Mathematics Higher Level Option 10: Series & Differential Equations

Citation preview

MATHEMATICS HIGHER SERIES

LEVEL

OPTION

10

&

EQUATIONS

DIFFERENTIAL

PREFACE This book has been specifically written to meet the demands of the Mathematics Higher Level Option 10: Series and Differential Equations. It presents an extensive and comprehensive coverage of the syllabus and in some sections goes beyond what is required of the student. However, this is for the teacher to decide how best to use these sections with their students. The text has been written in a conversational style so that students will find that they are not simply making reference to an encyclopedia filled with mathematical facts, but rather find that they are in some way participating in or listening in on a discussion of the subject matter. There is an extensive set of examples for students to refer to when learning the subject matter for the first time or during their revision period. There is an abundance of well-graded exercises for students to hone their skills. Of course, it is not expected that any one student work through every question in this text - such a task would be quite a feat. Again it is hoped then that teachers will guide the students as to which questions to attempt. The questions serve to develop routine skills, reinforce concepts introduced in the topic and develop skills in making appropriate use of the graphics calculator. Throughout the text the subject matter is presented using graphical, numerical, algebraic and

verbal means whenever appropriate. Classical approaches have been judiciously combined with modern approaches reflecting new technology - in particular the use of the graphics calculator.

A detailed solution manual has also been included as part of this text book. The solution manual has been made available on a disc, which can be found on the inside of the back cover. To open the files you will need to have Adobe® Acrobat® loaded on your system as well as Microsoft Excel® & Microsoft Word®. As always, we welcome and encourage teachers and students to contact IBID Press with feedback, not only on their likes and dislikes but suggestions on how the book can be improved as well as where errors and misprints occur. There will be updates on the IBID Press website in relation to errors that have been located in the book as well as other relevant material for this option (for example, alternative solutions to those presented in the solution manual) - so we suggest that you visit the IBID website at www.ibid.com.au. If you believe you have located an

error or misprint please email Fabio Cirrito at [email protected]. Fabio Cirrito, September 2006

iii

MATHEMATICS — HL (Option): Series and Differential Equations

CONTENTS 1

SEQUENCES

1.1

Sequences

1.1.1

1.2.1

LIMITS

1 1

Introduction

1.1.2

1.2

AND

.-

2

Visualising sequences — Graphs of sequences

6

Limits of Sequences

8

Introduction

122

8

Defining the limit of a sequence

123 124

11

More limit definitions Algebra of limits

18 22

13 13.1

Limit of a Function Sequences as real-valued functions

27 27

14

Improper Integrals

41

132

141

L’Hospital’s Rule — Limits of indeterminate forms

30

What are improper integrals?

142 143

41

Evaluating improper integrals Comparison test for improper integrals

42 45

2

SERIES

51

2.1

Series

51

2.1.1 2.1.2

2.1.3 2.14

2.15

2.16

Introduction Sum of a series

51 52

Criteria for convergence of a series

53

Necessary condition for series convergence

53

An important mathematical curiosity: The number e

The sufficient Test 1 Test 2 Test 3

56

condition for series convergence — The limit comparison test — D’ Alembert’s criterion (Ratio test) — The Integral Test

57 57 59 66

2.2 2.2.1 222

More Use of Integrals More on comparing infinite series and their integral counterpart Errors in approximating an infinite series

71 71 75

23

Which Test to Use?

79

3

ALTERNATING

31 3.1.1

Alternating Series Introduction

83 83

32 32.1

Other Types of Convergence Conditional and absolute convergence

86 86

SERIES AND

iv

CONVERGENCE

RADIUS

83

33 331 332 333 334

Power Series

34 341 342 343

Calculus with Power Series Power series as a function

Power series and radius of convergence

Convergence interval of a power series Radius of convergence or interval of convergence of a power series? Power series in (x — k) where k = 0

Use of power series to solve some tricky integrals

4.1 4.1.1 4.12 413 414 4.15 4.16 4.1.7 4138

EXPANSION:

