pointwise convergence, but the opposite is not necessarily true. This means that if a series of functions converges uniformly, it will also converge pointwise. However, if a series of functions converges pointwise, it does not necessarily converge uniformly. ∞

x2 is pointwise conver(1 + x2 )n n=0

Example 3.2.4 We saw that the series . gent to the function f defined by ⎧ ⎨1 + x2 , .f (x) = ⎩0,

if x = 0 if x = 0.

However, this series is not uniformly convergent on .[−1, 1]. In fact, lim

sup

lim

sup

n→+∞ x ∈ [−1,1]

=

n→+∞

.

=

lim

n→+∞

x ∈ [−1, 1] x = 0

sup x ∈ [−1, 1] x = 0

|f (x) − Sn (x)|   1 + x2 − (1 + x2 ) 1 −      1   lim  (1 + x2 )n  = n→+∞

  1   2 n+1 (1 + x ) sup

x ∈ [−1, 1] x = 0

1 = 1. (1 + x2 )n

The function f and some partial sums are illustrated in Fig. 3.6.

3.2. Pointwise and Uniform Convergence of Series of Functions

175

f (x)

Figure 3.6: An illustration of the sum function and some partial sums of Example 3.2.4

The following theorem allows us to test the uniform convergence of a series of functions without knowing its sum function. Theorem 3.2.1 It is a necessary and sufficient condition for the series

fn to be uniformly convergent on .D ⊂ R that   m     .∀δ > 0 ∃ p ∈ N : m > n > p ⇒  fr (x) < δ, ∀ x ∈ D.  

.

r=n+1

Theorem 3.2.2 (Weierstrass’ Test) If there exists a convergent numer ical series with positive terms, . an , such that |fn (x)| ≤ an , ∀ x ∈ D, ∀ n ∈ N,

.

then the series .

fn is uniformly convergent on D.

Proof: We know by Theorem 2.2.3 that . an converges if and only if ∀δ > 0 ∃ p ∈ N : m > n > p ⇒ |an+1 + · · · + am | < δ.

.

Let .δ > 0 be arbitrary. We have

Karl Theodor Wilhelm Weierstrass (1815–1897) was a German mathematician often referred to as the “father of modern analysis.” His work had a huge influence on the mathematics of the nineteenth and twentieth centuries, and many of his results are part of any Calculus course, such as, for example, the Weierstrass’ test for series and his example of a continuous function that is not differentiable at any point. (Source of image: Dibner Library of the History of Science and Technology, Smithsonian Libraries)

176

Series of Functions       fn+1 (x) + · · · + fm (x) ≤ fn+1 (x) + · · · + fm (x) ≤ an+1 + · · · + am

.

∀x ∈ D

= |an+1 + · · · + am |, since an > 0, ∀n ∈ N. Then ∃ p ∈ N : m > n > p ⇒ |fn+1 (x) + · · · + fm (x)| < δ,

.

  m     .∃ p ∈ N : m > n > p ⇒  fr (x) < δ.  

or still,

r=n+1

From Theorem 3.2.1 comes the desired result.

Example 3.2.5 Let k be a constant such that .|k| < 1. The series .

∞

fn (x),

n=1

where

2n 1 × 3 × · · · × (2n − 1) 2n  ·k sin(x) , 2 × 4 × · · · × 2n

f (x) =

. n

is uniformly convergent on .R. In fact,         fn (x) =  1 × 3 × · · · × (2n − 1) · k 2n sin(x) 2n   2 × 4 × · · · × 2n  .

≤

1 × 3 × · · · × (2n − 1) 2n · k , ∀x ∈ R, 2 × 4 × · · · × 2n

and the numerical series .

∞ 1 × 3 × · · · × (2n − 1) 2n ·k 2 × 4 × · · · × 2n n=1

is convergent. To verify this, we apply D’Alembert’s test: 1 × 3 × · · · × (2n − 1)(2n + 1) 2n+2 ·k 2n + 1 2 2 × 4 × · · · × 2n(2n + 2) · k = k 2 < 1. . lim = lim 1 × 3 × · · · × (2n − 1) 2n 2n + 2 ·k 2 × 4 × · · · × 2n

3.2. Pointwise and Uniform Convergence of Series of Functions   ∞  sin(nx)  1 1   ≤ 2 , .∀x ∈ R, and the series . Example 3.2.6 Since . is 2 n2  n n n=1 ∞ sin(nx) is uniformly convergent on .R. convergent, the series . n2 n=1 Note: The Weierstrass’ test is a sufficient but not necessary condition for the uniform convergence of a series of functions: there exist uniformly convergent series whose general term cannot be bounded as required by the Weierstrass’ test. It is important to note that this test implies absolute convergence of the series of functions. ∞

x2 + n is alternating, n2 n=1 and using Leibniz’s test, we can prove that it is convergent. However, it is not absolutely convergent because   2   x2 + n 1 n x + n  = . (−1) ≥ , ∀x ∈ R,   2 2 n n n Example 3.2.7 For each .x ∈ R, the series .

(−1)n

∞ 1 and the series . is divergent. This makes it impossible to use the Weiern n=1 strass’ test to determine the uniform convergence of the series. Instead, we have to study it directly. For each .x ∈ R, let .Sn (x) be the partial sum of order n of the series ∞ x2 + n .S(x) = (−1)n . By Corollary 1 of Theorem 2.3.2, n2 n=1

|S(x) − Sn (x)| ≤

.

x2 + n + 1 . (n + 1)2

If .X ⊂ R is a bounded set, that is, if there exists .M > 0 such that .|x| ≤ M , ∀x ∈ X, then

.

|S(x) − Sn (x)| ≤

.

M2 + n + 1 , ∀x ∈ X; (n + 1)2

therefore .(Sn ) converges uniformly to S on X, that is, the given series converges uniformly on any bounded subset of .R.

177

178

Series of Functions The importance of uniform convergence lies in the fact that a series that converges uniformly behaves similarly to the sum of a finite number of functions. For instance, when we add a finite number of continuous functions, the result is also a continuous function. Similarly, with a series of functions that converge uniformly, we obtain the following outcome: Theorem 3.2.3 If the functions .f1 , f2 , . . . , fn , . . . are continuous on D

and the series . fn converges uniformly to f on D, then f is continuous on D. Proof: Let .x0 be an arbitrary point of D. We want to prove that ∀δ > 0 ∃ ε > 0 : |x − x0 | < ε ⇒ |f (x) − f (x0 )| < δ.

.

We can write f (x) − f (x0 ) = f (x) − Sn (x) + Sn (x) − Sn (x0 ) + Sn (x0 ) − f (x0 ),

.

which implies that |f (x) − f (x0 )| ≤ |f (x) − Sn (x)| + |Sn (x) − Sn (x0 )| + |Sn (x0 ) − f (x0 )|.

.

Let .δ > 0. As the series converges uniformly to f , we know that ∃ p ∈ N : ∀n > p, |f (x) − Sn (x)|
0 : |x − x0 | < ε ⇒ |Sn (x) − Sn (x0 )|
0 : |x − x0 | < ε ⇒ |f (x) − f (x0 )| < δ.

.

The result follows from the arbitrariness of .x0 . ∞ cos(nx) √ . The terms of n3 + 1 n=1 the series are functions with domain .R. The function f is the sum of the series; therefore, its domain is the subset of .R where the series converges. We have the inequality    cos(nx)  1   . √ (3.1)  n3 + 1  ≤ √n3 + 1 , ∀x ∈ R,

Example 3.2.8 Consider the function .f (x) =

3.2. Pointwise and Uniform Convergence of Series of Functions since .| cos(x)| ≤ 1, .∀x ∈ R. Let us study the series of positive terms

∞

1 √ , comparing it with 3 n +1 n=1

.

∞ 1 , which we know is convergent. As the limit 3/2 n n=1

the series .

1  √ 3 n3 n +1 =1 . lim = lim 3 1 n +1 n3/2 is finite and different from zero, we conclude that the series

∞ .

√

1 n3

+1 n=1  ∞   cos(nx)  √  is convergent. By inequality (3.1), the series .  n3 + 1  converges, n=1

∞ cos(nx) √ .∀x ∈ R, so the series . is convergent, .∀x ∈ R, that is, the n3 + 1 n=1 domain of f is .R. To conclude that f is continuous on .R, it is sufficient that the following two conditions are satisfied:

cos(nx) (i) The functions .fn (x) = √ are continuous on .R. n3 + 1 ∞ cos(nx) √ is uniformly convergent to f on .R. n3 + 1 n=1

(ii) The series .

The functions .cos(nx) are continuous on .R, .∀n ∈ N, so the first condition is true. The second condition is a consequence of the inequality (3.1) and the Weierstrass’ test. As the series converges to f , this function is continuous. Note: If the sum of a series of functions is not continuous, this indicates that either the individual functions .f1 , .f2 , .. . . , .fn , . . . are not continuous or the series is not uniformly convergent. Therefore, if .f1 , f2 , . . . , fn , . . . are continuous functions and f is not continuous, we can be sure that the convergence is not uniform. ∞

x2 , on the interval (1 + x2 )n n=0 .[−a, a], .a > 0. We proved in Example 3.2.1 that this series converges Example 3.2.9 Let us consider the series

.

179

180

Series of Functions pointwise to the function

f (x) =

.

⎧ ⎨1 + x2 ,

if x = 0

⎩0,

if x = 0.

Since f is discontinuous at .x = 0 and .fn (x) = ∀n ∈ N, the series is not uniformly convergent.

x2 is continuous, (1 + x2 )n

.

Theorem 3.2.4 Let .a, b ∈ R, .a < b. If the functions .f1 , f2 , . . . , fn , . . . ∞ are continuous on .[a, b] and the series . fn converges uniformly to f on n=1

[a, b], then

.



b

f (x) dx =

.

a

∞  n=1

b

fn (x) dx. a

(The series is integrable term by term.) Proof: By Theorem 3.2.3, f is continuous on .[a, b]; therefore, it is integrable on .[a, b]. By hypothesis, the functions .f1 , f2 , . . . , fn , . . . are continuous on .[a, b], which implies that they are integrable on this interval. We intend to prove that the sequence .(Sn∗ ) of the partial sums of the series  b ∞  b . fn (x) dx has limit . f (x) dx, that is, n=1

a

a

   b    ∗  .∀δ > 0 ∃ p ∈ N : n > p ⇒ Sn − f (x) dx < δ.   a Let .δ > 0 be arbitrary.      b  b n  b      ∗    . S n − f (x) dx =  fi (x) dx − f (x) dx     a a a i=1

  n     b  b   = fi (x) dx − f (x) dx  a  a i=1

3.2. Pointwise and Uniform Convergence of Series of Functions        n n  b   b      = fi (x) − f (x) dx ≤ fi (x) − f (x) dx    a   a i=1



i=1

b

|Sn (x) − f (x)| dx.

= a

But the series .

∞

fn (x) converges uniformly to f on .[a, b]; therefore,

n=1

∃ p ∈ N : n > p ⇒ |f (x) − Sn (x)|
p ⇒

b

|Sn (x) − f (x)| dx
1; ∞ ∞ (n + 1) and . (−1)n (n + 1). if .|x| = 1, we have the divergent series . The series .

∞

n=0

n=0

n

(n + 1) x is uniformly convergent on every interval .[−r, r] if

n=0

∞   0 < r < 1 because .(n + 1)xn  ≤ (n + 1)rn and the series . (n + 1) rn is

.

n=0

3.2. Pointwise and Uniform Convergence of Series of Functions convergent. We can then write 

  ∞ ∞ 1 n = x = (n + 1) xn , 1−x n=0 n=0 .

⇔

∞ 1 = (n + 1) xn , (1 − x)2 n=0

|x| < 1.

|x| < 1

185

186

Series of Functions

3.3

Power Series

Definition 3.3.1 Let x0 ∈ R be arbitrary. A series of the form ∞ .

an (x − x0 )n

n=0

with an ∈ R, ∀n ∈ N, is called a power series of x − x0 . Notes: 1. By making y = x − x0 , a power series can always be reduced to the ∞ an y n . form n=0

2. To ensure accuracy, the power series should be written as a +

+∞

. 0

an (x − x0 )n .

n=1

At x = x0 , the sum of the series is a0 (all terms are 0 for n ≥ 1). This is the meaning of the series that appears in the definition, which we write in this way to simplify. When evaluating the indeterminate form a0 00 at x = x0 , it should be interpreted as equal to a0 .

Theorem 3.3.1 Let ∞

lim

1  n

|an |

= r.

If r ∈ R+ , the power series

an x is absolutely convergent at each point x ∈ ] − r, r[ and din

n=0

vergent at each point x ∈ ] − ∞, −r[ ∪ ]r, +∞[. If r = +∞, then the power series is absolutely convergent whatever x ∈ R. If r = 0, the series converges if x = 0 and diverges if x = 0.

Proof: Consider the series

∞

|an xn |. Let r ∈ R+ . Using the fact that

n=0

lim

.

 n

|an xn | = |x| lim

 n

|an |,

3.3. Power Series

187

 we can apply Corollary 1 of the Root Test to state that if |x| lim n |an | < 1 ∞ (i.e., if |x| < r), the series converges; therefore, the series an xn conn=0

verges absolutely.  If |x| lim n |an | > 1 (i.e., if |x| > r), then, by the reasoning used in   Corollary 1 of the Root Test, there exists a subsequence of |an xn | that takes values greater than or equal to 1, which implies that the sequence     |an xn | does not tend to zero; thus the sequence an xn does not tend ∞ to zero, and the series an xn diverges.  n

n=0

If r = 0, then lim |an | = +∞. It follows that, whatever x ∈ R, x = 0, ∞  |x| lim n |an | = +∞ > 1, and as before, the series an xn diverges. n=0

Obviously, if x = 0, all terms are zero; thus, the series is convergent.  If r = +∞, then lim n |an | = 0. Therefore, whatever x ∈ R, ∞  |x| lim n |an | = 0 < 1, and we can conclude that the series an x n n=0

converges absolutely.

Definition 3.3.2 Under the conditions of Theorem 3.3.1, r is called the radius of convergence of the series, and the interval ] − r, r[ is the interval of convergence.    an   = r is a positive real number, zero, or +∞,  Corollary 1 If lim  an+1  then the radius of convergence of the power series is r. Proof: It is sufficient to note that by Theorem 1.1.11,    an  |an | 1 1    = = r. . lim  an+1  = lim |an+1 | = lim  n n lim |an | |an |  n |an |; that  is, every power series has a  an   may not exist, as illustrated in radius of convergence. However, lim  an+1  the following example. Note that there always exists lim

188

Series of Functions

Example 3.3.1 Consider the series

∞

an x n =

n=0

.

an an+1

⎧ ⎨2n−1 , = ⎩ 1 , 2n+2

   an   does not exist, but r =  lim  an+1 

∞  n 3 + (−1)n xn . As n=0

if n is even if n is odd, 1  n

1 . 4 lim |an |  1 1 The interval of convergence of the series is − , . The series converges 4 4      1 1 1 1 absolutely on the interval − , and diverges on −∞, − ∪ , +∞ . 4 4 4 4 =

Example 3.3.2 Let us calculate the radius of convergence of the series ∞ xn : nn n=1 1 1 1   = = +∞. = .r = n 1 1 lim |an | n lim lim n nn The radius of convergence of the series is infinite; therefore, the series is absolutely convergent ∀x ∈ R.

Example 3.3.3 The series

∞

n! xn has a radius of convergence r = 0:

n=0

   an  n! 1  .r = lim   an+1  = lim (n + 1)! = lim n + 1 = 0, that is, the series only converges at x = 0.

Example 3.3.4 Consider the series    an   .r = lim   an+1  = lim

∞ xn . Taking into account that n n=1

1 n+1 n = 1, = lim n 1 n+1

3.3. Power Series

189

the interval of convergence is ] − 1, 1[. The series converges absolutely on the interval ] − 1, 1[ and diverges on ] − ∞, −1[ ∪ ]1, +∞[. Example 3.3.5 Consider the series The power series of y .

∞ (−1)n (x + 1)n . Let y = x + 1. n 2 n=0

∞ (−1)n n y 2n n=0

has radius of convergence    (−1)n       an   2n  = lim  .r = lim   = 2. n+1  an+1   (−1)    n+1 2 Then the power series of y converges absolutely if y ∈ ] − 2, 2[ and diverges if y ∈ ] − ∞, −2[ ∪ ]2, +∞[. The power series of x + 1 converges absolutely if x ∈ ] − 3, 1[ and diverges if x ∈ ] − ∞, −3[ ∪ ]1, +∞[. Example 3.3.6 Consider the series ∞

∞ n=0

(−1)n

x2n+1 . Let y = x2 . The (2n + 1)!

yn has radius of convergence (−1)n series (2n + 1)! n=0    (−1)n     (2n + 1)!  (2n + 3)!   .r = lim  (−1)n+1  = lim (2n + 1)! = lim (2n + 3)(2n + 2) = +∞;     (2n + 3)! therefore, the series is absolutely convergent for all y ∈ R+ 0 , and the original series is absolutely convergent for all x ∈ R (note that it is enough to multiply the power series of x by x). Note: The preceding theorem omits any reference to the convergence or divergence of the power series at the endpoints of the convergence interval ] − r, r[, r ∈ R+ . It is possible for the series to converge at both endpoints, converge at one and diverge at the other, or diverge at both. Therefore, we must always analyze the series at x = r and x = −r. In the case of Example 3.3.4, the convergence interval is ] − 1, 1[:

190

Series of Functions – If x = −1, we obtain the series vergent. – If x = 1, we obtain the series

∞ (−1)n , which is conditionally conn n=1

∞ 1 , which is divergent. n n=1

We conclude that the series diverges on ] − ∞, −1[∪[1, +∞[ and converges on [−1, 1[.

Theorem 3.3.2 If the radius of convergence of the series

∞

an xn is

n=0

r > 0 and if 0 < ρ < r, then the series is uniformly convergent on [−ρ, ρ]. Proof: By hypothesis, |an xn | ≤ |an | ρn = |an ρn |, ∀x ∈ [−ρ, ρ]. The series ∞ |an ρn | is convergent by Theorem 3.3.1, since ρ ∈ ] − r, r[. n=0

According to Weierstrass’ test, the series

∞

an xn is uniformly convergent

n=0

on [−ρ, ρ]. Corollary 1 A power series is uniformly convergent on every closed interval [a, b] contained within its interval of convergence and 

∞ b

.

an xn dx =

a n=0

∞ n=0

an

bn+1 − an+1 . n+1

Proof: If [a, b] ⊂ ] − r, r[, then there exists ρ > 0 such that [a, b] ⊂ [−ρ, ρ] ⊂ ] − r, r[ .

.

By the theorem, the series is uniformly convergent on [−ρ, ρ] and also on [a, b]. Then we can integrate the series term by term on [a, b]:  b  b ∞ ∞  b ∞ n n an x dx = an x dx = an xn dx a n=0

n=0

.

=

∞ n=0

a

an

n=0

bn+1 − an+1 . n+1

a

3.3. Power Series

191 

1

Example 3.3.7 Let us evaluate

f (x) dx, where f (x) = 0

∞

∞ n=0

(−1)n

x2n . (2n)!

yn has an infinite radius of convergence: Let y = x2 . The series (−1)n (2n)! n=0    (−1)n     (2n)!  (2n + 2)!  .r = lim   (−1)n+1  = lim (2n)! = lim (2n + 2)(2n + 1) = +∞,     (2n + 2)! which implies that the original series converges for all x in R; therefore, it is uniformly convergent on [0, 1] and integrable term by term on that interval: 

∞  1 x2n x2n dx = dx (−1) (−1)n (2n)! (2n)! 0 n=0 n=0 0 .  2n+1 1 ∞ ∞ x 1 (−1)n = . (−1)n = (2n)! 2n + 1 0 n=0 (2n + 1)! n=0 ∞ 1

n

Theorem 3.3.3 Every power series with radius of convergence r > 0 is term by term differentiable on the interval of convergence, that is, ∞  ∞ n . an x = n an xn−1 , ∀x ∈ ] − r, r[. n=0

Proof: Consider the series

n=1 ∞

an xn . It is pointwise convergent on ] − r, r[;

n=0

the functions (an xn ) = nan xn−1 are continuous on ] − r, r[, ∀n ∈ N. ∞ nan xn−1 is a power series with radius of convergence r: Moreover, n=0

.

1 1 1    = = = r; √ n n n n lim |nan | lim n |an | lim |an |

therefore, it is uniformly convergent on [a, b] ⊂ ] − r, r[. Thus, by Corollary 2 of Theorem 3.2.4, ∞  ∞ n . an x = n an xn−1 , ∀x ∈ ] − r, r[ . n=0

n=1

192

Series of Functions

∞ Note: If the power series n=0 an xn has radius of convergence r, then both the series of derivatives and the series of indefinite integrals have the same radius of convergence, r.

Example 3.3.8 Let us consider the power series

∞

(−1)n+1

n=1

∞

(x − 5)n . n 5n

yn (−1)n+1 n has radius of convergence Let y = x − 5. The series n5 n=1     .r = lim    

 (−1)n+1  n+1  n 5n  = lim (n + 1) 5 = 5, n 5n (−1)n+2   (n + 1) 5n+1

which implies the absolute convergence of the original series on the interval ]0, 10[. – If x = 0, we obtain the series

∞ −1 , which is divergent. n n=1

– If x = 10, we obtain the series convergent.

∞ (−1)n+1 , which is conditionally n n=1

Then, the original series converges absolutely on the interval ]0, 10[, converges conditionally at x = 10, and diverges on ] − ∞, 0] ∪ ]10, +∞[. The series of derivatives is ∞  ∞ n (x − 5)n−1 n+1 (x − 5) n+1 (−1) = (−1) n n 5n n 5n n=1 n=1 .

=

∞

(−1)n+1

n=1

(x − 5)n−1 . 5n

The interval of convergence of this series is ]0, 10[. – If x = 0, we obtain the series

∞ 1 which is divergent. 5 n=1

– If x = 10, we obtain the series

∞ (−1)n+1 which is divergent. 5 n=1

3.4. Taylor Series and Maclaurin Series

3.4

193

Taylor Series and Maclaurin Series

Let I be an interval and .f : I ⊂ R → R a function of class .C n on I. Let .x0 ∈ I. We know that there exists .0 < θ < 1 such that f (x) = f (x0 ) + f  (x0 )(x − x0 ) (x − x0 )2 (x − x0 )n−1 + · · · + f (n−1) (x0 ) + Rn (x), +f  (x0 ) 2! (n − 1)!

.

  (x − x0 )n where .Rn (x) = f (n) x0 +θ (x−x0 ) . This is the Taylor’s formula n! for f , of order n, with Lagrange remainder around the point .x0 . Suppose that .f ∈ C ∞ (I). The power series .

∞ f (n) (x0 ) (x − x0 )n n! n=0

is called the Taylor series of f at .x0 . If .x0 = 0 ∈ I, the Taylor series is called Maclaurin series and is written as follows: ∞ f (n) (0) n x . . n! n=0 Example 3.4.1 Let us determine the Maclaurin series of .f (x) = sin(x).  nπ  We know that .f ∈ C ∞ (R) and .f (n) (x) = sin x + , .∀n ∈ N. Then 2  nπ  (n) .f (0) = sin , and therefore, the Maclaurin series of f is 2 x−

.

∞ x5 x7 x2n+1 x3 + − + ··· = . (−1)n 3! 5! 7! (2n + 1)! n=0

We saw in Example 3.3.6 that this series is convergent on .R. Example 3.4.2 Consider the function .f (x) = (1 + x)α , .α ∈ R, .x > −1. By the rules of differentiation, .f ∈ C ∞ (] − 1, +∞[) and f (n) (x) = α(α − 1) . . . (α − n + 1)(1 + x)α−n , ∀n ≥ 1.

.

Therefore, .f (n) (0) = α(α − 1) . . . (α − n + 1), and its Maclaurin series is 1 + αx +

.

α(α −1) 2 α(α −1) . . . (α − n + 1) n x + ··· + x + ··· = 2! n!

Brook Taylor (1685–1731) was an English mathematician. His work titled Methodus Incrementorum Directa et Inversa (London, 1715) added a new branch to Mathematics known as “calculus of finite differences.” Among other applications, he used it to determine the shape of the movement of a vibrating string, reducing it to mechanical principles. This work also contains the famous formula known as Taylor’s Formula. (Source of image: Line engraving after Richard Earlom (1743–1822))

194

Series of Functions

.

Collin Maclaurin (1698– 1764) was a Scottish mathematician. In 1717, at the age of 19 years, he was appointed professor in Aberdeen. In 1719, he published the work Organic Geometry , a text that can be considered the most important of his works and contains, among others, an original method of generating conics. He developed Newton’s work in calculus, geometry, and gravitation. (Source of image: Pencil and chalk on paper by David Steuart Erskine (1742–1829), Scottish National Gallery)

=1+

∞ α(α − 1) . . . (α − n + 1) n x . n! n=1

If .α ∈ N, the series is reduced to the development of Newton’s binomial. Suppose that .α ∈ N0 . Then let us study the convergence of the series. The radius of convergence is     α(α − 1) . . . (α − n + 1)     1 n!  = lim (n + 1)! lim   α(α − 1) . . . (α − n + 1)(α − n) n! |α − n|   .   (n + 1)! n+1 = 1. = lim |α − n| Therefore, the series is absolutely convergent on .] − 1, 1[ and divergent on ] − ∞, −1[ ∪ ]1, +∞[. This series is usually referred to as the binomial series.

.

A fundamental question in the Taylor series development of an indefinitely differentiable function is as follows: Is there a neighborhood V of .x0 such that f (x) = f (x0 ) + f  (x0 )(x − x0 ) (x − x0 )2 (x − x0 )n−1 + · · · + f (n−1) (x0 ) + · · · , ∀x ∈ V ? +f  (x0 ) 2! (n − 1)!

.

In other words, does the Taylor series expansion of f at .x0 converge for all x ∈ V and sum up to .f (x)? In reality, the mere existence of the derivatives .f (n) (x0 ) for all natural values of n, although it allows writing the Taylor series of f at .x0 , does not guarantee that, in some neighborhood of .x0 , the equality is satisfied:

.

f (x) =

.

∞ f (n) (x0 ) (x − x0 )n , n! n=0

as can be seen in the following example. Example 3.4.3 Consider the function ⎧ ⎨e−1/x2 , .f (x) = ⎩0,

if x = 0 if x = 0.

(3.2)

3.4. Taylor Series and Maclaurin Series As .f (n) (0) = 0, .∀n ∈ N, the Maclaurin series of f is written as follows: 0 + 0x + 0x2 + · · · ,

.

which converges to the zero function on .R. Therefore, the only point where f equals the sum of the series is 0, since .f (x) = 0 if .x = 0. To ensure equality (3.2), what additional conditions must be imposed on a function f that is assumed to be indefinitely differentiable on a neighborhood of .x0 ? We can easily answer this question, by considering Taylor’s formula for f . Specifically, if .Sn (x) denotes the partial sum of order n of the Taylor series of f at .x0 ∈ I, then we have .Rn (x) = f (x) − Sn (x), which satisfies the following result: Theorem 3.4.1 It is a necessary and sufficient condition for an indefinitely differentiable function, .f : I → R, to be the sum of its Taylor series in a neighborhood, V , of .x0 ∈ I, that .

lim Rn (x) = 0, ∀x ∈ V.

In practice, sufficient conditions are used: Theorem 3.4.2 Let .f : I → R be an indefinitely differentiable function, and suppose that there are constants .M, k ≥ 0 such that, in a neighborhood, V , of .x0 ,    (n)  . f (x) ≤ M k n , ∀x ∈ V, ∀n ∈ N. Then, f is the sum of its Taylor series on V . Proof: We know that the expression of the Lagrange remainder, .Rn (x), is   (x − x0 )n , 0 < θ < 1. Rn (x) = f (n) x0 + θ (x − x0 ) n!

.



n k|x − x0 | , x∈V; . |Rn (x)| ≤ M n!  n k|x − x0 | since the series of general term . is convergent, this sequence n! has limit 0. The desired result is an immediate consequence of Theorem 3.4.1. Then

195

196

Series of Functions Corollary 1 If there exists .M ≥ 0 such that on the neighborhood V of .x0 ,    (n)  . f (x) ≤ M ∀x ∈ V, ∀n ∈ N, then f is the sum of its Taylor series on V .

Example 3.4.4 Consider the function .f (x) = sin(x). In Example 3.4.1, we found that the Maclaurin series of f converges absolutely on .R. We know that   sin θx + nπ 2 xn , 0 < θ < 1, .Rn (x) = n! from which  | sin θx + .0 ≤ |Rn (x)| = n!

nπ 2



|

|x|n ≤

|x|n . n!

|x|n = 0, .∀x ∈ R, as it is the general term of a convergent series, But .lim n! which implies that .

sin(x) =

∞ n=0

(−1)n

x2n+1 , ∀x ∈ R. (2n + 1)!

Figure 3.7: Some Taylor polynomials of the function .f (x) = sin(x)

In Fig. 3.7, we can see how the partial sums are increasingly better approximations of the function .sin(x), thus illustrating the convergence of n x2k+1 is denoted by .P2n+1 in the the series. The polynomial . (−1)k (2k + 1)! k=0 figure. The black line represents the graph of the function .sin(x).

3.4. Taylor Series and Maclaurin Series

197

Example 3.4.5 If .f (x) = ex , we obtain .f (n) (x) = ex , .∀n ∈ N. Then its Maclaurin series development is .

∞ ∞ f (n) (0) n xn x = . n! n! n=0 n=0

We know that this series is absolutely convergent on .R defining a function g on .R. We intend to prove that .f (x) = g(x), .∀ x ∈ R. For this purpose, we will show that the Lagrange remainder of the Maclaurin formula of the function f tends to 0 on .R. f (n) (θx) n eθ x n x = x , n! n!

Rn (x) =

.

0 < θ < 1,

which implies that, taking into account that .eθ x ≤ e|x| since f is an increasing function, e|x| n |x| . .0 ≤ |Rn (x)| ≤ n! e|x| n |x| is convergent, .∀x ∈ R; therefore, The series of general term . n! e|x| n |x| = 0, ∀ x ∈ R, n! which allows us to conclude that ∞ xn x , ∀ x ∈ R. .e = n! n=0 .

lim

The Maclaurin series developments in the first two examples were obtained directly from the formula: f  (0) 2 x + ··· , 2! by evaluating all derivatives of the function at zero. Since this process is quite laborious, it is not commonly employed. Instead, well-known developments are often utilized in practice. Furthermore, the next theorem guarantees that these developments are indeed the Taylor series. f (0) + f  (0) x +

.

Theorem 3.4.3 Every power series of .x − x0 is the Taylor series (around x ) of the function defined by it. In particular, the power series development of .x − x0 is unique.

. 0

198

Series of Functions Proof: By hypothesis, .f (x) =

∞

an (x − x0 )n in a neighborhood V of .x0 ,

n=0

which implies that .f (x0 ) = a0 . Differentiating, 

f (x) =

.

∞

n an (x − x0 )n−1 ,

n=1

and therefore, .f  (x0 ) = a1 . The nth order derivative is f (n) (x) = n! an + (n + 1) · · · 2 an+1 (x − x0 )

.

+(n + 2)(n + 1) · · · 3 an+2 (x − x0 )2 + · · · , and so .f (n) (x0 ) = n! an , .n ∈ N. Therefore, we have a =

. n

f (n) (x0 ) , ∀n ∈ N. n!

Example 3.4.6 Consider the function .f (x) = count that .

and that .

1 . Taking into ac2 + 3x

1 1 1   = · 2 + 3x 2 1 − − 23 x

∞ 1 = xn , 1 − x n=0

∀x ∈ ] − 1, 1[,

we can deduce that

n ∞ 1 1 3 1  = · . − x , 2 1 − − 32 x 2 n=0 2    3  2 and this equality is valid as long as .− x < 1, that is, .|x| < . Then, the 2 3 Maclaurin series of f is ∞ .

(−1)n

n=0

3n 2n+1

xn ,

|x|
−1,

into a Maclaurin series, thereby obtaining 1+

.

∞ α(α − 1) . . . (α − n + 1) n x , n! n=1

convergent on the interval .] − 1, 1[. Let g(x) = 1 +

.

∞ α(α − 1) . . . (α − n + 1) n x , |x| < 1. n! n=1

Next, we prove that .f (x) = g(x), .∀x ∈ ] − 1, 1[, that is, f is the sum of its Maclaurin series in that interval. Since a power series is differentiable term by term on the interval of convergence, we obtain g  (x) =

.

∞ α(α − 1) . . . (α − n + 1) n−1 x , (n − 1)! n=1

and multiplying by x x g  (x) =

.

∞ α(α − 1) . . . (α − n + 1) n x . (n − 1)! n=1

200

Series of Functions Then g  (x) + x g  (x) =

.

∞ ∞ α(α − 1) . . . (α − n + 1) n−1 α(α − 1) . . . (α − n + 1) n x x = + (n − 1)! (n − 1)! n=1 n=1 ∞ ∞ α(α − 1) . . . (α − n) n α(α − 1) . . . (α − n + 1) n x + x n! (n − 1)! n=0 n=1

∞ α(α − 1) . . . (α − n) α(α − 1) . . . (α − n + 1) + =α+ xn n! (n − 1)! n=1

∞ α(α − 1) . . . (α − n + 1) α − n + 1 xn =α+ (n − 1)! n n=1

=

.

∞ α(α − 1) . . . (α − n + 1) α n · x (n − 1)! n n=1   ∞ α(α − 1) . . . (α − n + 1) n x =α 1+ n! n=1

=α+

= α g(x), that is, (1 + x) g  (x) = α g(x). g(x) Let us consider the function . and calculate its derivative: (1 + x)α

 g(x) g  (x)(1 + x)α − α(1 + x)α−1 g(x) = (1 + x)α (1 + x)2α .   (1 + x)α−1 (1 + x) g  (x) − α g(x) = . (1 + x)2α .

(3.3)

By equality (3.3) the numerator of this fraction is zero; therefore,

 g(x) . = 0, (1 + x)α which implies that the function defined by .

g(x) (1 + x)α

is constant on .] − 1, 1[, that is, .g(x) = c (1 + x)α , for some constant c. Since .g(0) = 1, we obtain .c = 1, and we have g(x) = (1 + x)α , ∀x ∈ ] − 1, 1[ .

.

3.5. Introduction to Fourier Series

3.5

Introduction to Fourier Series

Taylor series are not the only series used to approximate and/or represent functions. The Fourier series, which we introduce in this section, are another example of how series are applied to the study of functions. First, we establish some definitions and results concerning periodic functions that will be useful in studying Fourier series (Fig. 3.8). Definition 3.5.1 We say that a function .f : R → R is periodic with period .T > 0 if .f (x + T ) = f (x), .∀x ∈ R.

Figure 3.8: A periodic function with period T

The following two propositions are direct consequences of the definition. Proposition 3.5.1 Let .f : R → R be a periodic function with period T . Then f is periodic with period kT , .∀k ∈ N. Proposition 3.5.2 Let f and g be periodic functions with period T , and .α, .β ∈ R. Then, the function .αf + βg is periodic with period T . Consider the functions a cos(kx) + bk sin(kx) , k = 1, 2, . . . ,

. k

where .ak , .bk .∈ R, ∀k ∈ N. These are periodic functions with period 2π .Tk = , .k = 1, 2, . . .. Therefore, .T = 2π = kTk is a common period k to all. As a result, sums of the form s (x) = A +

n 

. n

k=1

 ak cos(kx) + bk sin(kx) ,

201

202

Series of Functions where A is a constant, are periodic functions with period .T = 2π. These sums are usually referred to as trigonometric polynomials. Definition 3.5.2 A trigonometric series is a series of the form ∞

.

 a0  + an cos(nx) + bn sin(nx) . 2 n=1

The constants .a0 , .an , and .bn , .n ∈ N, are the coefficients of the trigonometric series. Two questions arise at this point: 1. What conditions ensure the convergence of a series of this kind? 2. What conditions must a function meet to be the sum of a trigonometric series, and how can the coefficients be determined? Let us begin by answering the final part of the second question, remarking that if the series converges, its sum is a periodic function with period .2π. Suppose that the trigonometric series converges and its sum is a function f , that is, ∞  a0  + an cos(nx) + bn sin(nx) . .f (x) = (3.4) 2 n=1 The objective is to find the coefficients .a0 , .an , and .bn , .n ∈ N. Suppose the function f is integrable and the series involved in this process is integrable term by term. By integrating the previous equality from .−π to .π, we obtain  π  π  π ∞   a0 dx + an cos(nx) + bn sin(nx) dx f (x) dx = −π −π 2 −π n=1  ∞ π   an cos(nx) + bn sin(nx) dx = πa0 + . = πa0 +

n=1 −π ∞

an

n=1



But .



and .

π





π −π



cos(nx) dx + bn

1 sin(nx) cos(nx) dx = n −π

π

sin(nx) dx .

−π

π =0

(3.5)

−π

π  1 sin(nx) dx = − cos(nx) = 0; n −π −π π

(3.6)

3.5. Introduction to Fourier Series therefore,



203

π

.

−π

f (x) dx = πa0 .

Solving for .a0 , we get a =

. 0

1 π



π

f (x) dx. −π

To calculate .an , for .n ≥ 1, we multiply both sides of equality (3.4) by .cos(mx) (.m ∈ N) and integrate term by term between .−π and .π:  π f (x) cos(mx) dx = −π



∞  a0  + an cos(nx) + bn sin(nx) cos(mx) dx 2 −π n=1  π ∞  π   a0 an cos(nx) + bn sin(nx) cos(mx) dx cos(mx) dx + .= 2 −π n=1 −π  π a0 cos(mx) dx+ = 2 −π

 π  π ∞ cos(nx) cos(mx) dx + bn sin(nx) cos(mx) dx . + an 

π

=

−π

n=1

−π

From the trigonometric equalities .

    1 cos (n + m)x + cos (n − m)x 2     1 sin (n + m)x + sin (n − m)x , sin(nx) cos(mx) = 2

cos(nx) cos(mx) =

it follows





π

cos(nx) cos(mx) dx =

.

−π

and



0,

if n = m

π,

if n = m

(3.7) (3.8)

(3.9)

π

sin(nx) cos(mx) dx = 0.

.

−π

From the previous equalities, we obtain  π . f (x) cos(mx) dx = am π. −π

(3.10)

204

Series of Functions Solving for .am and considering that .m = n, 1 .an = π



π

f (x) cos(nx) dx. −π

To evaluate .bn , for .n ≥ 1, we multiply both sides of equality (3.4) by .sin(mx) (.m ∈ N) and integrate term by term between .−π and .π: 

π

f (x) sin(mx) dx = −π

 ∞  a0  + = an cos(nx) + bn sin(nx) sin(mx) dx 2 −π n=1  ∞  π   a0 π an cos(nx) + bn sin(nx) sin(mx) dx sin(mx) dx + .= 2 −π n=1 −π  π a0 sin(mx) dx+ = 2 −π

 π  π ∞ cos(nx) sin(mx) dx + bn sin(nx) sin(mx) dx . + an 

π



−π

n=1

−π

From the trigonometric equality .

sin(nx) sin(mx) =

    1 cos (n − m)x − cos (n + m)x , 2

(3.11)

we conclude that 



π

sin(nx) sin(mx) dx =

.

−π

and we obtain



0,

if n = m

π,

if n = m,

π

.

−π

f (x) sin(mx) dx = bm π.

Solving for .bm and taking into account that .m = n, 1 .bn = π



π

f (x) sin(nx) dx. −π

(3.12)

3.5. Introduction to Fourier Series

205

Definition 3.5.3 Let f be a real-valued function of a real variable, periodic with period .2π, for which the coefficients .a0 , .an , and .bn , .n ∈ N, can be calculated as follows:  1 π a0 = f (x) dx π −π  π 1 f (x) cos(nx) dx . an = π −π  π 1 f (x) sin(nx) dx, bn = π −π and define the trigonometric series. This series is called the Fourier series of f . The coefficients are referred to as Fourier coefficients of f . Notes: 1. We will use the notation ∞

f (x) ∼

.

 a0  + an cos(nx) + bn sin(nx) 2 n=1

to mean that the series is the Fourier series of f , but we still do not know if it converges to f . 2. The Fourier series is well determined by the values of the coefficients .a0 , .an , .bn , .n ∈ N, calculated earlier. If we change the value of the function f at a finite number of points, the integrals that define the coefficients do not change. In particular, the value of the function at isolated points, including the endpoints of the interval, does not matter. As a result, we can define the function on the closed interval, the open interval, or the interval that is closed at one endpoint and open at the other. We selected the interval .[−π, π] for convenience. In fact, as the trigonometric functions involved are periodic with a period of .2π, any interval of amplitude .2π can be used. It is essential to keep in mind Proposition 3.5.3, which can be easily proved by a change of variables. Proposition 3.5.3 Let .f : R → R be a periodic function with period .2π. Then, for every a, .b ∈ R, 



a+2π

b+2π

f (x) dx =

.

a

f (x) dx. b

Joseph Fourier (1768– 1830) was a mathematician and physicist born in Auxerre, France. He ´ studied at the Ecole Royale Militaire of the Benedictine Order, where he became a professor. During 1795, Fourier was a student of Lagrange, Laplace, and Monge at ´ the Ecole Normal in Paris, and in that same year, he became a professor at ´ the Ecole Polytechnique. In 1797, he succeeded Lagrange in the chair of Analysis and Mechanics. His most well-known work is Th´ eorie analytique de la chaleur, published in 1822. Due to his research in partial differential equations, Fourier stated that “any” function could be represented by a series of elementary trigonometric functions—sines and cosines. This statement was so astonishing that scientists of his time did not believe it. (Source of image: Engraving by Am´ ed´ ee F´ elix Barth´ elemy Geille (1803–1843) after Julien-L´ eopold Boilly (1796–1874))

206

Series of Functions In the case where f is explicitly defined on the interval .[0, 2π] instead of [−π, π], it may be more convenient to compute the coefficients using the following formulas (see Example 3.5.6):

.

1 a0 = π .

1 an = π bn =

1 π



2π

f (x) dx 0



2π

f (x) cos(nx) dx 0



2π

f (x) sin(nx) dx. 0

A problem that often appears in applications is to develop a function f in a Fourier series with the function only defined on the interval .[−π, π] and without reference to the periodicity of f . As the formulas to compute the coefficients involve only the interval .[−π, π], we can calculate the Fourier series of the function. The trigonometric functions intervening in the Fourier series are periodic with period .2π. If the series is convergent, the sum function, g, will also be periodic with period .2π: g(x + 2π) = g(x), ∀x ∈ R.

.

When calculating the Fourier series of a function defined on .] − π, π], we are, in fact, determining the Fourier series of its periodic extension, whose definition we introduce next.

Proposition 3.5.4 Let .f : ]−π, π] → R. There exists a unique function .f˜ periodic with period .2π, called the periodic extension of f , that satisfies .f˜(x) = f (x), for all .x ∈ ] − π, π].

Given the expression of .f (x), the function .f˜ is well defined by f˜(x) = f (x), ∀x ∈ ] − π, π] .

f˜(x + 2π) = f˜(x), ∀x ∈ R.

In some situations, specifying the analytical expression of .f˜ may be useful.

3.5. Introduction to Fourier Series

Example 3.5.1 The periodic extension of .f (x) = x, defined on .] − π, π], is the function .f˜ represented in Fig. 3.9.

Figure 3.9: Periodic extension of .f (x) = x, .x ∈ ] − π, π]

The analytical expression of .f˜ is f˜(x) = x − 2kπ, x ∈ ](2k − 1)π, (2k + 1)π], k ∈ Z.

.

Example 3.5.2 The periodic extension of .f (x) = x2 , defined on the interval .] − π, π], is the function .f˜ represented in Fig. 3.10.