TAYLOR

AND

MACLAURIN

SERIES

Series Expansion

Introduction — Polynomials

Coefficients of a polynomial written in terms of its derivatives Advancing to approximating a function by a polynomial From the polynomial approximation to the expressions of a function as an infinite series. Taylor and Maclaurin series Trigonometric functions More Maclaurin expansions Generalising the Maclaurin series for a value of a: the Taylor series

Summary

99

102 102 104 107

Differentiating and integrating power series

SERIES

89 89 90 96

"3 113 113 114 116 119 122 124 125 125

4.2 42.1 422

Series Expansion of Combined Expressions Expansion of composite functions

132 132 133

43 43.1

Applications — Revisited

136 136

51 5.1.1

Taylor expansion involving multiplication

Two definite integrals

DIFFERENTIAL Introduction

EQUATIONS

139 139 139

Reviewing definitions

52 521

Geometrical Methods

141 141

53 531

Numerical Methods

146 146

54 54.1 542

Two New First Order d.ess | Homogeneous first order differential equations

The slope field

Euler’s method

Solving differential equations using the integrating factor

ANSWERS SOLUTION

MANUAL

- see disc

156 156 159 167

MATHEMATICS

- HL (Option): Series and Differential Equations

Dedication I dedicate this book Aram and Igor. The of inspiration. They the difficult times of to fully convey what

to my dearest wife Manola and to my beloved children three of them are the sunshine of my life and my source have also been a source of support and dedication during my illness and my present recovery. There are no words this has meant to me. Eduardo Balanovski

vi

Sequences and Limits - CHAPTER ¥}

1.1.1 INTRODUCTION {

2’

’4’

5 en }

{2,5,9,25,42,...}.

While both sets in (a) and (b) represent a sequence of numbers, set (a) gives the impression that

an ‘obvious’ pattern exists. That is, it seems to represent a well defined sequence of numbers. We say this because each number in this set can be defined by a rule that not only generates that

-

number but also tells us the position of that number in the sequence.

— ———

®

123

Wl

(a)

wiN

Consider the following sets

¥

. The first number in set (a), which we call the first term, is % . To indicate this we use the following notation; u,

= % , with the subscript

‘1’ indicating that it is the first term. The second term, = is

W

N

. .. . . 3 . written as #, 2 = = . Similarly,y the third term is ¥, 3% = 3= and so on. In fact, close observation leads us to deduce that the general term, called the nth term — denoted by «,, , is given by un

=

n m,Wheren

=

1,2,3,

e

r

u, =

n

n+1

, where n €7 +

The domain of this sequence is given by the values of n —i.e., the set of positive integers, while the range (i.e., the numbers in the sequence) is given by the set of rational numbers. Of course,

)

the range will depend on the rule, «,, . The three dots at the end of the sequence mean ‘and so on’.

\

That is, we are referring to an infinite sequence — which is what we will be dealing with in this

book. Because of this, the sequence {u,, u,, -

3, ...} is sometimes written as {u,} _, . .

.

.

o

We are now in the position to have the following definition:

In this course we will only be interested in sequences where u, = f(n),ie., the sequence is

given by a well defined function.

!

MATHEMATICS - HL (Option): Series and Differential Equations EXAMPI.Ei

The nth term of the sequence is given by u, = % .

The first term, when n =1, is given by u, = % ;

.

The second term, when n =2, 1s given by u,

.

..

The third term, when n =3, is

given by u;

’

’

2

=

E

3

=

The fourth term, when n = 4, is given by u,

= % .

3

= =

2 3

=

3

3

= 4_4_1

$T

1

3

2T

16

4

1131 Therefore the first four terms of the sequence are >34

®)

Withu, = n;l

we have,

u, = -11—1 =0; o

2 e

= 2-1_1,

2

= 3-1

3703

U,

2’

_ 2,

3’

= a-1_3

T4

T

123 Therefore the first four terms of the sequence are O, >34

(©)

Thistime u, = 2 sothat u;, = 2, uy, = 2, uy = 2, uy; = 2. The first four terms are 2,2, 2, 2.