Figure 3.10: Periodic extension of .f (x) = x2 , .x ∈ ] − π, π]

Its analytical expression is f˜(x) = (x − 2kπ)2 , x ∈ ](2k − 1)π, (2k + 1)π], k ∈ Z.

.

Note: If .f (−π + ) = f (π) and assuming that f is continuous on .] − π, π], its extension by periodicity is a continuous function on .R. If .f (−π + ) = f (π), the periodic extension is not continuous.

207

208

Series of Functions Before we study the properties of the Fourier series, let us examine some examples of how these series are calculated. Example 3.5.3 Consider the real-valued function of a real variable ⎧ ⎨1, if − π ≤ x < 0 .f (x) = ⎩x, if 0 ≤ x ≤ π. The graph of f is presented in Fig. 3.11. The Fourier coefficients of f can be calculated using integration by parts.

 0  π 1 f (x) dx = 1 dx + x dx π −π −π 0   0  2 π  x 1 π = + =1+ ; x π 2 2 −π 0

1 a0 = π

Figure 3.11: The graph of the function of Example 3.5.3

1 an = π



π





0



π

cos(nx) dx +

x cos(nx) dx

−π

0

  0 π  π sin(nx) sin(nx) sin(nx) dx + x − n n n 0 −π 0 π

 1 cos(nx) 1 cos(nπ) cos(0) (−1)n − 1 = = − ; = π n2 π n2 n2 πn2 0 1 = π

.

bn

1 = π 1 = π







0

sin(nx) dx + −π





π

x sin(nx) dx 0

0



π



π

cos(nx) dx + n 0 0  π

sin(nx) 1 cos(−nπ) cos(nπ) 1 = −π + − + π n n n n2 0

n n n (−1) π (−1) 1 1 (1 − π)(−1) − 1 − . = − + = π n n n πn cos(nx) − n

cos(nx) + −x n −π



The Fourier series of f is

∞ 1 π (−1)n − 1 (1 − π)(−1)n − 1 + + sin(nx) . . cos(nx) + 2 4 n=1 π n2 πn

3.5. Introduction to Fourier Series

209

With this example, it is easy to verify that the sum of the series is not equal to the function at all points of the interval .[−π, π]. In fact, at .x = −π and at .x = π, we obtain the numerical series ∞ 1 π 2 + + . , 2 4 n=1 π (2n − 1)2 and .f (−π) = 1 = f (π) = π. Example 3.5.4 Let us consider the function .f (x) = x defined on .[−π, π] (Fig. 3.12). The Fourier coefficients of f can be computed using integration by parts and equalities (3.5) and (3.6).  π  1 π 1 x2 a0 = x dx = = 0; π −π π 2 −π   π  π  sin(nx) 1 π 1 sin(nx) dx = 0; an = x cos(nx) dx = − x π −π π n n −π −π .   π  π  cos(nx) cos(nx) 1 π 1 bn = −x dx x sin(nx) dx = + π −π π n n −π −π

2 cos(−n π) 1 cos(nπ) −π = −π = (−1)n+1 . π n n n The Fourier series of f is .

∞

n=1

(−1)n+1

Figure 3.12: The graph of the function of Example 3.5.4

2 sin(nx). n

Example 3.5.5 Let us consider the function .f (x) = x2 defined on .[−π, π] (Fig. 3.13). Using the calculations from the previous example, we have  1 π 2 2π 2 ; a0 = x dx = π −π 3   π  π  1 π 2 1 sin(nx) 2 sin(nx) . an = x dx x cos(nx) dx = − 2x π −π π n n −π −π 4 = (−1)n 2 ; n  1 π 2 bn = x sin(nx) dx π −π   . π  π cos(nx) 1 2 cos(nx) dx = 0. + 2x = −x π n n −π −π

Figure 3.13: The graph of the function of Example 3.5.5

210

Series of Functions Thus, the Fourier series of f is ∞

.

π2 4 + (−1)n 2 cos(nx). 3 n n=1

In the following example, we have a periodic function with period .2π, defined on .]0, 2π[ (see Proposition 3.5.3).

Example 3.5.6 Consider the real-valued function of a real variable

f (x) =

 sin(x),

if

0 1. Let fn n∈N be a sequence of functions such that

22. Let

n=0

.|fn (x)|

≤ |bn |, ∀x ∈ R.

What can be said about the uniform convergence of ∞  fn ? the series of functions n=0

16. Find the values of x for which the series  ∞  x2n x2 is convergent. Justify the √ log 1 + √ n n n=1 answer. 17. Compute the radius of convergence of the series ∞  π n  x2n and indicate the + 2 − (−1)n cos n n=1 largest open interval on which the series converges. ∞ 

4n xn . Known + 5n 2 n=1 ing that the radius of convergence of this series is 5 , indicate, without resorting to additional calcu4 lations, whether the series converges or diverges at points 1 and 2. Investigate the convergence of the 5 series at the point . 4

18. Consider the power series

19. Find the real values of x for which the series ∞  an xn is absolutely convergent, conditionally n=0

23. Let (an ) be a sequence of positive terms such that ∞  the series an (x − 1)n is conditionally convern=1

gent at x = −1. a) What is the radius of convergence of the power ∞  series an (x − 1)n ? Justify. n=1

b) Show that x = −1 is the only point at which the series is conditionally convergent. 24. Consider the series

n=1

=

b) Let f be the sum of the series. Find the set of points where f is differentiable and determine the analytical expression of f  . Justify the answer given.

Justify the answer.

25. Consider the series 1,

if n is even

2,

if n is odd.

1 (x − 2)n . n 5n

a) Find the real values of x for which the series is absolutely convergent, conditionally convergent, and divergent. Justify.

convergent, and divergent, where

.an

∞ 

∞  n+1 (−1)n  . log(x) n + 1 n=0

a) Find the real values of x for which the series is absolutely convergent, conditionally convergent, and divergent. Justify.

232

Series of Functions a)

26. Consider the real-valued function f of a real variable, which is defined by

.f (x)

=

n=0

1

0

(−1)n x2n . 22n (n!)2

 d)

b) Show that x2 f  (x) + xf  (x) + x2 f (x) = 0. Justify the answer given.

30.

a) Test for convergence the series

∞ 

an and

an 3n . Give a reason for the answer.

n=1

sin(an ) . Justify. an

28. Let f be a real-valued function of a real variable, defined by .f (x)

=

∞ 

cos(t2 ) dt

x + x2 (1 − x)2

i) x arctan(x) j)

x

arctan(t) dt

2x (1 + x2 )2

k) log(2 + x3 )  1+x l) log 1−x

e)

1 (2x + 5)2

f)

2 + e2x 3 + 2x

m) log(4 − x2 )

a) Determine Maclaurin series of the function .f (x)

=

x . (1 − x)2

Indicate for which values of x the development is valid. Justify the answer.

n=1

b) Calculate lim

h)

1 − e−tx dt t

27. Let (an ) be a sequence of real numbers such that ∞  the series an 2n is conditionally convergent. n=1

x

0

0

a) Determine the domain of f .

∞ 

g)

0

 c)

∞ 



5x − 1 x2 − x − 2  x 2 et dt b)

b) Using the series of derivatives, calculate, for each x, the sum of the series.

an xn ,

Hint: Use term by term differentiation. b) Using item a), calculate the sum of the series ∞  n . n 3 n=0 31. Let f be a real-valued function of a real variable defined by ⎧ ⎨ sin(x) , if x = 0 x .f (x) = ⎩ 1, if x = 0.

n=0

such that f  (x) = f (x), ∀x ∈ R, and f  (0) = 1. a) For each n ∈ N, compute an . Hint: Use the method of Mathematical Induction. b) Using item a), identify the function f . 29. Develop the following functions in a Maclaurin series and indicate the interval on which the expansion is valid. Explain the findings for each question and provide a reason for the answers.

a) Expand the function f in a Maclaurin series. Indicate for which values of x the development is valid. Explain the reasoning behind the response. b) Use item a) to obtain the value of f (n) (0), for all n ∈ N. 32. Consider the real-valued function f of a real variable defined by .f (x)

2

= ex + cos(2x).

3.6. Solved Exercises

233

a) Compute the Maclaurin series of f , indicating the values of x for which the development is valid. b) Using the expansion obtained in item a), indicate the value of f (17) (0). 33. Using only the Maclaurin expansions of the functions 1 , find f (x) = sin(x), g(x) = ex , and h(x) = 1−x the sum of the following series: ∞ 

n=1

37. Expand in a power series of x − 3 the real-valued function f of a real variable defined by =

1 . x2

Determine the values of x for which the development is valid. Provide a clear explanation of the answer.

38. Develop in a power series of x + 1 the real-valued function of a real variable f defined by

1 2n n

34. Consider the function f : R → R defined by .f (x)

.f (x)

b) Using item a) compute the sum of the series ∞ 

(−1)n−1

n=1

π 2n 42n



22n 1 − n (2n)!

 .

35. Consider the function f : R → R defined by 2x .f (x) = sin(2x) + . (1 − x2 )2

b) Use item a) to calculate the sum of the series

. n=0



n+1 (−1)n + 2n (2n + 1)! 2

1 . x2 + x − 6

39. Consider the function  x log(t) − t + 1 dt. .f (x) = t−1 1 Find the power series of x − 1 of f , indicating the values of x for which the development is valid. 40. Consider the function  x arctan(t − 2) − t + 2 .f (x) = dt. (t − 2)2 2 Expand f in a power series of x − 2, indicating the values of x for which the development is valid. 41. Let f be a function of class C ∞ in a neighborhood of the origin that satisfies the following conditions:

a) Express f in a power series of x indicating the values of x for which the development is valid.

∞ 

=

Indicate the values of x for which the expansion is valid. Justify.

= log(x2 + 1) + cos(2x).

a) Find the Maclaurin series of f and indicate the values of x for which the development is valid.

+

1 + log(4 − x). 4−x

Hint: Use term by term differentiation.

(−1)n

.1

=

Determine the values of x for which the development is valid. Give a clear explanation of the answer.

n

∞  2n b) n! n=0

c)

.f (x)

.f (x)

1 a) (−1) (2n + 1)! n=0

∞ 

36. Represent in a power series of x − 3 the real-valued function f of a real variable defined by

 .

.f (0)

= 1 and f  (x) = f (x) + x, ∀x ∈ R.

Find the Maclaurin development of f . 42. Find the Fourier series relative to the periodic functions of period 2π, defined on ] − π, π[ as follows:

π, if − π < x ≤ 0 a) 0, if 0 < x < π

234

Series of Functions

b)

c)

0,

if

−π 1. – If x = 0, then x x n n n=1 n=1 Therefore, the domain of f is the set ]1, +∞[ ∪ {0}. ∞ 

1 . To conclude that f is continuous on ]1, +∞[, it is sufficient nx to prove that g is continuous in this set since f (x) = x g(x). If we show that g is continuous on [a, +∞[, for every a > 1, then g is continuous on ]1, +∞[. For this, it is enough (see Theorem 3.2.3) that the following two conditions are met: 1 (i) The functions gn (x) = x are continuous on ]a, +∞[. n ∞  1 is uniformly convergent to g on ]a, +∞[. (ii) The series x n n=1

b) Let us consider the function g(x) =

n=1

238

Series of Functions The functions gn (x) = e−x log(n) are continuous, since the exponential is continuous on R. Let us look at the second condition. Let a > 1 be a real constant. The inequality    1  ≤ 1 , nx  na

. 

∀x ∈ [a, +∞[, ∀n ∈ N,

∞  1 is convergent, is sufficient to conclude, by Weierstrass’ test, na n=1 ∞  1 on [a, +∞[. As the series converges to g, this function is the uniform convergence of the series x n n=1 continuous on [a, +∞[. Since a is an arbitrary constant greater than 1, we also have g continuous on ]1, +∞[.

combined with the fact that the series

4. The terms of the series

∞  n=1

1 √

(nx + 1)

n3 + 1

are continuous functions on [1, 2]. As

    1 1 1 1 1 1   √ ·√ ·√ ≤ ≤ √ , =  (nx + 1) n3 + 1  |nx + 1| n+1 n3 + 1 n3 + 1 n3

.

and the series of positive terms

∞  n=1

∞ 

∀x ∈ [1, 2], ∀n ∈ N,

1 3 √ is convergent because it is a p-series with p = , by Weierstrass’ test, 2 n3

1 √ converges uniformly to f , on [1, 2]. It follows from Theorem 3.2.3 that f is the series n3 + 1 (nx + 1) n=1 continuous on [1, 2]; by Theorem 3.2.4, the series is integrable term by term on this interval and 



2

2

f (x) dx = 1

.

1

=

∞ 

n=1

√

n=1

=

∞  n=1

 2 ∞  1 1 1 √ √ dx dx = 3 + 1 1 nx + 1 (nx + 1) n3 + 1 n n=1

∞ 

n



1 n3

√

+1 1

n3 + 1

log(nx + 1) n  log

2n + 1 n+1

2 = 1

∞  log(2n + 1) − log(n + 1) √ n n3 + 1 n=1

 .

∞  cos(nx) are functions of domain R. The function f is the sum of the series; therefore, n2 n=1 its domain is the subset of R where the series converges. Since we have

5. The terms of the series

   cos(nx)  ≤ 1 , n2  n2

. 

∀x ∈ R, ∀n ∈ N,

because | cos(nx)| ≤ 1, ∀x ∈ R, ∀n ∈ N, and the series of positive terms series with p = 2, by Weierstrass’ test, the series Theorem 3.2.3, given that the functions

∞  1 is convergent as it is a p2 n n=1

∞  cos(nx) converges uniformly to f , on R. Consequently, by n2 n=1

cos(nx) are continuous on R, f is continuous on R; by Theorem 3.2.4, n2

3.6. Solved Exercises

239

 π the series is integrable term by term on every closed bounded interval of R. In particular, f is integrable on 0, , 2 and  π/2  π/2  π/2  ∞ ∞  cos(nx) 1 f (x) dx = dx = cos(nx) dx n2 n2 0 0 0 n=1 n=1 .

=

 π ∞ ∞ ∞ ∞     sin( nπ ) 1 sin(nx) 2 (−1)n−1 (−1)n 2 = = = , 2 3 3 n n n (2n − 1) (2n + 1)3 0 n=1 n=1 n=1 n=0

because . sin

6. The terms of the series f (x) =

 nπ 2

⎧ ⎪ ⎨0, = −1, ⎪ ⎩ 1,

if n is even if n = 3, 7, 11, . . . if n = 1, 5, 9, . . .

∞  e−nx are functions of domain R. As n4 + 5 n=0

 −nx   e  1 1 ≤ ≤ 4, 4 n + 5 n4 + 5 n

. 

∀x ∈ [0, 1], ∀n ∈ N0 ,

since |e−nx | = e−nx ≤ e0 = 1, ∀x ∈ [0, 1], ∀n ∈ N0 , and the series of positive terms as it is a p-series with p = 4. By Weierstrass’ test, the series

∞  1 is convergent 4 n n=1

∞  e−nx converges uniformly to f , on [0, 1]. 4+5 n n=0

e−nx are continuous on [0, 1], f is continuous on n4 + 5 [0, 1]. Theorem 3.2.4 shows that the series is integrable term by term on this interval. We have Therefore, by Theorem 3.2.3 since the functions fn (x) =



  1 ∞ ∞   e−nx 1 1 1 + dx = + e−nx dx 4 4 5 n=1 n + 5 5 n=1 n + 5 0 0  −nx 1 ∞ ∞   1 1 − e−n e 1 1 + − . = = + 4 5 n=1 n + 5 n 5 n=1 n (n4 + 5) 0 

1

1



f (x) dx = 0 .

7.

a) The series of functions defined, for x ≥ 0, by ∞   .

x+

n=1

1 n

n+ x

n

is of positive terms. Let us study this series by applying Cauchy’s Root Test:  lim .

n





1 x+ n

1 = lim x + n





n+ x

n

= lim

  n

x+

1 n

x

n

n

  x  1 n 1 n x+ x+ n n

   x 1 1 n2 = lim x + = x. x+ n n

240

Series of Functions By Cauchy’s Root Test, the series converges if x < 1 and diverges if x > 1. It remains to see what happens  1 ∞   1 n+ n if x = 1. In this case, we obtain the numerical series , which is divergent because its 1+ n n=1 general term is not a null sequence. In fact, . lim

   1   1 1 n+ n 1 n 1 n 1+ = lim 1 + = e. 1+ n n n

b) Let 0 < α < 1 and x ∈ [0, α].    x   x  α  1 n+ n  1 n+ n 1 n+ n  . x + ≤ α+ , ∀n ∈ N. = x+   n n n

As shown in item a), the numerical series

 α ∞   1 n+ n is convergent. By Weierstrass’ test, the original α+ n n=1

series converges uniformly on [0, α]. 8. We will use Weierstrass’ test to prove that the series of functions

∞ 

an fn (x) converges uniformly on R. Since

n=1

|fn (x)| ≤

1 , ∀n ∈ N, ∀ x ∈ R, and (an ) is a sequence of positive terms, we have n .|an fn (x)|

We prove that the series of positive terms

≤

an , ∀n ∈ N, ∀ x ∈ R. n

∞  an is convergent by applying D’Alembert’s test: n n=1

an+1 an+1 n 1 n an+1 n+1 · . lim a = lim = = lim n (n + 1) an n+1 an 2 n because, by hypothesis, lim the series of functions

∞ 

an+1 1 = . As the limit is less than 1, the series is convergent. By Weierstrass’ test, an 2

an fn (x) converges uniformly on R.

n=1

9. Since the functions fn : [0, 1] → R are continuous, if the convergence was uniform, we would have, by Theorem 3.2.4,  1 ∞ ∞  1  . fn (x) dx = fn (x) dx. 0 n=1

But



∞ 1 

.

n=1



and

∞   . n=1

0

1

fn (x) dx =

0 n=1 1

fn (x) dx =

0

f (x) dx = 0 0

∞   1 n=1

n

−

1 n+1

Therefore, the series of functions is not uniformly convergent on [0, 1].

 = 1.

3.6. Solved Exercises 10.

241

∞  arccot(nx) √ are functions with domain R. The function f is the sum of the n5 n=1 series; therefore, its domain is the subset of R where the series converges. The terms of the series satisfy

a) The terms of the series

   arccot(nx)   ≤ √π , √  n5 n5

. 

∀x ∈ R, ∀n ∈ N,

since |arccot(nx)| ≤ π, ∀x ∈ R. The series of positive terms

∞  n=1

p-series with p =  ∞    arccot(nx)    √   n5 n=1 domain of f is R.

(3.14)

1 √ is convergent because it is a n5

∞  π 5 √ . Therefore, the series is also convergent. By inequality (3.14), the series 2 n5 n=1 ∞  arccot(nx) √ converges, ∀x ∈ R. Then the series is convergent, ∀x ∈ R, that is, the n5 n=1

b) To conclude that f is continuous on R, it is sufficient, according to Theorem 3.2.3, that the following two conditions are satisfied: (i) The functions fn (x) = (ii) The series

arccot(nx) √ are continuous on R. n5

∞  arccot(nx) √ is uniformly convergent to f on R. n5 n=1

The functions hn (x) = arccot(nx) are continuous on R, which implies the first condition. The second condi∞  π √ tion is a consequence of the previous item: Inequality (3.14) and the fact that the series is convern5 n=1 ∞  arccot(nx) √ . gent are sufficient, by Weierstrass’ test, to conclude the uniform convergence of the series n5 n=1 As the series converges to f , this function is continuous. c) To conclude that f is differentiable on R, it is sufficient, by Corollary 2 of Theorem 3.2.4, that the following conditions are fulfilled: (i) The functions fn (x) are continuous on R. ∞  arccot(nx) √ is pointwise convergent to f on R. n5 n=1  ∞  ∞   n arccot(nx)  √ √ = − is uniformly convergent on R. (iii) The series n5 (1 + n2 x2 ) n5 n=1 n=1

(ii) The series

  The first condition is satisfied because the functions arccot(nx) = −

n are continuous on R. We 1 + n 2 x2 proved the second condition in the previous item. The third is again an application of Weierstrass’ test. We have the inequality     n n 1   √ ≤ √ . − = √ , ∀x ∈ R, ∀n ∈ N,  (1 + n2 x2 ) n5  n5 n3

 ∞   arccot(nx)  1 3 √ √ is convergent because it is a p-series with p = , so the series 2 n3 n5 n=1 n=1 is uniformly convergent. We can conclude that the series converges to f  , that is, and the series

∞ 

242

Series of Functions .f



(x) =

 ∞  ∞   arccot(nx)  n √ , √ = − 5 2 x2 ) n 5 n (1 + n n=1 n=1

which means that f is differentiable on R. 11. If the series

∞ 

an cos(nx) converges uniformly on R, then it converges pointwise on R. In particular, the series

n=0

is convergent at x = 0. At this point, we obtain the numerical series

∞ 

an , which is convergent.

n=0 ∞ 

Conversely, suppose that the series

an is convergent. We have

n=0 . |an

By Weierstrass’ test the series

∞ 

cos(nx)| ≤ |an | = an ,

∀x ∈ R, ∀n ∈ N.

an cos(nx) is uniformly convergent on R.

n=0

12. We know that the series

+∞ 

(bn − bn+1 ) converges, that is, the sequence Sn =

n=1

n 

(bk − bk+1 ) = b1 − bn+1

k=1

converges. Therefore, there exists b = lim bn = b1 − lim Sn , so the sequence (bn ) is bounded, that is, there exists M > 0 such that |bn | ≤ M, ∀n ∈ N. Based on this fact, we have the following inequalities: .|bn fn (x)|

= |bn | |fn (x)| ≤ |bn | an ≤ M an , ∀x ∈ R, ∀n ∈ N.

Thus, by Weierstrass’ test, the series R. Since, by hypothesis, the 13.

a) Consider the series

∞ 

bn fn (x) is uniformly convergent n=1 functions fn are continuous on R, ∀n ∈ N, the

∞ 

on R, defining a function f of domain function f is continuous on its domain.

(−1)n (1 − x) xn , x ∈ [0, 1].

n=0

– If x = 1, all terms are zero. Therefore the series converges. ∞  (1 − x) (−x)n is geometric with ratio r = −x, it converges because x ∈ [0, 1[. – If x = 1, as the series n=0

In this case,

∞  .

(1 − x) (−x)n = (1 − x) ·

n=0

Then the series

∞ 

1 1−x = . 1 − (−x) 1+x

(−1)n (1 − x) xn converges pointwise to the function f (x) =

n=0

convergence will be uniform if .

lim

sup |f (x) − Sn (x)| = 0,

n→+∞ x∈[0,1]

where Sn denotes the sequence of partial sums of the original series. – If x = 1, |f (1) − Sn (1)| = 0.

1−x on [0, 1]. This 1+x

3.6. Solved Exercises

243

– If x = 1, Sn (x) = (1 − x) ·

1−x 1 − (−x)n+1 = · (1 − (−x)n+1 ), and therefore, 1 − (−x) 1+x

 1 − x |f (x) − Sn (x)| =  1+x   1−x . =  1+x  1 − x =  1+x

  1−x  · 1 − (−x)n+1  1+x      · 1 − 1 − (−x)n+1    1−x · (−x)n+1  = · xn+1 . 1+x −

1−x · xn+1 . 1+x The function g is continuous on [0, 1]; therefore, by Weierstrass’ Theorem, it has a maximum and a minimum on this interval. As g is nonnegative and g(0) = g(1) = 0, the maximum is attained at some point within the interior of the interval. The function g is differentiable on ]0, 1[, so this point corresponds to a zero of the first derivative.     1−x xn  · xn+1 = − .g (x) = · (n + 1)x2 + 2x − (n + 1) , 2 1+x (1 + x)

Let g(x) = |f (x) − Sn (x)| =

and on the interval ]0, 1[, .g



(x) = 0 ⇔ (n + 1)x2 + 2x − (n + 1) = 0 ⇔ x =

−1 +

Then  .

−1 +

sup |f (x) − Sn (x)| = g x∈[0,1]



1 + (n + n+1

and, as lim

n→+∞

−1 +



1 + (n + 1)2 n+1  , ≤ −1 + 1 + (n + 1)2 1+ n+1 1−



1 + (n + 1)2 n+1  = 0, −1 + 1 + (n + 1)2 1+ n+1 1−

.

−1 +

1)2



 1 + (n + 1)2 . n+1

we can conclude that .

that is, the series

∞ 

lim

sup |f (x) − Sn (x)| = 0,

n→+∞ x∈[0,1]

(−1)n (1 − x) xn converges uniformly to the function f , on the interval [0, 1].

n=0

b) Let us consider the series

∞ 

(1 − x) xn , x ∈ [0, 1].

n=0

– If x = 1, all terms are zero. Hence, the series converges. ∞  – If x = 1, as the series (1 − x) xn is geometric with ratio r = x, the series converges because n=0

x ∈ [0, 1[. In this case,

∞  . n=0

(1 − x) xn = (1 − x) ·

1 = 1. 1−x

244

Series of Functions Then, the series

∞ 

(1 − x) xn , x ∈ [0, 1], converges pointwise to the function

n=0

.f (x)

=

0, 1,

if x = 1 if x ∈ [0, 1[.

Let fn (x) = (1 − x)xn . The functions fn are continuous on [0, 1], ∀n ∈ N0 , because they are polynomial functions. If the series was uniformly convergent on the interval [0, 1], f would be continuous on this interval. ∞  Since f is not continuous at x = 1 because lim f (x) = 1 and f (1) = 0, the series (1 − x) xn does x→1−

n=0

not converge uniformly to f on [0, 1]. 14.

∞ 

π n xn πn . Setting an = , the radius of convergence of 2 n log (n) n log2 (n) n=2 this series is, by Corollary 1 of Theorem 3.3.1     πn       2  an  n + 1 log2 (n + 1) 1 n log (n)    = lim  .r = lim  · = ,  = lim  n+1 2   π an+1 π n π log (n)    (n + 1) log2 (n + 1) 

a) Let us study the power series of x,

  1 1 which implies, by the same theorem, that the series is absolutely convergent if x ∈ − , and divergent π π     1 1 , +∞ . if x ∈ −∞, − ∪ π π 1 At x = ± , we have to study the corresponding numerical series. π  n ∞ ∞   1 πn 1 1 – If x = , we obtain the series = . Let us study the convergence of 2 2 π π n log (n) n log (n) n=2 n=2 this series using the Integral Test. 1 . We have: x log2 (x) (i) f (x) > 0, ∀x ≥ 2. (ii) f is continuous on [2, +∞[. (iii) f is decreasing on [2, +∞[ because x log2 (x) is positive and increasing.  +∞ ∞  1 1 So the numerical series converges if and only if the improper integral dx 2 2 n log (n) x log (x) 2 n=2 converges. We have Let f (x) =

 .

+∞ 2

 x −2 1 1 dt log(t) dt = lim x→+∞ 2 t log2 (t) x→+∞ 2 t x    1 1 1 1 = − − = lim = lim + , x→+∞ x→+∞ log(t) 2 log(x) log(2) log(2)

1 dx = x log2 (x)



x

lim

that is, the improper integral is convergent, so the same happens to the series. As it is a series of positive terms, it converges absolutely.

3.6. Solved Exercises

245

  ∞ ∞   1 n πn 1 1 − , which is alternating. – If x = − , we obtain the series = (−1)n 2 2 π π n log (n) n log (n) n=2 n=2 As we just saw, the series of modules is convergent. Then the series is absolutely convergent.     1 1 ∪ , +∞ and is absolutely convergent if Conclusion: The original series diverges if x ∈ −∞, − π π   1 1 x∈ − , . π π Remark: To calculate the lim

log2 (n + 1) = log2 (n)

 lim

log(n + 1) log(n)

2 . we proceed as follows: Consider the func-

log(x + 1) tion g(x) = ; we use L’Hˆ opital’s Rule applied to the calculation of the limit log(x) ∞ arises: where the indeterminate form of type ∞  .

lim

x→+∞

log(x + 1)   log(x)

 =

lim

x→+∞

lim

x→+∞

log(x + 1) , log(x)

1 x x+1 = 1. = lim x→+∞ x + 1 1 x

log(x + 1) log2 (n + 1) = 1, which implies that lim = 1. log(x) log2 (n) ∞  2n nn 2n nn xn . Setting an = , the radius of converb) Let us study the power series of x, n+1 (n + 1) (n + 1)n+1 n=1 gence is by Corollary 1 of Theorem 3.3.1     2n nn       n+1  an  2n nn (n + 2)n+2 (n + 1)     = lim  n+1 r = lim   = lim  2 (n + 1)n+1  an+1 (n + 1)n+1 2n+1 (n + 1)n+1   .   (n + 2)n+2  n   n 1 1 n + 2 n+2 1 1 = lim · = · ·e= . 2 n+1 n+1 2 e 2       1 1 1 1 , +∞ . Then, the series is absolutely convergent if x ∈ − , and divergent if x ∈ −∞, − ∪ 2 2 2 2 Therefore,

lim

x→+∞

Let us study the convergence of the numerical series corresponding to the endpoints of the interval of convergence.  n ∞ ∞   2n nn nn 1 1 , we obtain the series of positive terms · = . n+1 2 (n + 1) 2 (n + 1)n+1 n=1 n=1 We will study this series by comparing it with the harmonic series, which is divergent. The limit

– If x =

nn n+1 n   1 n n n (n + 1)n+1 = . lim = lim · = lim 1 n+1 n+1 n+1 e n is finite and different from zero; therefore, by Corollary 2 of the General Comparison Test, the series is divergent.

246

Series of Functions   ∞ ∞   2n nn nn 1 1 n – If x = − , we obtain the series · − = (−1)n , which is altern+1 2 (n + 1) 2 (n + 1)n+1 n=1 n=1 n n nating, since > 0, ∀n ∈ N. We proved in the previous case that the series of modules is (n + 1)n+1 divergent. Therefore, we need to verify whether the following conditions hold to apply Leibniz’s test: n  1 nn n 1 = × 0 = 0. (i) lim = lim · (n + 1)n+1 n+1 n+1 e nn (ii) > 0, ∀n ≥ 1. (n + 1)n+1  n n nn 1 (iii) The sequence of general term is decreasing, as it is the product = · n+1 (n + 1) n+1 n+1 of two decreasing positive sequences. Therefore, the series is convergent. It is conditionally convergent since the series of modules diverges.  Conclusion: The power series of x is absolutely convergent if x ∈     1 1 1 x = − , and divergent if x ∈ −∞, − ∪ , +∞ . 2 2 2 c) By Corollary 1 of Theorem 3.3.1, the power series of x,

∞ 

 1 1 , conditionally convergent if − , 2 2

an xn , with an =

n=1

convergence   3n log(n)      7n n2  = lim    3n+1 log(n + 1) n+1   7n+1 (n + 1)2

  an .r = lim  a

 Then, the series converges absolutely if x ∈

7 7 − , 3 3

3n log(n) , has a radius of 7n n2

    7(n + 1)2 log(n) 7  = .  = lim  3n2 log(n + 1) 3  



 and diverges if x ∈

−∞, −

7 3



 ∪

 7 , +∞ . 3

At the endpoints of the convergence interval, we obtain two numerical series.   ∞ ∞   7 7 n 3n log(n) log(n) – If x = − , we have the series · − = (−1)n which is alternating. Let us n 2 3 7 n 3 n2 n=1 n=1 ∞ ∞   log(n) 1 , by comparison with the series , which is start by studying the series of modules, 2 3/2 n n n=1 n=1 3 convergent because it is a p-series with p = . Since 2 log(n) n2 = lim log(n) = 0, . lim 1 n1/2 n3/2 by Corollary 3 of the General Comparison Test, the series of modules of

∞  n=1

(−1)n

∞  log(n) is convergent. As it is the series n2 n=1

log(n) , this series is absolutely convergent. n2

3.6. Solved Exercises

247 ∞

 log(n) 7 , we obtain the series , which we proved in the previous case to be convergent. As 3 n2 n=1 it is of positive terms, it is absolutely convergent.       7 7 7 7 , +∞ . Conclusion: The series converges absolutely if x ∈ − , and diverges if x ∈ −∞, − ∪ 3 3 3 3 – If x =

∞  2 × 5 × · · · × (3n − 1) n 2 × 5 × · · · × (3n − 1) x . Setting an = , (n + 2)! (n + 2)! n=1 the radius of convergence is by Corollary 1 of Theorem 3.3.1

d) Let us study the power series of x,

 2 × 5 × · · · × (3n − 1)      (n + 2)!  = lim   2 × 5 × · · · × (3n − 1)(3n + 2)  n+1   (n + 3)!

  an .r = lim  a

     = lim n + 3 = 1 .  3n + 2 3  



     1 1 1 1 − , and diverges if x ∈ −∞, − ∪ , +∞ . 3 3 3 3 Let us study the behavior of the numerical series at the endpoints of the convergence interval. Then, the series converges absolutely if x ∈

1 – If x = − , we obtain the alternating series 3 .

  ∞ ∞   2 × 5 × · · · × (3n − 1) 2 × 5 × · · · × (3n − 1) 1 n = (−1)n · − . (n + 2)! 3 3n (n + 2)! n=1 n=1

Let us begin by studying the series of modules

.

∞ ∞   2 × 5 × · · · × (3n − 1) = bn , 3n (n + 2)! n=1 n=1

using Raabe’s test, ⎛ ⎞ 2 × 5 × · · · × (3n − 1)  n ⎜ ⎟ bn 3 (n + 2)! ⎟ lim n − 1 = lim n ⎜ ⎝ 2 × 5 × · · · × (3n − 1)(3n + 2) − 1⎠ bn+1 . 3n+1 (n + 3)!   3n + 9 7n 7 = lim n − 1 = lim = > 1, 3n + 2 3n + 2 3 

showing the series is convergent. Then,

∞ 

(−1)n bn is absolutely convergent.

n=1 ∞  1 2 × 5 × · · · × (3n − 1) , we obtain , which is the series of modules of the previous series 3 3n (n + 2)! n=1 that we have seen to be convergent. As it is of positive terms, it is absolutely convergent.

– If x =

      1 1 1 1 and diverges if x ∈ −∞, − ∪ , +∞ . Conclusion: The series converges absolutely if x ∈ − , 3 3 3 3

248

Series of Functions ∞ ∞   log(n) n log(n) n log(n) x = x . Setting an = , the radius of n n n n=1 n=2 convergence of this series is by Corollary 1 of Theorem 3.3.1    log(n)        an   n  = lim   = lim n + 1 · log(n) = 1; .r = lim     log(n + 1) an+1 n log(n + 1)     n+1

e) Let us study the power series of x,

therefore, the series converges absolutely if x ∈ ] − 1, 1[ and diverges if x ∈ ] − ∞, −1[ ∪ ]1, +∞[. Regarding the endpoints of the convergence interval, we need to study the respective numerical series. – If x = −1, we obtain the alternating series

∞ 

(−1)n

n=1

log(n) . Let us start by studying the series of n

∞  log(n) , using the Integral Test. modules, n n=1 log(x) Let f (x) = , x ≥ 1: x (i) If x > 1, then log(x) > 0, therefore, f (x) > 0, ∀x > 1. (ii) f is continuous, ∀x ≥ 1. (iii) We can study the monotonicity of f by analyzing its derivative. We have

.f



x· (x) =

1 − log(x) 1 − log(x) x = , x2 x2

and thus, f  (x) < 0, ∀x > e; therefore, f is decreasing on ]e, +∞[. The conditions of the Integral Test are fulfilled on the interval [3, +∞[. Then the numerical series  +∞ ∞  log(n) log(x) and the improper integral dx are both convergent or both divergent. Since n x 3 n=3 

+∞ 3

.

 x log(t) 1 dt = lim log(t) dt x→+∞ 3 x→+∞ 3 t t  x log2 (t) = lim x→+∞ 2 3   2 log2 (3) log (x) − = +∞, = lim x→+∞ 2 2

log(x) dx = x



x

lim

the improper integral is divergent; the same happens to the series. Because the series of modules is divergent, let us apply Leibniz’s test to the alternating series: log(n) log(x) ∞ (i) lim = 0 (just note that lim is an indeterminate form of type that can be x→+∞ n x ∞   log(x) 1 = 0). = lim lifted using L’Hˆ opital’s Rule: lim x→+∞ x→+∞ x (x) log(n) > 0, ∀n > 1. (ii) n   log(x) log(n) (iii) We saw that the function f (x) = is decreasing on ]e, +∞[, which implies that x n is decreasing for n ≥ 3.

3.6. Solved Exercises

249

Then, the series is convergent. As the series of modules is divergent, the original series is conditionally convergent. ∞  log(n) – If x = 1, we obtain the series which, as we saw before, is divergent. n n=1 Conclusion: The power series is absolutely convergent if x ∈ ] − 1, 1[, conditionally convergent if x = −1, and divergent if x ∈ ] − ∞, −1[ ∪ [1, +∞[. Remark: The series

∞ ∞   log(n) log(n) and are both divergent because they only differ in a finite n n n=1 n=3

number of terms. ∞ 

3 n2 3 n2 xn . Setting an = n 3 , the radius of conver3 (n + 2) 4 (n + 2) n=0 gence of this series is, by Corollary 1 of Theorem 3.3.1,     3 n2         n 3  an  3n2 4n+1 (n + 1)3 + 2 4 (n + 2)   = lim  = lim r = lim     3 (n + 1)2 an+1  3 (n + 1)2 4n (n3 + 2)   .  4n+1 (n + 1)3 + 2   2 (n + 1)3 + 2 n = 4. = lim 4 n+1 n3 + 2

f) Let us study the power series of x,

4n

The series converges absolutely if x ∈ ] − 4, 4[ and diverges if x ∈ ] − ∞, −4[ ∪ ]4, +∞[. Regarding the endpoints of the convergence interval, we have to study the respective numerical series. ∞ ∞   3 n2 3 n2 – If x = 4, we obtain the series 4n = . We will study this series by comparing n 3 4 (n + 2) n3 + 2 n=0 n=0 it with the harmonic series, which is divergent. The limit 3 n2 3 n3 n3 + 2 =3 = lim 3 . lim 1 n +2 n is finite and different from zero; therefore, the series is divergent by Corollary 2 of the General Comparison Test. ∞ ∞   3 n2 3 n2 n – If x = −4, we obtain the series = (−1)n 3 (−4) , which is alternating since n (n3 + 2) 4 n +2 n=0 n=0 3 n2 > 0, ∀n ∈ N. The series of modules is the series we studied for the case x = 4, which we n3 + 2 proved to be divergent. Being an alternating series, we apply Leibniz’s test: 3 n2 (i) lim 3 = 0. n +2 2 3n > 0, ∀n ≥ 1. (ii) 3 n +2   3 n2 (iii) The sequence is decreasing, because n3 + 2 .

3 (n + 1)2 3 n2 −n4 − 2n3 − n2 + 4n + 2 − 3 =3 < 0, ∀n ≥ 2. (n + 1)3 + 2 n +2 ((n + 1)3 + 2)(n3 + 2)

250

Series of Functions Then, the series is convergent. As the series of modules is divergent, the series is conditionally convergent. Conclusion: The power series is absolutely convergent if x ∈ ] − 4, 4[, conditionally convergent if x = −4, and divergent if x ∈ ] − ∞, −4[ ∪ [4, +∞[. ∞  1 × 4 × · · · × n2 (n!)2 (n!)2 n = . By Corollary 1 of Theorem 3.3.1, the power series of x, x , (2 n)! (2 n)! (2 n)! n=1 has radius of convergence

g) Let an =

  (n!)2      = lim   (2 n)!    (n + 1)! 2 n+1   (2 n + 2)!

  an .r = lim  a

    (2n + 2)(2n + 1)  = 4.  = lim  (n + 1)2  

The series converges absolutely if x ∈ ] − 4, 4[ and diverges if x ∈ ] − ∞, −4[ ∪ ]4, +∞[. Let us study the numerical series corresponding to the endpoints of the interval. – If x = 4, we obtain the series

∞ ∞   (n!)2 n 4 = bn . Let us study this series using D’Alembert’s test1 : (2 n)! n=1 n=1

 . lim

2 (n + 1)!

4n+1 bn+1 2n + 2 (2 n + 2)! = 1. = lim = lim (n!)2 n bn 2n + 1 4 (2 n)!

Since the limit is 1, but for values greater than 1, taking into account the Note after D’Alembert’s test, we can conclude that the series is divergent. ∞ ∞   (n!)2 (n!)2 n (−4)n = 4 . The series of – If x = −4, we obtain the alternating series (−1)n (2 n)! (2 n)! n=1 n=1 ∞  (n!)2 n 4 is divergent. Since we establish the divergence of the series of modules by modules (2 n)! n=1 D’Alembert’s test, we can affirm that the alternating series is divergent (refer to the Note on page 105 for further details). Conclusion: The power series converges absolutely if x ∈ ] − 4, 4[ and diverges if x ∈ ] − ∞, −4] ∪ [4, +∞[.   n   n 1 1 xn . Let an = nn sin . The radius of sin 2 2 n n n=1 convergence of this series is, by Theorem 3.3.1,

h) Let us study the power series of x,

.r

=

∞ 

1  = lim n |an |

nn

 lim n

1 1   = +∞.   n = 1 1 lim n sin nn sin 2 n 2 n

Moreover, the series is absolutely convergent on R. 1 In

general, this test is not useful for studying the endpoints of a power series because the limit is always 1.

3.6. Solved Exercises

251 ∞ 

1 1  xn . Let an =  . The   n n arctan(n) − π n n arctan(n) − π 2 2 2 2 n=1 radius of convergence of this series is, by Corollary 1 of Theorem 3.3.1,   1             an   2(n + 1) π2 − arctan(n + 1) 2n n arctan(n) − π2     = lim   = lim .r = lim  = 2.   1 an+1  n π2 − arctan(n)    2n+1 (n + 1) arctan(n + 1) − π   2

i) Let us study the power series of x,

Then, the series is absolutely convergent if x ∈ ] − 2, 2[ and diverges if x ∈ ] − ∞, −2[ ∪ ]2, +∞[. At the endpoints of the convergence interval, we obtain two numerical series. ∞ ∞   1 1  2n =  . Let us calculate   – If x = 2, we have the series π n n arctan(n) − 2 n arctan(n) − π2 2 n=1 n=1 the limit of its general term: 1  = −1. . lim  n arctan(n) − π2 Since the general term of the series is not a null sequence, by Theorem 2.2.1, the series is divergent. ∞ ∞   (−2)n 1  =   , which is (−1)n+1  π – If x = −2, we obtain the series π n n arctan(n) − 2 n − arctan(n) 2 2 n=1 n=1 1  > 0, ∀n ∈ N. The limit of the general term does not alternating because an =  π n 2 − arctan(n) exist, as the subsequence of even order terms has limit −1 and the subsequence of odd order terms has limit 1. Then, the series is divergent. Conclusion: The power series converges absolutely if x ∈ ] − 2, 2[ and diverges if x ∈ ] − ∞, −2] ∪ [2, +∞[. ∞ ∞   3n n! 3 × 6 × · · · × 3n n 3n n! x = xn . If an = , then the radius (n + 3)! (n + 3)! (n + 3)! n=1 n=1 of convergence of this series is, by Corollary 1 of Theorem 3.3.1,

j) Consider the power series

3n n!   n+4 1 (n + 3)!  = lim = lim = ;  3n+1 (n + 1)! 3(n + 1) 3 n+1 (n + 4)!       1 1 1 1 therefore, the series converges absolutely if x ∈ − , and diverges if x ∈ −∞, − ∪ , +∞ . 3 3 3 3 Let us study the numerical series at the endpoints of the convergence interval. ∞ ∞   1 n! 1 – If x = , we obtain the series = . Let us study this series by 3 (n + 3)! (n + 3)(n + 2)(n + 1) n=1 n=1 ∞  1 which we know to be convergent (p-series with p = 3): comparing it with 3 n n=1   an .r = lim  a

1 n3 (n + 3)(n + 2)(n + 1) = 1. . lim = lim 1 (n + 3)(n + 2)(n + 1) 3 n Since this value is finite and different from zero, the series is convergent by Corollary 2 of the General Comparison Test. As it is a series of positive terms, it is absolutely convergent.

252

Series of Functions ∞  n! 1 . As the series of modules is convergent, – If x = − , we obtain the alternating series (−1)n 3 (n + 3)! n=1 the alternating series is absolutely convergent.

      1 1 1 1 Conclusion: The series converges absolutely if x ∈ − , and diverges if x ∈ −∞, − ∪ , +∞ . 3 3 3 3 ∞ 

n+1 n+1 xn . Let an = (−1)n √ . The radius of 2n + 5 2n + 5 e e n=0 convergence of this series is, by Corollary 1 of Theorem 3.3.1,   n+1    √  (−1)n √      an  (n + 1) e2n+2 + 5 e2n+2 + 5 n+1 e2n + 5     = lim  .r = lim  = e;  = lim (n + 2)√e2n + 5 = lim n + 2 · n + 2 an+1  e2n + 5 n+1  (−1)  √   2n+2 e +5

k) Let us study the power series of x,

(−1)n √

then, the series is absolutely convergent if x ∈ ] − e, e[ and divergent if x ∈ ] − ∞, −e[ ∪ ]e, +∞[. At the endpoints of the convergence interval, we obtain two numerical series that we will proceed to study. ∞ 

– If x = −e, we obtain the series

(−1)n √

n=0

n+1 e2n

. lim

+5

(−e)n =

∞  (n + 1) en √ . As e2n + 5 n=0

(n + 1) en √ = +∞, e2n + 5

the general term is not a null sequence; therefore, by Theorem 2.2.1, the series diverges. ∞ ∞   n+1 (n + 1) en – If x = e, we obtain the series (−1)n √ (−1)n √ en = . As the limit of the 2n e +5 e2n + 5 n=0 n=0 general term does not exist, the series is divergent. Conclusion: The power series converges absolutely if x ∈ ] − e, e[ and diverges if x ∈ ] − ∞, −e] ∪ [e, +∞[. l) Let us study the power series of x,

∞  1 × 3 × 5 × · · · × (2n − 1) 1 n · x . Setting 2 × 4 × 6 × · · · × 2n n n=1

.an

=

1 × 3 × 5 × · · · × (2n − 1) 1 · , 2 × 4 × 6 × · · · × 2n n

the radius of convergence of this series is, by Corollary 1 of Theorem 3.3.1,     1 × 3 × 5 × · · · × (2n − 1) 1     ·    an  2 × 4 × 6 × · · · × 2n n  = lim (n + 1)(2n + 2) = 1.  = lim  .r = lim     1 × 3 × 5 × · · · × (2n − 1)(2n + 1) 1 an+1 n(2n + 1)   ·   2 × 4 × 6 × · · · × 2n(2n + 2) n+1 The series converges absolutely if x ∈ ] − 1, 1[ and diverges if x ∈ ] − ∞, −1[ ∪ ]1, +∞[. Let us study the numerical series obtained at the endpoints of the convergence interval. – If x = −1, we obtain the alternating series ∞  . n=1

(−1)n

1 × 3 × 5 × · · · × (2n − 1) 1 · . 2 × 4 × 6 × · · · × 2n n

3.6. Solved Exercises

253

Let us start by studying the series of modules, .

∞ ∞   1 × 3 × 5 × · · · × (2n − 1) 1 · = an , 2 × 4 × 6 × · · · × 2n n n=1 n=1

using Raabe’s test     an (n + 1)(2n + 2) 3n + 2 3 − 1 = lim = > 1, . lim n − 1 = lim n an+1 n(2n + 1) 2n + 1 2 which shows that the series is convergent. As it is the series of modules of

∞ 

(−1)n an , this series is

n=1

absolutely convergent. ∞ 

1 × 3 × 5 × · · · × (2n − 1) 1 · , which we have seen to be convergent. 2 × 4 × 6 × · · · × 2n n As it is of positive terms, it is absolutely convergent.

– If x = 1, we obtain the series

n=1

Conclusion: The power series converges absolutely if x ∈ [−1, 1] and diverges if x ∈ ] − ∞, −1[ ∪ ]1, +∞[. 15.

a) Let us study the power series of x − 1, ∞  

n 2n +1 n=1 Theorem 3.3.1, series

.r

=

2n y n . Setting an

1  = lim n |an |

 lim n

∞  

2n n (x − 1)n . Let y = x − 1, and consider the 2n + 1 n=1 2n  n = , the radius of convergence of this series is, by 2n + 1

1 n 2n + 1

2n =

 lim

1 n 2n + 1

2 =

1 = 4. n2 lim 4n2 + 4n + 1

The series converges absolutely if y ∈ ] − 4, 4[ and diverges if y ∈ ] − ∞, −4[ ∪ ]4, +∞[. As y = x − 1, we 2n ∞   n can conclude that the series (x − 1)n is absolutely convergent if x ∈ ] − 3, 5[ and divergent 2n + 1 n=1 if x ∈ ] − ∞, −3[ ∪ ]5, +∞[. At the endpoints of the convergence interval, we have to study the corresponding numerical series. – If x = −3, we obtain the series ∞   . n=1

n 2n + 1

2n (−4)n =

∞ 

 (−1)n 4n

n=1

As

 . lim

2n 2n + 1

n 2n + 1 2n =

2n =

∞ 

 (−1)n

n=1

2n 2n + 1

2n .

1 , e

the general term does not have limit; therefore, by Theorem 2.2.1, the series is divergent. – If x = 5, we obtain the series ∞   . n=1

n 2n + 1

2n 4n =

∞   n=1

2n 2n + 1

2n ,

whose general term does not tend to zero, as we saw earlier; therefore, the series is divergent.

254

Series of Functions In summary, the power series of x − 1 is divergent if x ∈ ] − ∞, −3] ∪ [5, +∞[ and absolutely convergent if x ∈ ] − 3, 5[. ∞ 

b) Let us study the power series of x − 2, ∞ 

n=2

(−1)n

(x − 2)n . Let y = x − 2, and consider the series n log(n)

yn (−1)n . Setting an = , the radius of convergence of this series is, by Corollary 1 of (−1) n log(n) n log(n) n=2 Theorem 3.3.1,     (−1)n          an  n + 1 log(n + 1) n log(n)    = lim  = lim · = 1; .r = lim     (−1)n+1 an+1  n log(n)    (n + 1) log(n + 1)  n

therefore, the series converges absolutely if y ∈ ]−1, 1[ and diverges if y ∈ ]−∞, −1[ ∪ ]1, +∞[. As y = x−2, ∞  (x − 2)n the series (−1)n is absolutely convergent if x ∈ ]1, 3[ and divergent if x ∈ ] − ∞, 1[ ∪ ]3, +∞[. n log(n) n=2 Let us study the numerical series obtained at the endpoints of the convergence interval. – If x = 1, we obtain the series

∞ 

(−1)n

n=2

∞  (−1)n 1 = . Let us study it using the Integral n log(n) n log(n) n=2

Test. Let f (x) =

1 : x log(x) (i) As x ≥ 2 ⇒ log(x) > 0, we have f (x) > 0, ∀x ≥ 2. (ii) f is continuous on [2, +∞[. (iii) f is decreasing on [2, +∞[ because x log(x) is increasing and positive.  +∞ ∞  1 1 Then the numerical series and the improper integral dx are both convern log(n) x log(x) 2 n=2 gent or both divergent. Since  .

+∞ 2

1 dx = x log(x)

 2

 =

x

lim

x→+∞

lim

x→+∞

1 dt = lim x→+∞ t log(t)

  x = log log(t) 2

lim



x→+∞

x

2



1 t

log(t)

dt

    log log(x) − log log(2) = +∞,

the improper integral is divergent, the same happening to the series. ∞  1 (−1)n . We apply Leibniz’s test since we already – If x = 3, we obtain the alternating series n log(n) n=2 know that the series of modules is divergent: 1 = 0. (i) lim n log(n) 1 (ii) > 0, ∀n ≥ 2. n log(n) 1 (iii) We saw earlier that the function f (x) = is decreasing on [2, +∞[, which implies that x log(x)   1 is decreasing for n ≥ 2. n log(n)

3.6. Solved Exercises

255

Then, we conclude that the series is convergent. As the series of modules is divergent, the series is conditionally convergent. Conclusion: The power series of x − 2 is absolutely convergent if x ∈ ]1, 3[, conditionally convergent if x = 3, and divergent if x ∈ ] − ∞, 1] ∪ ]3, +∞[. ∞ 

n (x + 4)n . We can rewrite it in the form n (2n2 − 1) (−3) n=1 ∞ ∞   n n n (−1)n n . Let y = x + 4, and consider the series (−1)n n (x + 4) y n . If 2 − 1) 2 − 1) 3 (2n 3 (2n n=1 n=1 n an = (−1)n n , the radius of convergence of this series is, by Corollary 1 of Theorem 3.3.1, 3 (2n2 − 1)   n       (−1)n n 2 2 − 1)   an   3 (2n  = lim   = lim 3n(2n + 4n + 1) = 3. .r = lim     n+1 an+1 (n + 1)(2n2 − 1)  (−1)n+1    3n+1 2(n + 1)2 − 1 

c) Let us study the power series of x + 4,

The series converges absolutely if y ∈ ] − 3, 3[ and diverges if y ∈ ] − ∞, −3[ ∪ ]3, +∞[. As y = x + 4, ∞  n the series (x + 4)n is absolutely convergent if x ∈ ] − 7, −1[ and divergent if (−1)n n 2 − 1) 3 (2n n=1 x ∈ ] − ∞, −7[ ∪ ] − 1, +∞[. At the endpoints of the convergence interval, we have to study the respective numerical series. ∞ 

(−1)n

n

∞ 

n . 2−1 − 1) 2n n=1 n=1 It is of positive terms, and we can compare it with the harmonic series, which is divergent. Since the limit n n2 1 2n2 − 1 = lim = . lim 1 2n2 − 1 2 n

– If x = −7, we obtain the series

3n (2n2

(−3)n =

is finite and different from zero, the series is divergent by Corollary 2 of the General Comparison Test. ∞ ∞   n n – If x = −1, we obtain the alternating series 3n = . We apply (−1)n n (−1)n 2 2 3 (2n − 1) 2n − 1 n=1 n=1 Leibniz’s test since the series of modules is divergent: n (i) lim = 0. 2n2 − 1 n > 0, ∀n ≥ 1. (ii) 2n2 − 1 n+1 n 2n2 + 2n + 1 (iii) − = − < 0, ∀n ∈ N, which implies that 2 2n2 − 1 (2n2 + 4n + 1)(2n2 − 1) 2(n + 1) − 1 n is decreasing. 2n2 − 1 We can conclude that the series is convergent. Since the series of modules is divergent, the series is conditionally convergent. Conclusion: The power series of x + 4 is absolutely convergent if x ∈ ] − 7, −1[, conditionally convergent if x = −1 and divergent if x ∈ ] − ∞, −7] ∪ ] − 1, +∞[.

256

Series of Functions d) Let us study the power series of x − e, ∞  n=2

∞ 

n=2

en (x − e)n . Let y = x − e, and consider the series log(n)

en en y n . If an = , the radius of convergence of this series is, by Corollary 1 of Theorem 3.3.1, log(n) log(n)   an .r = lim  a

n+1

   = lim 

en 1 log(n + 1) 1 log(n) = lim · = , en+1 e log(n) e log(n + 1)

    1 1 and divergent if y ∈ −∞, − ∪ , +∞ . As e e   1 1 is absolutely convergent if x ∈ e − , e + and divergent if e e 

so the series is absolutely convergent if y ∈ ∞  en (x − e)n y = x − e, the series log(n)  n=2    1 1 ∪ e + , +∞ . x ∈ −∞, e − e e

1 1 − , e e



Let us study the numerical series corresponding to the endpoints of the interval.   ∞ ∞   1 en (−1)n 1 n . It is an alternating series because – If x = e − , we get the series = − e log(n) e log(n) n=2 n=2 ∞  1 1 > 0, ∀n > 1. Let us consider the series of modules, , and study it by an = log(n) log(n) n=2 comparing it with the harmonic series. As the limit 1 n log(n) = lim . lim = +∞ 1 log(n) n by Corollary 4 of the General Comparison Test, the series test:

∞  n=2

1 is divergent. Let us apply Leibniz’s log(n)

1 = 0. log(n) 1 > 0, ∀n ≥ 2. (ii) log(n)    (iii) The sequence log(n) is increasing; therefore, (i) lim

 1 is a decreasing sequence for n ≥ 2. log(n) Then, the series is convergent. Since the series of modules is divergent, the alternating series is conditionally convergent.  n ∞ ∞   en 1 1 1 = which, as we have seen, is divergent. – If x = e + , we get the series e log(n) e log(n) n=2 n=2

  1 1 , conditionally converConclusion: The power series of x − e is absolutely convergent if x ∈ e − , e + e e     1 1 1 gent if x = e − and divergent if x ∈ −∞, e − ∪ e + , +∞ . e e e

3.6. Solved Exercises

257

e) Let us consider the series ∞  . n=1

 π  π  π   n n ∞ cos ∞ cos   1 1 2n 2n 2n = = , (2n − 1)2 (x − 1)n (2n − 1)2 x − 1 (2n − 1)2 x − 1 n=1 n=2 cos

π  ∞  cos 2n We study the power series of y, y n . Note that if n > 1, then (2n − 1)2   n=2 π  π cos 2n π π 0< . The radius of convergence of this series < ; therefore, cos > 0. Let an = 2n 2 2n (2n − 1)2 is, by Corollary 1 of Theorem 3.3.1,

and let y =

1 . x−1

       cos π  2n   π     (2n − 1)2   an  cos 2n (2n + 1)2    = lim    = lim .r = lim  = 1.   π π  cos  an+1 (2n − 1)2 cos 2(n+1)  2(n+1)     (2n + 1)2  Therefore, the series converges absolutely if y ∈ ] − 1, 1[ and diverges if y ∈ ] − ∞, −1[ ∪ ]1, +∞[. If y = ±1, we obtain two numerical series we will study. π  ∞  cos 2n – If y = 1, we obtain the series , which is of positive terms. We are going to compare it (2n − 1)2 n=2 ∞  1 with , which we know to be convergent, as it is a p-series with p = 2. The limit 2 n n=1 cos

. lim



π 2n



π  n2 cos 2n 1 (2n − 1)2 = lim = 1 (2n − 1)2 4 n2

is finite and different from zero; therefore, by Corollary 2 of the General Comparison Test, the series is convergent. As it is of positive terms, it is absolutely convergent. π  ∞  cos 2n (−1)n , whose series of modules is the one – If y = −1, we obtain the alternating series (2n − 1)2 n=2 we studied for the case y = 1, which we proved to be convergent. Therefore, the series is absolutely convergent. Conclusion: The power series of y is divergent if y ∈ ] − ∞, −1[ ∪ ]1, +∞[ and absolutely convergent if y ∈ [−1, 1]. Concerning the original series, we have, because y =

.|y|

  ≤ 1 ⇔ 

1 , x−1

 1  ≤ 1 ⇔ |x − 1| ≥ 1 ⇔ x ≥ 2 ∨ x ≤ 0, x − 1

which implies that the series converges absolutely if x ∈ ] − ∞, 0] ∪ [2, +∞[ and diverges if x ∈ ]0, 2[ \{1}, 1 . x−1

because 1 does not belong to the domain of

258

Series of Functions f) Let us study the power series of 2x2 − 5, ∞  n=1

∞  (2x2 − 5)n . Let y = 2x2 − 5 and consider the series n 3n+1 n=1

1 yn . n 3n+1

Setting an =

1 , the radius of convergence of this series is, by Corollary 1 of Theorem 3.3.1, n 3n+1     1        an  n+1 n 3   = lim 3(n + 1) = 3;   = lim  .r = lim    1 an+1 n    (n + 1) 3n+2 

therefore, the series converges absolutely if y ∈ ] − 3, 3[ and diverges if y ∈ ] − ∞, −3[ ∪ ]3, +∞[. At the endpoints of the interval of convergence, it is necessary to study the respective numerical series. ∞ ∞ ∞   1 (−1)n 1  (−1)n n = , which is the – If y = −3, we obtain the series (−3) = n 3n+1 3n 3 n=1 n n=1 n=1 alternating harmonic series that we know to be conditionally convergent. ∞ ∞  1 1  1 = , which is divergent. – If y = 3, we obtain the series 3n 3 n=1 n n=1 But |y| < 3 ⇔ |2x2 − 5| < 3 ⇔ −3 < 2x2 − 5 < 3 ⇔ 1 < x2 < 4 .   ⇔ x ∈ ] − 2, 2[ ∩ ] − ∞, −1[ ∪ ]1, +∞[ ⇔ x ∈ ] − 2, −1[ ∪ ]1, 2[. In addition, .2x

2

− 5 = 3 ⇔ x = −2 ∨ x = 2

and .2x

2

− 5 = −3 ⇔ x = −1 ∨ x = 1.

Conclusion: The original series is absolutely convergent if x ∈ ] − 2, −1[ ∪ ]1, 2[, conditionally convergent if x = −1 or x = 1 and divergent if x ∈ ] − ∞, −2] ∪ ] − 1, 1[ ∪ [2, +∞[. √ √ n+1− n √ √ (x + 2)n . We can rewrite the series in the form n+1+ n n=1 ∞ ∞   1 1   (x + 2)n . Let y = x + 2 and consider the series yn . 2n + 1 + 2 n(n + 1) 2n + 1 + 2 n(n + 1) n=1 n=1 1  Let an = . The radius of convergence of this series is, by Corollary 1 of Theorem 3.3.1, 2n + 1 + 2 n(n + 1)

g) Let us study the power series of x + 2,

  an .r = lim  a

n+1

   = lim 

∞ 

1   2n + 3 + 2 (n + 1)(n + 2) 2n + 1 + 2 n(n + 1)  = 1; = lim 1 2n + 1 + 2 n(n + 1)  2n + 3 + 2 (n + 1)(n + 2)

therefore, the series absolutely converges if y ∈ ] − 1, 1[ and diverges if y ∈ ] − ∞, −1[ ∪ ]1, +∞[. As ∞  1  (x + 2)n is absolutely convergent if y = x + 2, we can conclude that the series 2n + 1 + 2 n(n + 1) n=1 x ∈ ] − 3, −1[ and divergent if x ∈ ] − ∞, −3[ ∪ ] − 1, +∞[. Let us study the numerical series at the endpoints of the interval of convergence.

3.6. Solved Exercises

259

– If x = −1, we obtain the series

∞  n=1

1  . As it is of positive terms and the limit 2n + 1 + 2 n(n + 1)

1  2n + 1 + 2 n(n + 1) n 1  = lim = . lim 1 4 2n + 1 + 2 n(n + 1) n is finite and different from zero, the series is divergent because the harmonic series is divergent (see Corollary 2 of the General Comparison Test). ∞ 

(−1)n  . It is an alternating series, and we apply n=1 2n + 1 + 2 n(n + 1) Leibniz’s test since the series of modules is divergent: 1  = 0. (i) lim 2n + 1 + 2 n(n + 1) 1  (ii) > 0, ∀n ≥ 1. 2n + 1 + 2 n(n + 1) 1 1   (iii) − < 0, which implies that the sequence is 2n + 3 + 2 (n + 1)(n + 2) 2n + 1 + 2 n(n + 1) decreasing.

– If x = −3, we obtain the series

Then, the alternating series is convergent. It is conditionally convergent because the series of modules is divergent. Conclusion: The power series of x + 2 is absolutely convergent if x ∈ ] − 3, −1[, conditionally convergent if x = −3 and divergent if x ∈ ] − ∞, −3[ ∪ [−1, +∞[.

h) Let us consider the series

 n ∞  log2 (n) 2x 2x · and set y = 2 . We are going to study the power 2 2 n x + 1 x +1 n=1

series of y,

∞  log2 (n) n y . The radius of convergence of this series is, by Corollary 1 of Theorem 3.3.1, n2 n=1

with an =

log2 (n) , n2   log2 (n)      n2  = lim    log2 (n + 1) n+1   (n + 1)2

  an .r = lim  a

    (n + 1)2 log2 (n)  = 1.  = lim 2  n log2 (n + 1)  

The series is absolutely convergent if y ∈ ] − 1, 1[ and diverges if y ∈ ] − ∞, −1[ ∪ ]1, +∞[. About the original series, we have .|y|

  < 1 ⇔ 

 2x  2x < 1 ⇔ −1 < 2 < 1 ⇔ x = −1 ∧ x = 1. x2 + 1  x +1

Let us study the numerical series at x = −1 and x = 1.

260

Series of Functions ∞ ∞   log2 (n) log2 (n) = , which is of positive terms, and we will study 2 n n2 n=1 n=2 ∞  1 it by comparing it with the convergent series . The limit n3/2 n=1

– If x = 1, we obtain the series

log2 (n) log2 (n) n2 =0 . lim = lim √ 1 n n3/2 allows us to conclude, by Corollary 3 of the General Comparison Test, that the series is convergent. As it is of positive terms, it is absolutely convergent. ∞  log2 (n) – If x = −1, we obtain the series (−1)n , which is alternating. As we have seen, the series of n2 n=2 modules is convergent, and the alternating series is absolutely convergent. Conclusion: The original power series is absolutely convergent on R. ∞

1  1 1 √ , · . It is important to note that the set where the x n=1 2n n + 1 xn series converges does not include the point x = 0 since the functions are not defined at that point. Rather, ∞  1 1 1 √ y n . By setting an = n √ , we can apply we can set y = , and consider the series n x 2 n + 1 2 n+1 n=1 Corollary 1 of Theorem 3.3.1 to determine the radius of convergence of this series:

i) Let us study the power series of

  an .r = lim  a

n+1

   = lim 

1 √ √ 2 n+2 n+1 = lim √ = 2. 1 n+1 √ n+1 2 n+2 2n

The series converges absolutely if y ∈ ] − 2, 2[ and diverges if y ∈ ] − ∞, −2[ ∪ ]2, +∞[. Since y = ∞ 

1 , x

1 1 1 √ ∈ ] − 2, 2[ and diverges if · n converges absolutely if x x n + 1 n=1     1 1 1 ∈ ] − ∞, −2[ ∪ ]2, +∞[, that is, it is absolutely convergent if x ∈ −∞, − , +∞ and divergent ∪ x 2 2   1 1 if x ∈ − , \ {0}. 2 2 1 1 Let us study the numerical series obtained at x = − and x = . 2 2 ∞ ∞   1 1 1 √ = – If x = , we obtain the divergent series √ . n 2 n + 1 n=1 n=2 we conclude that the series

2n

∞  1 1 – If x = − , we obtain the alternating series (−1)n √ . We will apply Leibniz’s test since the 2 n+1 n=1 series of modules is divergent: 1 (i) lim √ = 0. n+1 1 > 0, ∀n ≥ 1. (ii) √ n+1

3.6. Solved Exercises (iii) √

261

1 1 −√ = n n+1

√ √   n− n+1 1 is decreasing. < 0, ∀n ∈ N, which implies that √ √ √ n n n+1

It follows that the series is convergent. As the series of the modules is divergent, the series is conditionally convergent.  Conclusion: The power series is absolutely convergent if x ∈   1 1 1 convergent if x = − , and divergent if x ∈ − , \ {0}. 2 2 2 ∞   2

j) Let us study the power series ∞   n=1

2 1 + 3 2n



n y n . If an =

.r

=

n=1

3

2 1 + 3 2n

1  = lim n |an |

+ n

1 2n

−∞, −

1 2



 ∪

 1 , +∞ , conditionally 2

n (2x − 3)n .

Let y = 2x − 3, and consider the series

, the radius of convergence of this series is, by Theorem 3.3.1,

 lim n

1 2 1 + 3 2n

n =

 lim

3 1  = . 2 1 2 + 3 2n

      3 3 3 3 , +∞ . However, The series converges absolutely if y ∈ − , and diverges if y ∈ −∞, − ∪ 2 2 2 2    ∞   3 9 2 1 n n + , (2x − 3) is absolutely convergent if x ∈ y = 2x − 3, so the series and divergent 3 2n 4 4   n=1   9 3 ∪ , +∞ . if x ∈ −∞, 4 4 Regarding the endpoints of the interval of convergence, we have to study the respective numerical series. – If x =

    ∞  ∞    2 1 n 3 n 9 3 n , we obtain the series + = . As 1+ 4 3 2n 2 4n n=1 n=1     1 3 3 n 3 4n 4 . lim 1 + = lim 1+ = e4 , 4n 4n

the general term is not a null sequence; therefore, by Theorem 2.2.1, the series diverges.      ∞  ∞   1 n 3 2 3 n 3 n + = (−1)n 1 + . As the limit of the – If x = , we obtain the series − 4 3 2n 2 4n n=1 n=1 general term does not exist, the series is divergent.  Conclusion: The power series of x is divergent if x ∈   3 9 x∈ , . 4 4 k) Let us study the power series of

x2 ,

−∞,

 2 ∞  23n (n − 1)! n=2

(2n)!

3 4



 ∪

 9 , +∞ and absolutely convergent if 4

x2n . Let y = x2 . The radius of convergence of

262

Series of Functions the series

 2 ∞  23n (n − 1)! (2n)!

n=2

y n is, by Corollary 1 of Theorem 3.3.1, and setting an =

 2 23n (n − 1)! (2n)!

,

   23n (n − 1)!2          (2n)!  an  (2n + 2)(2n + 1) 1  = lim   = lim .r = lim  = .  23n+3 (n!)2  an+1  23 n2 2    (2n + 2)!        1 1 1 1 , +∞ . Since The series is absolutely convergent if y ∈ − , and diverges if y ∈ −∞, − ∪ 2 2 2 2  2   ∞  23n (n − 1)! 1 1 2n 2 y = x , the series x is absolutely convergent if x ∈ − √ , √ and divergent if (2n)! 2 2   n=2 1 1 x ∈ −∞, − √ ∪ √ , +∞ . 2 2 At the endpoints of the interval of convergence, we obtain two numerical series.  2 ∞ ∞   23n (n − 1)! 1 1 · n = – If x = − √ , we obtain the series bn . Using Raabe’s test, (2n)! 2 2 n=1 n=1  . lim

n

bn bn+1

2 ⎞ ⎛ 2n  2 (n − 1)!    ⎟ ⎜ (2n + 2)(2n + 1) 3 (2n)! ⎟ ⎜ = lim n − 1 − 1 = lim n ⎜ − 1 = > 1, ⎟ ⎠ ⎝ 22n+2 (n!)2 4n2 2 (2n + 2)!

so the series is convergent. As it is of positive terms, it is absolutely convergent.  2 ∞  22n (n − 1)! 1 – If x = √ , we again obtain the series , which we have seen to be absolutely (2n)! 2 n=1 convergent.  Conclusion: The series is divergent if x ∈   1 1 x ∈ −√ , √ . 2 2

 ∞   2n − 1 2n

l) Let us study the power series of x − 3,  ∞   2n − 1 2n n=0

3n + 1

 y n . If an =

n=0 2n

2n − 1 3n + 1

1  = lim n |an |

1 −∞, − √ 2

3n + 1



 ∪

1 √ , +∞ 2

 and is absolutely convergent if

(x − 3)n . Let y = x − 3, and consider the series

, the radius of convergence of this series is, by Theorem 3.3.1,

1 9 2 = . 4 2n − 1 2n − 1 lim lim n 3n + 1 3n + 1       9 9 9 9 and diverges if y ∈ −∞, − ∪ The series is absolutely convergent if y ∈ − , , +∞ . Since 4 4 4 4    ∞   3 2n − 1 2n 21 , (x − 3)n is absolutely convergent if x ∈ y = x − 3, the series and divergent if 3n + 1 4 4    n=0  3 21 , +∞ . x ∈ −∞, ∪ 4 4 .r

=



1

2n =



3.6. Solved Exercises

263

At the endpoints of the interval of convergence, we obtain two numerical series.      ∞  ∞   2n − 1 2n 6n − 3 2n 3 9 n – If x = , we have the series = (−1)n . As − 4 3n + 1 4 6n + 2 n=0 n=1  . lim

6n − 3 6n + 2

2n

5

= e− 3 ,

the limit of the general term does not exist, and, by Theorem 2.2.1, the series diverges.     ∞  ∞    2n − 1 2n 9 n 6n − 3 2n 21 , we obtain the series = . The general term is not – If x = 4 3n + 1 4 6n + 2 n=0 n=1 a null sequence, so the series is divergent.     3 21 Conclusion: The power series of x − 3 is divergent if x ∈ −∞, ∪ , +∞ and absolutely convergent 4 4   3 21 if x ∈ , . 4 4 π π ∞ ∞   sin n sin n 2n (x−1) (x−1)2n . Let y = (x−1)2 , = m) Let us study the power series of x−1, 4n (n + 1) 4n (n + 1) n=1 n=2   ∞ π  sin n and consider the series y n . The radius of convergence of this series is, by Corollary 1 of 4n (n + 1) n=2 π sin n Theorem 3.3.1 and setting an = n , 4 (n + 1)   an .r = lim  a

n+1

   = lim 

   sin π     n  n   π  n+1  4 (n + 1)  sin n (n + 2) 4   π   = lim  π  = 4; n (n + 1)   sin 4 sin   n+1 n+1   n+1 4 (n + 2) 

the series is absolutely convergent if y ∈ ] − 4, 4[ and divergent if y ∈ ] − ∞, −4[ ∪ ]4, +∞[. π ∞  sin n (x − 1)2n is absolutely convergent if x ∈ ] − 1, 3[ and divergent As y = (x − 1)2 , the series 4n (n + 1) n=2 if x ∈ ] − ∞, −1[ ∪ ]3, +∞[. At the endpoints of the interval of convergence, we obtain two numerical series. π π ∞ ∞   sin n sin n 2n (−2) . Let us study it by comparison with – If x = −1, we get the series = 4n (n + 1) n+1 n=2 n=2 ∞  1 , which is convergent (p-series with p = 2): 2 n n=1 sin . lim

π

n π sin n n n+1 = lim = π. · 1 1 n+1 n n2

Since this limit is finite and different from zero, the series is convergent by Corollary 2 of the General Comparison Test. Being of positive terms, it is absolutely convergent.

264

Series of Functions – If x = 3, we again obtain the series

∞  n=2

sin

π n

4n (n + 1)

22n =

∞  sin n=2

π n

n+1

, which we have seen to be

absolutely convergent. Conclusion: The power series of x − 1 is divergent if x ∈ ] − ∞, −1[ ∪ ]3, +∞[ and absolutely convergent if x ∈ [−1, 3].   ∞  x2n x2 is of positive terms. For it to be convergent, its general term must be a null √ log 1 + √ n n n=1 sequence (see Theorem 2.2.1). But  

x2 x2n 0, if |x2 | ≤ 1 log 1 + √ . lim √ = n n +∞, if |x2 | > 1;

16. The series

therefore, the series can only converge if |x| ≤ 1. We have .0

    x2n x2 x2n x2n x2 x2n+2 ≤ √ 1+ √ = √ + ≤ √ log 1 + √ , ∀x ∈ R; n n n n n n

let us study the power series

∞ ∞   x2n x2n+2 . √ and n n n=1 n=1

1 Let an = √ . The radius of convergence of the first series is, by Corollary 1 of Theorem 3.3.1, n   an . lim  a

n+1

   = lim 

1 √ √ n+1 n = 1; = lim √ 1 n √ n+1

therefore, the series is absolutely convergent if x ∈ ] − 1, 1[. 1 Let bn = . The radius of convergence of the second series is, by Corollary 1 of Theorem 3.3.1, n   bn . lim  b

n+1

   = lim 

1 n+1 n = lim = 1; 1 n n+1

therefore, the series is absolutely convergent if x ∈ ] − 1, 1[.   ∞  x2n x2 1+ √ is convergent if x ∈ ] − 1, 1[, which implies, by the General We can conclude that the series √ n n n=1 Comparison Test, that the original series is convergent if x ∈ ] − 1, 1[.   ∞  1 1 For x2 = 1, that is, x = 1 or x = −1, we obtain the series . Comparing it with the √ log 1 + √ n n n=1 harmonic series, which is divergent,   1 1    √n √ log 1 + √ √ 1 1 n n = lim n log 1 + √ . lim = 1, = lim log 1 + √ 1 n n n

3.6. Solved Exercises

265

we conclude, by Corollary 2 of the General Comparison Test, that the series is divergent. The original series converges if x ∈ ] − 1, 1[ and diverges if x ∈ ] − ∞, −1] ∪ [1, ∞[.

17. Let y = x2 . The series

∞  

cos

π

n=1

convergence .

1  = lim n |an |

!

lim n

n

+ 2 − (−1)n

n

n  π + 2 − (−1)n y n setting an = cos n has radius of

1 1 1  π = . π n = n 4 lim cos + 2 − (−1) + 2 − (−1)n cos n n 

   1 1 ∪ , +∞ . 4 4     ∞  π n  1 1 1 1 + 2 − (−1)n cos The series (x2 )n is absolutely convergent if x2 ∈ − , , that is, x ∈ − , , n 4 4 2 2 n=1 and this is the largest open interval on which the series converges.

The series is absolutely convergent if y ∈

1 1 − , 4 4





and diverges if y ∈

−∞, −

∞ 

4n 5 xn . If the radius of convergence of the series is , then, by Theon + 5n 2 4 n=1       5 5 5 5 rem 3.3.1, the series is absolutely convergent on the interval − , , +∞ . and diverges on −∞, − ∪ 4 4 4 4   5 5 But 1 ∈ − , , therefore, the series is absolutely convergent at this point. The series is divergent at x = 2 4 4    5 5 because 2 ∈ −∞, − ∪ , +∞ . 4 4

18. Consider the power series of x,

5 , the series may be convergent or divergent as it is one of the endpoints of the interval of convergence. 4  n ∞ ∞   5 4n 5n Let us study the series = . Its general term is not a null sequence. In fact, n + 5n n + 5n 2 4 2 n=1 n=1

At x =

. lim

2n

5n 1 = 1; = lim  n 2 + 5n +1 5

therefore, by Theorem 2.2.1, the series is divergent. 19. The series

∞ 

an xn is a power series of x. We have

n=0 .

√ n

an =

1, √ n

2,

if n is even if n is odd,

√ √ n which implies that lim n an = 1, because lim 2 = 1. Then the radius of convergence of this series is, by Theorem 3.3.1, 1 .r = = 1; √ lim n an therefore, the series is absolutely convergent if x ∈ ] − 1, 1[ and divergent if x ∈ ] − ∞, −1[ ∪ ]1, +∞[.

266

Series of Functions If x = −1, we obtain the series

∞ 

(−1)n an , whose general term does not have a limit; therefore, by Theo-

n=1

rem 2.2.1, the series is divergent. ∞  an . As lim an does not exist, the series is divergent. If x = 1, we obtain the series n=1

Conclusion: The power series is absolutely convergent if x ∈ ] − 1, 1[ and diverges if x ∈ ] − ∞, −1] ∪ [1, +∞[. 20. Let us consider the power series

∞ 

an xn . If, by hypothesis, the series

n=1

∞ 

an 5n converges, then either the power

n=1

series converges absolutely on R or there exists r ∈ R+ such that the series converges absolutely if x ∈ ] − r, r[ ∞  and diverges if x ∈ ] − ∞, −r[ ∪ ]r, +∞[. In the first case, it is evident that the series an (−2)n is absolutely n=1

convergent. In the second case, if the series converges at x = 5, then r ≥ 5. As −2 ∈ ] − 5, 5[ ⊂ ] − r, r[ , the ∞  series an (−2)n is absolutely convergent. n=1

21. Suppose the radius of convergence of the power series of x − 3 is r > 0. In that case, the series is absolutely convergent on the interval ]3 − r, 3 + r[, which is the largest open interval where the series is convergent. The ∞  series may also be convergent at the endpoints of the interval. If the series an (x − 3)n converges at x = 0, n=0

then 0 ∈ [3 − r, 3 + r], which implies that r ≥ 3. But in that case, 1 ∈ ]3 − r, 3 + r[, so the series cannot diverge at x = 1. If the radius of convergence of the series is +∞, then the series converges absolutely at all points of R, with no points where it can be divergent. 22. If the power series of x, that is, the series

∞ 

∞ 

bn xn , has a radius of convergence R > 1, then it is absolutely convergent if x = 1,

n=0

bn is absolutely convergent. If, for each n ∈ N, |fn (x)| ≤ |bn |, ∀x ∈ R, Weierstrass’ test

n=0

allows us to conclude that the series of functions 23.

a) Let us consider a power series

∞ 

∞ 

fn (x) is uniformly convergent on R.

n=0

an (x − x0 )n with a radius of convergence r > 0. By Theorem 3.3.1, the

n=1

series is absolutely convergent if x ∈ ]x0 − r, x0 + r[ and divergent if x ∈ ] − ∞, x0 − r[ ∪ ]x0 + r, +∞[, which implies that it can only be conditionally convergent at the endpoints of the interval of convergence, that is, at x = x0 − r or x = x0 + r. ∞  If the series an (x − 1)n is conditionally convergent at x = −1, then x0 − r = −1. Since x0 = 1, we n=1

have r = 2. b) From the previous item, we know that the interval of convergence of the series is ] − 1, 3[, so the series ∞  can only be conditionally convergent at x = −1 or x = 3. If x = 3, we get the series an 2n . As, n=1

by hypothesis, (an ) is a sequence of positive terms, this is a series of positive terms that, if convergent, ∞  will be absolutely convergent. Therefore, the only point where the series an (x − 1)n is conditionally convergent is x = −1.

n=1

3.6. Solved Exercises 24.

267

∞  1 1 n (x−2) . Let y = x−2, and consider the series yn . n n 5 n 5n n=1 n=1 1 The radius of convergence of this series is, by Corollary 1 of Theorem 3.3.1 and setting an = , n 5n

a) Let us study the power series of x−2,

∞ 

   1      n 5n   (n + 1) 5n+1  = lim   = lim = 5;    1 n 5n n+1    (n + 1) 5n+1 

  an . lim  a

the series is absolutely convergent if y ∈ ] − 5, 5[ and diverges if y ∈ ] − ∞, −5[ ∪ ]5, +∞[. As y = x − 2, the ∞  1 series (x − 2)n is absolutely convergent if x ∈ ] − 3, 7[ and divergent if x ∈ ] − ∞, −3[ ∪ ]7, +∞[. n 5n n=1 If x = 7, we get the series

∞ 

∞  1 1 , which is divergent. 5n = n n5 n n=1

n=1

If x = −3, we get the series

∞  n=1

∞  1 (−1)n (−5)n = , which is conditionally convergent. n n5 n n=1

Conclusion: The power series of x − 2 is absolutely convergent if x ∈ ] − 3, 7[, conditionally convergent if x = −3 and divergent if x ∈ ] − ∞, −3[ ∪ [7, +∞[. b) Let f (x) =

∞  n=1

1 (x − 2)n , ∀x ∈ ] − 3, 7[. By Theorem 3.3.3, we have n 5n 

f  (x) =

∞  n=1

.

1 (x − 2)n n 5n

 =

 ∞  ∞   1 1 n (x − 2) = n (x − 2)n−1 n n n 5 n 5 n=1 n=1

   ∞ ∞ ∞    1  x−2 n 1 1 x−2 n n−1 = (x − 2) = = , ∀x ∈ ] − 3, 7[. 5n 5 5 5 n=0 5 n=1 n=0 x−2 . Then 5

This series is geometric with r =

.

 ∞   x−2 n n=0

5

=

5 1 = x−2 7 − x 1− 5

and, finally, .f

25. Let us consider the series



(x) =

1 5 1 · = , 5 7−x 7−x

∞  n+1 (−1)n  . log(x) n + 1 n=0

∀x ∈ ] − 3, 7[.

268

Series of Functions ∞  (−1)n n+1 y . The radius of convergence of this series n+1 n=0 n (−1) , is, by Corollary 1 of Theorem 3.3.1 and setting an = n+1    (−1)n       n+1   an  n+2    = lim = lim  = 1; . lim   (−1)n+1  an+1  n+1    n+2 

a) Let y = log(x), and let us study the power series

so the series is absolutely convergent if y ∈ ] − 1, 1[ and diverges if y ∈ ] − ∞, −1[ ∪ ]1, +∞[. Since ∞  n+1 (−1)n  y = log(x), we conclude that the series log(x) is absolutely convergent if log(x) ∈ ] − 1, 1[ n + 1 n=0

and divergent if log(x) ∈ ] − ∞, −1[ ∪ ]1, +∞[, that is, it is absolutely convergent if x ∈ ]e−1 , e[ and divergent if x ∈ ]0, e−1 [ ∪ ]e, +∞[, because the domain of log(x) is R+ . ∞ ∞ ∞ ∞     (−1)n (−1)2n+1 1 1 (−1)n+1 = = − = − , – If x = e−1 , we obtain the series n+1 n+1 n+1 n n=0 n=0 n=0 n=1 which is divergent. ∞ ∞ ∞    (−1)n (−1)n+1 (−1)n =− =− , which is the alternating – If x = e, we obtain the series n + 1 n + 1 n n=0 n=0 n=1 harmonic series, therefore conditionally convergent.

Conclusion: The original series is absolutely convergent if x ∈ ]e−1 , e[, conditionally convergent if x = e and divergent if x ∈ ]0, e−1 ] ∪ ]e, +∞[. ∞  n+1 (−1)n  b) The function is the sum of a power series with interval of convergence ]e−1 , e[, so log(x) n + 1 n=0 it is differentiable term by term on this interval (see Theorem 3.3.3), that is, if x ∈ ]e−1 , e[, we have  ∞   ∞  ∞  (−1)n   n+1 n+1   n (−1)n  1  log(x) log(x) . = = (−1)n log(x) . n + 1 n + 1 x n=0 n=0 n=0

But the series

∞ 

 n (−1)n log(x) is geometric with ratio r = − log(x); therefore,

n=0

.

so

 .

Then .

∞  n 1  1 1 , (−1)n log(x) = · x n=0 x 1 + log(x)

∞  n+1 (−1)n  log(x) n + 1 n=0

 =

1 1 · . x 1 + log(x)

 ∞  n+1   1 (−1)n  1 log(x) · dx = log 1 + log(x) + C. = n + 1 x 1 + log(x) n=0

Let us determine the value of the constant C. For x = 1 (note that this value belongs to the interval of convergence), we obtain .0

=

∞    n+1 (−1)n  = log 1 + log(1) + C ⇔ C = 0; log(1) n + 1 n=0

3.6. Solved Exercises

269

thus, .

26.

∞  n+1   (−1)n  log(x) = log 1 + log(x) . n + 1 n=0

a) The domain of f is the subset of R where the series is convergent. Let us consider the power series of x2 , ∞ ∞   (−1)n (−1)n x2n . Let y = x2 . The radius of convergence of the series y n is, by Corollary 1 2n 2 2n 2 2 (n!) 2 (n!) n=0 n=0 (−1)n of Theorem 3.3.1 and setting an = 2n , 2 (n!)2    (−1)n      22n (n!)2    = lim   = lim 22 (n + 1)2 = +∞.    n+1 (−1)n+1        22n+2 (n + 1)! 2 

  an . lim  a

Then, the series converges absolutely for all y ∈ R. As y = x2 , the series convergent for all x ∈ R; therefore the domain of f is R.

∞  n=0

(−1)n x2n is absolutely (n!)2

22n

b) As the power series is differentiable term by term on its interval of convergence (see Theorem 3.3.3), then, for all x ∈ R,   ∞ ∞  (−1)n  (−1)n 2n 2n−1  2n .f (x) = x = x 2n 2 2 (n!) 22n (n!)2 n=0 n=1 and  .f



∞  (−1)n 2n 2n−1 x 22n (n!)2 n=1

(x) =

 =

∞  (−1)n 2n(2n − 1) 2n−2 x . 22n (n!)2 n=1

Then .x

2

f (x) = x2

.x f



(x) = x

∞ ∞   (−1)n (−1)n 2n x = x2n+2 2n (n!)2 2n (n!)2 2 2 n=0 n=0

∞ ∞   (−1)n 2n 2n−1 (−1)n 2n 2n x = x 2n (n!)2 2 22n (n!)2 n=1 n=1

and .x

2

f  (x) = x2

∞ ∞   (−1)n 2n(2n − 1) 2n−2 (−1)n 2n(2n − 1) 2n x = x . 2n 2 2 (n!) 22n (n!)2 n=1 n=1

270

Series of Functions Therefore, x2 f  (x) + xf  (x) + x2 f (x) = =

=

∞ ∞ ∞  (−1)n 2n(2n − 1) 2n  (−1)n 2n 2n  (−1)n x + x + x2n+2 22n (n!)2 22n (n!)2 22n (n!)2 n=1 n=1 n=0

  ∞  (−1)n 2n(2n − 1) + 2n 22n (n!)2

n=1

x2n +

=

∞ ∞  (−1)n 4n2 2n  (−1)n−1 2n x +  2 x 2n 2 2n−2 2 (n!) (n − 1)! 2 n=1 n=1

=

∞ ∞  (−1)n 4n2 2n  (−1)n 4 2n x −  2 x 2n (n!)2 2n 2 (n − 1)! n=1 n=1 2

.

=

∞  n=1

∞  (−1)n 4 (−1)n 4 2n 2n  2 x −   x = 0. 2n (n − 1)! 2 22n (n − 1)! n=1 2

27. Let (an ) be a sequence of real numbers such that the series a) If the series

∞  (−1)n x2n+2 2n 2 (n!)2 n=0

∞ 

∞ 

an 2n is conditionally convergent.

n=1

an 2n is conditionally convergent, then the radius of convergence of the series

n=1

∞ 

an xn

n=1

is 2, which implies that the power series of x is absolutely convergent if x ∈ ] − 2, 2[ and divergent if ∞  x ∈ ] − ∞, −2[ ∪ ]2, +∞[. It follows that the series an is absolutely convergent (x = 1) and the series ∞ 

n=1 n

an 3 is divergent (x = 3).

n=1

b) The series

∞ 

an is convergent; therefore, by Theorem 2.2.1, lim an = 0. Then lim

n=1

28. Let f (x) =

∞ 

sin(an ) = 1. an

an xn .

n=0

a) We know, by Theorem 3.4.3, that the power series of x is unique and is equal to the Maclaurin series of f , ∞  f (n) (0) n f (n) (0) that is, f (x) = x , ∀n ∈ N0 . Therefore, an = . We will prove, by induction, that n! n! n=0 .f

(n)

(x) = f (x), ∀n ∈ N, ∀x ∈ R.

By hypothesis, f  (x) = f (x), ∀x ∈ R, which proves the equality for n = 1. Assume that f (n) (x) = f (x), ∀x ∈ R. We prove that f (n+1) (x) = f (x), ∀x ∈ R. .f

(n+1)

    (x) = f (n) (x) = f (x) = f (x), ∀x ∈ R.

Then f (n) (0) = f (0) = f  (0) = 1 and an =

f (n) (0) 1 = , ∀n ∈ N. n! n!

3.6. Solved Exercises

271

b) Taking into account part a), we have f (x) = of ex , that is, f (x) = ex . 29.

a) Let f (x) =

∞  xn , ∀x ∈ R. However, this series is the Maclaurin series n! n=0

5x − 1 . As x2 − x − 2 = (x + 1)(x − 2), we have x2 − x − 2

.f (x)

=

5x − 1 A B A(x + 1) + B(x − 2) (A + B)x + A − 2B = + = = . x2 − x − 2 x−2 x+1 (x − 2)(x + 1) (x − 2)(x + 1)

Then .

therefore, .f (x)

=

A+B =5 A − 2B = −1

⇔

A=3 B = 2;

2 3 2 3 2 3 1 + + =− + =− · . x−2 x+1 2−x 1 − (−x) 2 1− x 1 − (−x) 2

1 1 is the sum of a geometric series with ratio r = −x and the first term equal to 1 and x 1 − (−x) 1− 2 x and the first term equal to 1, we have is the sum of a geometric series with ratio r = 2

Since

.

∞ ∞   1 = (−x)n = (−1)n xn 1 − (−x) n=0 n=0

and

∞   x n . x = 2 1− n=0 2 x   These equalities are valid if | − x| < 1 and   < 1, respectively, that is, |x| < 1 and |x| < 2. We can write 2 the Maclaurin series expansion of f :  ∞ ∞ ∞    3   x n 3 .f (x) = − +2 (−1)n xn = 2 (−1)n − n+1 xn , 2 n=0 2 2 n=0 n=0 .

1

which is valid if |x| < 1.  x 2 2 et dt. The function g(t) = et is continuous on R, so the domain of f is R. We know, b) Let f (x) = 0

from Example 3.4.5, that ex =

∞  xn , ∀x ∈ R; therefore, n! n=0 .e

t2

=

∞ ∞ 2n   (t2 )n t = , n! n! n=0 n=0

∀t ∈ R.

As it is a power series, by Corollary 1 of Theorem 3.3.2, it is integrable term by term on every closed bounded interval of R. Therefore,  x  x  x ∞ 2n ∞  x 2n ∞   2 t t 1 dt = dt = f (x) = et dt = t2n dt n! n! 0 0 0 n=0 n! n=0 0 n=0 .

=

 x ∞ ∞ ∞    1 t2n+1 1 x2n+1 x2n+1 · = , = n! 2n + 1 n! 2n + 1 n!(2n + 1) 0 n=0 n=0 n=0

with the expansion valid on R.

272

Series of Functions c) We know, from Example 3.4.5, that .e

x

∞  xn , n! n=0

=

∀x ∈ R.

Therefore, for all x ∈ R and all t ∈ R, .e

−tx

=

∞ ∞   (−tx)n (−1)n tn xn = . n! n! n=0 n=0

Then 

1

f (x) = 0

.

 =

1 − e−tx dt = t

∞ 1 

−

0 n=1



1

1−

∞  (−1)n tn xn n! n=0

t

0

(−1)n tn−1 xn dt = n!





1

−

dt =

∞  (−1)n tn xn n! n=1

t

0

∞ 1  0 n=1

dt

(−1)n+1 xn n−1 t dt. n!

For each x ∈ R \ {0}, the power series of t has an infinite radius of convergence:    (−1)n xn  |x|n     n+1 n! n!  = lim = lim = +∞, . lim  n+1  (−1)n+1 xn+1  |x| |x|     (n + 1)! (n + 1)! and if x = 0, all terms of the series are zero. The interval of convergence of the series is R; therefore, by Corollary 1 of Theorem 3.3.2, the series is integrable term by term on every closed interval of R. In particular, we have on [0, 1] ∞  1  (−1)n+1 xn n−1 (−1)n+1 xn n−1 t t dt = dt n! n! 0 n=1 n=1 0    ∞ ∞ ∞    (−1)n+1 xn 1 n−1 (−1)n+1 xn tn 1 (−1)n+1 n x . t dt = = = n! n! n n · n! 0 0 n=1 n=1 n=1



f (x) = .

∞ 1 

This development is valid on R.  x d) Let f (x) = arctan(t) dt. By the Fundamental Theorem of Integral Calculus, f  (x) = arctan(x). 0





1 1 and is the sum of a geometric series with ratio r = −x2 1 + x2 1 + x2 and the first term equal to 1, we have Considering that

arctan(x)

=

.

∞ ∞    n 1 − x2 = = (−1)n x2n 2 1+x n=0 n=0

if, and only if, |x2 | < 1, that is, |x| < 1. Then  arctan(x) = 0

x

1 dt = 1 + t2

.

=



∞ x 

0 ∞  n=0

(−1)n t2n dt =

n=0

(−1)n



t2n+1 2n + 1

x

∞   n=0 ∞ 

= 0

n=0

x

 (−1)n t2n dt

0

(−1)n

x2n+1 2n + 1

3.6. Solved Exercises

273

because power series are integrable term by term on every interval contained on its interval of convergence (see Corollary 1 of Theorem 3.3.2). By the same theorem, we have    x  x  ∞ ∞  x  t2n+1 t2n+1 dt = dt arctan(t) dt = (−1)n (−1)n 2n + 1 2n + 1 0 0 0 n=0 n=0   ∞ ∞ x n 2n+2 n 2n+2  (−1)  (−1) t x = · = . 2n + 1 2n + 2 2n + 1 2n +2 0 n=0 n=0 =

∞  n=0

(−1)n x2n+2 . (2n + 1)(2n + 2)

Therefore, .f (x)

=

∞  n=0

(−1)n x2n+2 ; (2n + 1)(2n + 2)

this development is valid on ] − 1, 1[.   2 1 =− , we can write e) Taking into account that 2x + 5 (2x + 5)2 .f (x)

=

1 1 =− (2x + 5)2 2



1 2x + 5

 .

We know that

1 n ∞  ∞  1  1 2n 2 5 = = . = (−1)n n+1 xn , − x 2 2x + 5 5 n=0 5 5 n=0 1+ x 5      2  5 5 and this development is valid if − x < 1, that is, x ∈ − , . It is a power series, so it is differentiable 5 2 2 term by term on its interval of convergence (see Theorem 3.3.3), that is,  .

1 2x + 5



 =

∞ 

(−1)n

n=0

Then

2n

xn 5n+1

 =

∞   n=0

(−1)n

2n

xn 5n+1



=

∞  n=1

(−1)n

n2n n−1 x , ∀x ∈ 5n+1

∞ ∞  n2n n2n−1 1 1  =− (−1)n n+1 xn−1 = (−1)n+1 n+1 xn−1 2 (2x + 5) 2 n=1 5 5 n=1   ∞ n  2 (n + 1) n 5 5 = (−1)n x , ∀x ∈ − , . n+2 5 2 2 n=0

f (x) = .

f) Let f (x) =

2 + e2x . We know that 3 + 2x

 ∞  ∞  2x n 2  2 2n+1 xn 2 1 − = = (−1)n n+1 , = · 3 + 2x 3 1 + 2x 3 n=0 3 3 n=0 3      2x  3 3 and this development is valid if −  < 1, that is, x ∈ − , . 3 2 2 .



 5 5 − , . 2 2

274

Series of Functions We also know that .e

2x

∞ ∞   (2x)n 2n xn = , ∀x ∈ R. n! n! n=0 n=0

=

Then .f (x)

=

  ∞ ∞ ∞    (−1)n 2 2n+1 xn 2n xn 2 1 + e2x = = xn , (−1)n n+1 + 2n + n+1 3 + 2x 3 n! 3 n! n=0 n=0 n=0

  3 3 development valid on − , . 2 2 g) By Example 3.4.4, we have the Maclaurin series representation of the function sin(x): . sin(x)

∞ 

=

(−1)n

n=0

x2n+1 , (2n + 1)!

∀x ∈ R.

Then, by Theorem 3.3.3, . cos(x)

that is, . cos(x)

=

 ∞     x2n+1 = sin(x) = , (−1)n (2n + 1)! n=0 ∞ 

(−1)n

n=0

∀x ∈ R,

∞  (2n + 1) x2n x2n = , (−1)n (2n + 1)! (2n)! n=0

∀x ∈ R.

Therefore, for every t ∈ R,  2 2n ∞ ∞    2 t4n n t . cos t (−1) (−1)n = = , (2n)! (2n)! n=0 n=0 and

 .f (x)

x

=



∞ x 

cos(t2 ) dt =

0

0

n=0

(−1)n

t4n dt. (2n)!

The interval of convergence of the series is R; therefore, by Corollary 1 of Theorem 3.3.2, the series is integrable term by term on every closed interval of R. We have 

∞ x 

f (x) = 0 .

=

(−1)n

n=0

∞  n=0

(−1)n (2n)!



  ∞   (−1)n x 4n t4n dt = t dt (2n)! (2n)! 0 n=0

t4n+1 4n + 1

x = 0

∞  n=0

(−1)n x4n+1 . (2n)!(4n + 1)

Thus, the power series development of x is .f (x)

=

∞  n=0

(−1)n x4n+1 , ∀x ∈ R. (2n)!(4n + 1)

3.6. Solved Exercises

275 

h) Let us consider the following derivative:

1 1−x

 =

1 . Based on this, we can express f (x) as (1 − x)2

follows: .f (x)

=

x + x2 = (x + x2 ) (1 − x)2

We know that .



1 1−x

 .

∞  1 = xn . 1−x n=0

Moreover, this development is valid on the interval ] − 1, 1[. According to Theorem 3.3.3, this power series is differentiable term by term on its interval of convergence, that is,  .

1 1−x



 =



∞ 

xn

=

n=0

∞ 

(xn ) =

n=0

∞ 

n xn−1 , ∀x ∈ ] − 1, 1[.

n=1

Then  f (x) = (x + x2 )

=

.

∞ 

1 1−x

n xn +

n=1

= x+

∞ 



= (x + x2 )

n xn+1 =

(2n + 1) xn+1 =

n=1

∞ 

∞ 

n xn−1 = x

n=1

n=1 ∞ 

∞ 

(n + 1) xn+1 +

n=0 ∞ 

n xn−1 + x2

n=1 ∞ 

∞ 

n xn−1

n=1

n xn+1

n=1

∀x ∈ ] − 1, 1[.

(2n + 1) xn+1 ,

n=0

  i) Let f (x) = x arctan(x). We know that arctan(x) =

1 , and this function is the sum of a geometric 1 + x2 series with ratio r = −x2 and the first term equal to 1. Therefore,

.

∞ ∞   1 = (−x2 )n = (−1)n x2n 2 1+x n=0 n=0

if and only if | − x2 | < 1, that is, if and only if |x| < 1. Then, if x ∈ ] − 1, 1[,  arctan(x) = 0

x

1 dt = 1 + t2

.

=



∞ x 

0 ∞  n=0

(−1)n t2n dt =

n=0

(−1)n



t2n+1 2n + 1

x

∞   n=0 ∞ 

= 0

x

 (−1)n t2n dt

0

(−1)n

n=0

x2n+1 2n + 1

because power series, by Corollary 1 of Theorem 3.3.2, are integrable term by term on every closed interval contained in their interval of convergence. Therefore,

.f (x)

=x

∞  n=0

and this development is valid on ] − 1, 1[.

(−1)n

∞  x2n+1 x2n+2 (−1)n = , 2n + 1 2n +1 n=0

276

Series of Functions  j) Taking into account that

1 1 + x2



2x , we can write (1 + x2 )2   2x 1 .f (x) = =− . 2 2 2 (1 + x ) 1+x =−

1 is the sum of a geometric series with ratio r = −x2 and the first term equal to 1. 1 + x2

We know that Thus,

.

∞   n 1 − x2 ; = 2 1+x n=0

this development is valid if | − x2 | < 1, that is, if x ∈ ] − 1, 1[. This is a power series that, by Theorem 3.3.3, is differentiable term by term on its interval of convergence, that is,      ∞ ∞  ∞    1 n 2n . = (−1) x = (−1)n 2n x2n−1 , ∀x ∈ ] − 1, 1[. (−1)n x2n = 2 1+x n=0 n=0 n=1 Then .f (x)

=

∞ ∞   2x =− (−1)n 2n x2n−1 = (−1)n+1 2n x2n−1 , ∀x ∈ ] − 1, 1[. 2 2 (1 + x ) n=1 n=1

      k) Let f (x) = log 2 + x3 . We have f  (x) = log 2 + x3 =

3x2 and 2 + x3 n ∞  ∞  1 1 1 (−1)n 3n x3 1  . = · = x ; −   = 3 3 x 2+x 2 2 n=0 2 2n+1 n=0 1− − 2    x3  √ this development is valid if, and only if, −  < 1, that is, if and only if |x| < 3 2. Thus, 2 .f



(x) = 3x2

∞ ∞   (−1)n 3n 3 (−1)n 3n+2 x = x , n+1 2 2n+1 n=0 n=0

√ √ if and only if x ∈ ] − 3 2, 3 2[.  x f  (t) dt = f (x) − f (0) = f (x) − log(2), we have Since 0



x

f (x) =

f  (t) dt + log(2) = log(2) +

0



∞ x  0

.



∞ 

= log(2) +

n=0

3 (−1)n 2n+1



x

n=0

3 (−1)n 3n+2 t dt 2n+1

 t3n+2 dt

0

because, by Corollary 1 of Theorem 3.3.2, a power series is integrable term by term on its interval of convergence. Then  x ∞ ∞   3 (−1)n t3n+3 3 (−1)n x3n+3 f (x) = log(2) + = log(2) + n+1 2 3n + 3 2n+1 3n + 3 0 n=0 n=0 .

= log(2) +

∞  n=0

and the development is valid on ] −

(−1)n x3n+3 , 2n+1 (n + 1)

√ 3

2,

√ 3

2[.

3.6. Solved Exercises

277  1+x , and calculate its derivative: 1−x

! l) Let f (x) = log





1+x  1−x !  ! 2 1+x 1+x  !  2 1 1+x (1 − x)2 1−x 1−x  = ! = = .f (x) = log = ! . 1+x 1−x 1 − x2 1+x 1+x 2 1−x 1−x 1−x Knowing that have

1 is the sum of a geometric series with ratio r = x2 and the first term equal to 1, we 1 − x2 .

∞  1 = x2n , |x2 | < 1, 2 1−x n=0

and we can write the power series expansion of x of f  : .f



(x) =

∞ 

x2n ;

n=0

the development is valid on ] − 1, 1[. As, by Corollary 1 of Theorem 3.3.2, a power series is term by term integrable on its interval of convergence, we have  x  ∞ ∞  x ∞  2n+1 x ∞    x2n+1 t .f (x) − f (0) = f (x) = t2n dt = t2n dt = = . 2n + 1 0 2n + 1 0 n=0 n=0 0 n=0 n=0 The development of f in a power series of x is .f (x)

=

∞  x2n+1 , 2n +1 n=0

which is valid on the interval ] − 1, 1[.   m) Considering that log(4 − x2 ) = −

1 2x x 1 =− · and that the function g(x) = is 4 − x2 2 1 − ( x2 )2 1 − ( x2 )2  2 x and the first term equal to 1, we can write the sum of a geometric series with ratio r = 2 ∞   ∞  x2n+1 x  x 2n  · .f (x) = − =− ; 2 n=0 2 22n+1 n=0

   x 2  < 1, that is, x ∈ ] − 2, 2[. This is a power series that, by Corollary 1 of this development is valid if  2  Theorem 3.3.2, is term by term integrable on its interval of convergence, that is,  .f (x)−log(4)

=− 0

∞ x  n=0

∞  x 2n+1 ∞   t2n+1 t 1 dt = − dt = − x2n+2 , ∀x ∈ ]−2, 2[. 22n+1 22n+1 22n+1 (2n + 2) n=0 0 n=0

Then .f (x)

= log(4) −

∞  n=0

this development is valid on ] − 2, 2[.

1 x2n+2 ; 22n+1 (2n + 2)

278

Series of Functions 

30.

a) Since

1 1−x

 =

1 , we can write (1 − x)2 .f (x)

We know that Therefore,

=

x =x (1 − x)2



1 1−x

 .

1 is the sum of a geometric series with ratio r = x and the first term equal to 1. 1−x ∞  1 = xn ; 1−x n=0

.

this development is valid on ]−1, 1[. It is a power series that, by Theorem 3.3.3, is term by term differentiable on its interval of convergence, that is,  .

1 1−x



 =

 xn

=

n=0

Then .f (x)

∞ 

=

∞ 

(xn ) =

n=0

∞ 

n xn−1 , ∀x ∈ ] − 1, 1[.

n=1

∞ ∞   x =x n xn−1 = n xn , ∀x ∈ ] − 1, 1[. 2 (1 − x) n=1 n=1

b) Rewriting the general term of the numerical series, we get

.

By item a), with x =

 n ∞ ∞   n 1 = n . n 3 3 n=0 n=1

1 , 3  n   1 1 . n =f =  3 3 n=1

1 3

∞ 

⎧ ⎨ sin(x) , 31. Let f (x) = x ⎩1,

1 1− 3

2 =

3 . 4

if x = 0 if x = 0.

a) We know that . sin(x)

=

∞ 

(−1)n

n=0

therefore, .

x2n+1 , (2n + 1)!

∞  sin(x) x2n = , (−1)n x (2n + 1)! n=0

∀x ∈ R;

∀x ∈ R \ {0}.

Observing that for x = 0, the previous series is convergent and has sum 1, we conclude that .f (x)

=

∞  n=0

(−1)n

x2n , (2n + 1)!

∀x ∈ R.

3.6. Solved Exercises

279

b) Given the uniqueness of the power series development (see Theorem 3.4.3), the series obtained in item a) f (n) (0) is the Maclaurin series of f , so, for each n ∈ N, the coefficient an of xn in the previous series is , n! that is, if ⎧ k ⎪ ⎨ (−1) , if n = 2k, k ∈ N 0 .an = (2k + 1)! ⎪ ⎩0, if n = 2k + 1, k ∈ N0 , then .f

32.

(n)

⎧ k ⎨ (−1) , (0) = n! an = 2k + 1 ⎩ 0,

if n = 2k, k ∈ N if n = 2k + 1, k ∈ N0 .

2

a) Let f (x) = ex + cos(2x). Taking into account that . cos(x)

=

∞ 

(−1)n

x2n , ∀x ∈ R, (2n)!

(−1)n

(2x)2n , ∀x ∈ R. (2n)!

n=0

we can write . cos(2x)

=

∞  n=0

Knowing that .e

x

=

we have .e

x2

=

∞  xn , ∀x ∈ R, n! n=0

∞  (x2 )n , ∀x ∈ R. n! n=0

Therefore, the power series development of f is .f (x)

=

 ∞ ∞ ∞     1 (x2 )n (2x)2n 22n + = + (−1)n (−1)n x2n , n! (2n)! n! (2n)! n=0 n=0 n=0

which is valid on R. b) By Theorem 3.4.3, the power series of x of a function f is unique. Therefore, the series we obtained in the previous item is the Maclaurin series of f , that is, .

 ∞ ∞    1 f (n) (0) n 22n x = + (−1)n x2n . n! n! (2n)! n=0 n=0

Considering that in this series, all terms of odd order are zero, all derivatives of odd order of f at x = 0 are zero. In particular, f (17) (0) = 0. 33.

a) We know that sin(x) =

∞  n=0

(−1)n

∞  x2n+1 1 , ∀x ∈ R. If x = 1, we obtain = sin(1). (−1)n (2n + 1)! (2n + 1)! n=0

b) Upon substituting x = 2 in the equality ex =

∞ ∞   xn 2n , which holds true on R, we obtain e2 = . n! n! n=0 n=0

280

Series of Functions c) We have the following equality, valid on ] − 1, 1[ :

∞  1 = xn . By Corollary 1 of Theorem 3.3.2, we 1−x n=0

can integrate this series term by term obtaining

.

− log(1 − x) + log(1) =

∞ ∞   xn+1 xn = . n+1 n n=0 n=1

1 If x = − , we arrive at the following result: 2

.

34.

− log

  ∞ ∞   (− 12 )n 1 3 = = (−1)n n . 2 n 2 n n=1 n=1

a) Let f (x) = log(x2 + 1) + cos(2x), and calculate its derivative: .f



(x) =

2x − 2 sin(2x). x2 + 1

Taking into account that . sin(x)

=

∞ 

(−1)n

n=0

x2n+1 , ∀x ∈ R, (2n + 1)!

we can write . sin(2x)

=

∞ 

(−1)n

n=0

Knowing that

(2x)2n+1 , ∀x ∈ R. (2n + 1)!

1 is the sum of a geometric series with ratio r = −x2 and the first term equal to 1, we 1 + x2

have .

∞  1 = (−1)n x2n , | − x2 | < 1. 2 1+x n=0

Then, the power series expansion of x of f  is f  (x) = 2x

∞ 

(−1)n x2n − 2

n=0 .

=

=

∞  n=0 ∞  n=0

which is valid on ] − 1, 1[.

∞ 

(−1)n

n=0

(−1)n 2 x2n+1 −  2− (−1) n

∞ 

(−1)n

n=0

22n+2 (2n + 1)!

(2x)2n+1 (2n + 1)!

22n+2 x2n+1 (2n + 1)!

 x2n+1 ,

3.6. Solved Exercises

281

By Corollary 1 of Theorem 3.3.2, we have 

 22n+2 t2n+1 dt (2n + 1)! 0 n=0   x ∞  22n+2 (−1)n 2 − t2n+1 dt = (2n + 1)! 0 n=0   ∞  2 22n+2 = − (−1)n x2n+2 2n + 2 (2n + 2)! n=0   ∞  1 22n+2 = (−1)n − x2n+2 . n+1 (2n + 2)! n=0 

f (x) − f (0) = f (x) − 1 =

.

x

∞ 

(−1)n

2−

The power series expansion of f in terms of x is

.1

+

∞ 

 (−1)n

n=0

1 22n+2 − n+1 (2n + 2)!

 x2n+2 = 1 +

∞ 

 (−1)n−1

n=1

1 22n − n (2n)!

 x2n ,

which is valid on the interval ] − 1, 1[. b) Let x =

π . Substituting this value into the expansion obtained in item a), we get the series 4 .1

∞ 

+

(−1)n−1

n=1

This series has sum f

35.

π 4

= log

π 2n 42n



1 22n − n (2n)!

.

     π π 2 π 2 + 1 + cos 2 +1 . = log 4 4 4

a) We can write .f (x)



= sin (2x) +

2x = sin (2x) + (1 − x2 )2



1 1 − x2

 .

Considering that . sin(x)

=

∞ 

(−1)n

n=0

x2n+1 , ∀x ∈ R, (2n + 1)!

we get . sin(2x)

=

∞  n=0

Since

(−1)n

(2x)2n+1 , ∀x ∈ R. (2n + 1)!

1 is the sum of a geometric series with ratio r = x2 and the first term equal to 1, we have 1 − x2

.

∞  1 = x2n , |x2 | < 1. 2 1−x n=0

282

Series of Functions As, by Theorem 3.3.3, power series are differentiable term by term on their interval of convergence,  ∞  ∞   (2x)2n+1 + f (x) = (−1)n x2n (2n + 1)! n=0 n=0 = .

∞  n=0 ∞ 

(−1)n

∞  (2x)2n+1 + 2nx2n−1 (2n + 1)! n=1

∞  (2x)2n+1 + 2(n + 1)x2n+1 (2n + 1)! n=0 n=0  ∞   22n+1 = + 2(n + 1) x2n+1 , (−1)n (2n + 1)! n=0

=

(−1)n

development valid on ] − 1, 1[. 1 b) Let x = . Substituting this value into the development obtained in item a), we get 2   2n+1  ∞  ∞    1 (−1)n 22n+1 n+1 . + 2(n + 1) = + 2n (−1)n ; (2n + 1)! 2 (2n + 1)! 2 n=0 n=0 therefore, the sum of this series is f

36. Let f (x) =

  1 = sin(1) +  2

1 1 1− 4

2 = sin(1) +

1 + log(4 − x). Taking into account that log(4 − x) = − 4−x  x 1 1 − dt. .f (x) = 4−x 4 − t 3

We know that .



x 3

16 . 9

1 dt, we can write 4−t

∞  1 1 = = (x − 3)n ; 4−x 1 − (x − 3) n=0

this development is valid if |x − 3| < 1, that is, x ∈ ]2, 4[. It is a power series so, by Corollary 1 of Theorem 3.3.2, it is integrable term by term on its interval of convergence, that is,   x  x  ∞ ∞  x ∞   1 (x − 3)n+1 n dt = − . log(4 − x) = − (t − 3) (t − 3)n dt = − dt = − . 4 − t n+1 3 3 n=0 n=0 3 n=0 Then f (x) = .

∞ ∞   (x − 3)n+1 1 (x − 3)n − + log(4 − x) = 4−x n+1 n=0 n=0

= 1+

∞  n=1

37. Let f (x) =

(x − 3)n −

∞ ∞   (x − 3)n n−1 =1+ (x − 3)n , n n n=1 n=1

  1 1 1 . Considering that = − , we can write x2 x2 x       1 1 1 1 .f (x) = − =− =− . x 3+x−3 3 1 + x−3 3

∀x ∈ ]2, 4[.

3.6. Solved Exercises

283

Since .

1 1+

we also have .f (x)

=−

1 3



∞  

=

x−3 3

−

n=0

∞ 

(−1)n

n=0

x−3 3

(x − 3)n 3n

n =

∞ 

(−1)n

n=0



 =

∞ 

(x − 3)n , 3n

(−1)n+1

n=0

(x − 3)n 3n+1

 ;

this development is valid if |x − 3| < 3, that is, x ∈ ]0, 6[. It is a power series that, by Theorem 3.3.3, is differentiable term by term on its interval of convergence, that is, .f (x)

∞  

=

(−1)n+1

n=0

(x − 3)n 3n+1

 =

∞ 

(−1)n+1

n=1

n (x − 3)n−1 . 3n+1

The development is valid on the interval ]0, 6[. 38. Let f (x) =

1 . As x2 + x − 6 = (x − 2)(x + 3), we get x2 + x − 6 .f (x)

=

1 A B A(x + 3) + B(x − 2) (A + B)x + 3A − 2B = + = = , x2 + x − 6 x−2 x+3 (x − 2)(x + 3) (x − 2)(x + 3)

so

.

therefore, f (x) =

.

A+B =0 3A − 2B = 1

⇔

⎧ 1 ⎪ ⎨A = 5 1 ⎪ ⎩B = − ; 5

1 1 1 1 1 1 1 1 · − · =− · − · 5 x−2 5 x+3 5 2−x 5 3+x

= −

1 1 1 1 · − · 5 3 − (x + 1) 5 2 + (x + 1)

= −

1 · 15

1 1 1  . · − x+1 x+1 10 1− 1− − 3 2

x+1 1 and the first term equal to 1 and is the sum of a geometric series with ratio r = x+1 3 1− 3 1 x+1  is the sum of a geometric series with ratio r = −  and the first term equal to 1, we have x+1 2 1− − 2

Knowing that

.

and 

.

1−

 ∞   x+1 n 1 = x+1 3 n=0 1− 3

   ∞  ∞   x+1 n x+1 n 1  = − = (−1)n , x+1 2 2 n=0 n=0 − 2

284

Series of Functions      x + 1   < 1 and − x + 1  < 1, respectively. If x ∈ ] − 3, 1[, we can write the development of f equalities valid if   3  2  in powers of x + 1: .f (x)

=−

     ∞  ∞ ∞  x+1 n 1 (−1)n+1 1  x+1 n 1  1 − (−1)n = − n+1 (x + 1)n . n+1 15 n=0 3 10 n=0 2 5 2 3 n=0 



1 1 1 and = is the sum of a geometric series with ratio t t 1 + (t − 1) r = −(t − 1) and the first term equal to 1, we have

39. Let g(t) = log(t). Since

log(t)

.

=

∞ ∞    n 1 − (t − 1) = = (−1)n (t − 1)n 1 + (t − 1) n=0 n=0

if, and only if, |t − 1| < 1, that is, t ∈ ]0, 2[. Then  log(t) − log(1) = log(t) =

∞ t  1 n=0

.

=

∞ 



(−1)n

n=0

∞  

(−1)n (x − 1)n dx = t 1)n+1

(x − n+1

n=0 ∞ 

=

1

t

 (−1)n (x − 1)n dx

1

(−1)n

n=0

(t − 1)n+1 , n+1

because, by Corollary 1 of Theorem 3.3.2, power series are integrable term by term on every interval contained on their interval of convergence. Again, by Corollary 1 of Theorem 3.3.2, we have ∞ 

 f (x) − f (1) = f (x) =

x

(−1)n

n=0

∞ 

(t − 1)n+1 − (t − 1) n+1 t−1



x

dt =

(−1)n

n=1

(t − 1)n+1 n+1

t−1    x ∞ ∞ x  (t − 1)n (t − 1)n = dt = dt (−1)n (−1)n n+1 n+1 1 n=1 1 n=1   ∞ ∞ x  (−1)n (t − 1)n+1  (−1)n (x − 1)n+1 = = · n + 1 n + 1 n+1 n+1 1 n=1 n=1 1

.

=

dt

1

∞  (−1)n (x − 1)n+1 . (n + 1)2 n=1

Therefore, .f (x)

=

∞  (−1)n (x − 1)n+1 . (n + 1)2 n=1

Moreover, this development is valid on ]0, 2[.   40. Let g(t) = arctan(t − 2). Taking into account that arctan(t − 2) =

1 1 and is the sum 1 + (t − 2)2 1 + (t − 2)2 2 of a geometric series with ratio r = −(t − 2) and the first term equal to 1, we have .

∞ ∞    n 1 − (t − 2)2 = = (−1)n (t − 2)2n 2 1 + (t − 2) n=0 n=0

3.6. Solved Exercises

285

if, and only if, |(t − 2)2 | < 1, that is, t ∈ ]1, 3[. Then, by Corollary 1 of Theorem 3.3.2, 

∞ t 

arctan(t − 2) − arctan(0) = arctan(t − 2) =

2 n=0

.

=

∞ 

∞  

(−1)n (x − 2)2n dx = 

(−1)n

n=0

(x − 2)2n+1 2n + 1

t

n=0 ∞ 

= 2

t

 (−1)n (x − 2)2n dx

2

(−1)n

n=0

(t − 2)2n+1 . 2n + 1

By the corollary mentioned above, we have ∞ 



x

(−1)n

n=0

f (x) =

 dt =

(t − 2)2

2

∞ 

(t − 2)2n+1 − (t − 2) 2n + 1

x

n=1

2

(−1)n

(t − 2)2n+1 2n + 1

(t − 2)2

dt

 ∞  x  (t − 2)2n−1 (t − 2)2n−1 (−1)n dt = dt 2n + 1 2n + 1 2 n=1 2 n=1  x ∞ ∞ n 2n n 2n   (−1) (−1) (t − 2) (x − 2) = · = 2n + 1 2n 2n + 1 2n 2 n=1 n=1 

=

.

=

∞ x 

∞  n=1

(−1)n

(−1)n (x − 2)2n . 2n (2n + 1)

Therefore, .f (x)

=

∞  n=1

(−1)n (x − 2)2n . 2n (2n + 1)

Moreover, this development is valid on ]1, 3[. ∞  f (n) (0) n x . Therefore, we need to calculate f (n) (0), ∀n ∈ N. We will n! n=0

41. We know that the Maclaurin series is

prove, by induction, that f (n) (x) = f  (x), ∀n > 2, ∀x ∈ R.

By hypothesis, f  (x) = f (x) + x, ∀x ∈ R, which implies that f  (x) = f  (x) + 1 and f  (x) = f  (x), ∀x ∈ R which proves the equality for n = 3. Suppose that f (n) (x) = f  (x), ∀x ∈ R. We prove that f (n+1) (x) = f  (x), ∀x ∈ R. .f

(n+1)

    (x) = f (n) (x) = f  (x) = f  (x) = f  (x).

Then, f  (0) = f (0) + 0 = 1 and f (n) (0) = f  (0) = f  (0) + 1 = 2, so the Maclaurin series of f is

.

42.

a) Let f (x) =

π, 0,

∞ ∞   f (n) (0) n 2 n x =1+x+ x . n! n! n=0 n=2

if − π < x ≤ 0 if 0 < x < π.

286

Series of Functions

(b) The periodic extension, f˜, of the function f

(a) The function f

Figure 3.29: Graphs of exercise 42 a)

The graph of f can be seen in Fig. 3.29. By Definition 3.5.3, the Fourier coefficients are

.

a0 =

1 π

an =

1 π

bn

1 = π



π

f (x) dx = −π



1 π



0 −π



0

π cos(nx) dx = −π



" #0 π dx = x −π = π; sin(nx) n

0



0

= 0; −π

cos(nx) π sin(nx) dx = − n −π

0 −π

⎧ ⎨0, cos(−n π) (−1)n − 1 1 = = =− + ⎩− 2 , n n n n

if n is even if n is odd.

Thus, the Fourier series of f is .

∞    π 2 − sin (2n − 1)x . 2 2n − 1 n=1

Since f is a piecewise C 1 function, by Theorem 3.5.1 the series converges to f˜(x) if x = kπ, k ∈ Z, and π takes the value , if x = kπ, k ∈ Z. In Fig. 3.30, we see an illustration of the sum of the series. 2

Figure 3.30: The sum of the series of exercise 42 a)

b) Let us consider the function f (x) =

(a) The function f

−π − ; 2 3 divergent if .a ≤ − 2 m) Absolutely convergent

25.

n) Divergent

b) Convergent

o) Divergent p) Absolutely convergent q) Absolutely convergent

26.

.

8 7

27. 32

330

Answers to Proposed Exercises

Chapter 3 x +x3

1. Convergent on .] − ∞, 0[, .S = e 2

 3 −1 , 1 − ex

divergent on .[0, +∞[   1 2. Convergent on . , 1 ,  3  1 divergent on . − ∞, ∪ [1, +∞[ 3 4. Absolutely convergent on .] − ∞, −1[ ∪ ]1, +∞[ and divergent on .] − 1, 1[, .∀α ∈ R if .x = 1: absolutely convergent if .α < −1 and divergent if .α ≥ −1 if .x = −1: absolutely convergent if .α < −1, conditionally convergent if .−1 ≤ α < 0 and divergent if .α ≥ 0 6.

a) Convergent

7.

a) .]1, +∞[

9.

b)

b) Is continuous

∞  1 − cos(n) n3 n=1

.

a) Converges uniformly √

√  ∞  2 4n5 + 1 − 2n5 + 1 b) . n4 n=1

11.

b)

∞   1

n=1

13.

n

 − log

n+1 n

a) .R

l) Absolutely convergent on .]1, 7[ Conditionally convergent at .x = 7 Divergent on .] − ∞, 1] ∪ ]7, +∞[



b)

∞ sin  . n=1

14.

f) Absolutely convergent on .[−2, 8] Divergent on .] − ∞, −2[ ∪ ]8, +∞[ √ √ g) Absolutely convergent on .] − 2, 2[\{0} √ Conditionally convergent at .x = − 2 and at √ .x = 2 √ √ Divergent on .] − ∞, − 2[ ∪ ] 2, +∞[ ∪ {0}   1 1 h) Absolutely convergent on . 2− , 2+ π π     1 1 ∪ 2 + , +∞ Divergent on . −∞, 2 − π π   5 1 i) Absolutely convergent on . − , − 6    6 5 1 Divergent on . −∞, − ∪ − , +∞ 6 6 √ √ j) Absolutely convergent on .] − 2 − 1, 2 − 1[ √ √ Divergent on .] − ∞, − 2 − 1] ∪ [ 2 − 1, +∞[ k) Absolutely convergent on .] − 1, 3[ Divergent on .] − ∞, −1] ∪ [3, +∞[

10.

.

e) Absolutely convergent on .] − 5, −1[ Conditionally convergent at .x = −5 Divergent on .] − ∞, −5[ ∪ [−1, +∞[



1 n2



1 + (nx)2

a) Divergent on .R \ {0}

  1 b) Absolutely convergent on . − , +∞ 2   1 \ {−1} Divergent on . −∞, − 2 c) Absolutely convergent on .]0, +∞[ Conditionally convergent at .x = 0 Divergent on .] − ∞, 0[ \{−1} d) Absolutely convergent on .] − 2, 2[ Conditionally convergent at .x = 2 Divergent on .] − ∞, −2] ∪ ]2, +∞[

m) Absolutely convergent on .] − ∞, −1[ ∪ ]3, +∞[ Conditionally convergent at .x = −1 Divergent on .] − 1, 3] \ {1}   1 5 n) Absolutely convergent on . , 3 3     5 1 ∪ , +∞ Divergent on . −∞, 3 3 o) Absolutely convergent on .] − 2, 0[ Conditionally convergent at .x = −2 Divergent on .] − ∞, −2[ ∪ [0, +∞[ p) Absolutely convergent on .[−4, 0] Divergent on .] − ∞, −4[ ∪ ]0, +∞[ q) Absolutely convergent on .]0, 2e[ Divergent on .] − ∞, 0] ∪ [2e, +∞[ r) Absolutely convergent on .] − 2, 6[ Conditionally convergent at .x = 6 Divergent on .] − ∞, −2] ∪ ]6, +∞[

Answers to Proposed Exercises

331

s) Absolutely convergent on .] − 1, 5[ Divergent on .] − ∞, −1] ∪ [5, +∞[ t) Absolutely convergent on .[0, 4] Divergent on .] − ∞, 0[ ∪ ]4, +∞[ 15.

17.

a) Absolutely convergent on .R b) Absolutely convergent on .] − a, a[, if .a ≥ b .]

16.

o)

a)

. n=0 .

c)

.

e)

log(a)

xn , for all .x ∈ R

n!

.

1 x2n+1 , for all .x ∈ R (2n + 1)!

∞  (−1)n 22n+1 2n+2 , .∀x ∈ R x (2n + 1)! n=0

.

b) .−π 20.

a)

.

∞   (−1)n 22n+1

n=0

(2n + 2)!

−

1 n+1

 x2n+2 , .|x| < 1

b) .f (15) (0) = 0; .f (16) (0) = −215 − 2 × 15! ∞  −22n+1 (x + 3)n , .− 19 < x < − 45 , .r = 4 7n+1 n=0

21.

.

22.

.

∞  n=0

23.

∞ 

∞  .

7 4

−1 (x − 3)2n , .1 < x < 5 22n+2

(−1)n (n + 1) xn , .|x| < 1, .r = 1

n=0

.

∞  3n n x , for all .x ∈ R j) . n! n n=1  ∞   1 1 − n+1 xn , .|x| < 1 k) . 3 n=0 ∞  . ∞ 

24.

∞  . n=0

(x − 1)n , .0 < x < 2 1 xn , .|x| < 2 2n+1

25. .log(2) +

∞  n=0

(2n + 1) x , .|x| < 1

(−1)n+1 x2n−1 , .∀x ∈ R (2n)!(2n − 1) n=1  ∞   1 1 − n+1 xn , .|x| < 1 n) . 2 n=0

∞  . n=0

n

n=0

m)

a)

∞ 

.

(−1)n x2n+1 , for all .x ∈ R n!(2n + 1) n=0   ∞  1 (−1)n+1 i) . − 1 xn , .|x| < 1 3 2n+1 n=0

l)

19.

∞  (−1)n 2n+1 , .|x| < 1 x 2n +1 n=0 ∞ 

∞  xn+1 , .∀x ∈ R n! n=0

.

b) .2 e

.

1 x2n , for all .x ∈ R (2n)! n=0

n ∞  log(2) (−1)n + n+1 g) . xn , .|x| < 2 n! 2 n=0 h)

a)

n

∞  (−1)n 2n x , .|x| < |a| a2n+2 n=0

n=0

f)

18.

∞  x2n+1 , for all .x ∈ R n! n=0

b)

d)

∞  (−1)n 3n+4 , .|x| < 1 x n+1 n=0

.

− b, b[, if .b > a

∞ 

∞  n−3 n x , .|x| < 5 5n+1 n=0

.

26.

(−1)n (x − 2)n+1 , .0 < x < 4 (n + 1)2n+1

∞  e3 (−1)n + 3 logn (3) (x − 1)n , .∀x ∈ R n! n=0

.

.

27. .2(x − 1) + 3(x − 1)2 + 4

∞  n=0

.0