That is, each term has a value of 2. In fact, every term in this sequence will be 2 — this is known as a constant sequence

It is important to make appropriate use of a graphics calculator when working with sequences. To do this we need to first set the calculator mode to seq (i.e., sequence mode) and Dot (this is for

when we want to plot the sequence). Once that is completed, by pressing the equation editor we can enter the sequence rule and proceed as required.

Sci

Eng

0123456789 unc

Degree

Par

Conhected athi

Pol

Horiz

Floti Flotz Flots nMin=1

uin =1

uinMing=

glny=

vinhMina=

G-T

iR =

winMind=

Sequences and Limits - CHAPTER

"}

Using the examples in Example 1.1 we have: Step 1:

Step 2:

Ploti Platz Flots rMin=1 UCRIBRA L2

TRELE SETUR ThlStart=1 aTbl=11

uCaMin>Ell

Indernt:

=

Derend:

- Efi”}é” )= winMiny=

Step 1:

Step 3:

RE

[REE

Ask

Ask

|

n_|utid TR S z £

:;

£

7

=1

3EF

'.:I.SEZE

Eags

Enter the expression for the sequence. If you wish, you may also enter the value of u; (ie.,u(nrMin)=0.5)

Step 2:

To view the sequence using the Table format you need to first set up the Table

Step 3:

View the entries by calling up the TABLE.

parameters, In this case, set TblStart to 1 (as this represents n = 1) and then increase the values of n by 1 (i.e., set ATbl

=1).

It is also possible to generate the sequence by using the following process: Step 1:

Press 2nd STAT (this will bring up the LIST screen).

Step 2: Step 3:

Move cursor to OPS and then select seq(. Enter the expression for the sequence using the following syntax: seq(expression, variable, term to start from, last term).

Then store the sequence as a list.

Step 1: 1

Step 2: oPS

MATH

ile 3Lz 4:ly SiLsg 5! RESID

Using u,, = (-1)"*1

Step 3:

E@E t ortD( dimg Fill¢ sedl foumSumC Tlalist(

we have:

MATH

sed(n 2™l 1885 £.5 .9 375 Ang+L1 £.5 .5 .373 a1

u, = (-1)?

uy, = (-1)3 uy = (-1)* uy = (1) Therefore the first four terms of the sequenceare 1,-1,1,-1

(b)

Using the graphics calculator we have:

So the first four terms of the sequence are; 1,-2, 3, 4.

s 1 .25 .25

MATHEMATICS ©)

— HL (Option): Series and Differential Equations

Again, making use of the graphics calculator we have: Givi

seqdCl+((-12"n)n

325

analsd?

1vugthefimtfimrwnnsas0,35,1.

8

2

1.5

Ans+L1 EB 1.0

.GE66566.

.6666666..

+

2

X

|

|

W

1

1.25%

+

—

o

uy = J3+4

2 a.

uy = J1+4 =5, u, = 2+4

=

With u, = J/n+4 we have:

I

(b)

3

?2,

Wi oo

That is, the first four terms are 1,

Wl

I

W

=

1 i66666666?

= ./6;

ug = JA+4 = 8.

That is, the first four terms are Jg, JB, J7.48.

(¢)

. Withu,

. AT = sin-—- we have:

o1 Uy = Sing = 55U 2’ Uy = sin%

That is, the first four terms are %, g,

. 21 fi = Sine = 6 2

=1 and u, = siniy—t = fi 6

2

1, —“g .

So far we have only considered sequences whose rules were written explicitly in terms of . However, it is possible to define sequences implicitly. The process of evaluating successive terms in these sequences does not require too much more effort. The following examples illustrate this.

Sequences and Limits - CHAPTER

.

.

1

Using the expression u, n=1u

=-andu,=

n=2u

=1andu;

n=3u

—éandu

T3

g

1

| = u, + —o U, = i,n =1,2,3, ... wehave: 1

1

u,+§—1

= §+§

1

= 142+2—2

=

1

=1;

1 _ 5

1+Z=

Z;

ra =2, AT=u BT 478 81 1,511 2774 8

Therefore, the first four terms are =, 1, =, —.

(b)

Using the expression u,,; = %u,,—l,ul

=2,n=123,...

1 = iu,—l

1 = §x2—1

=0;

1 n=2,u2=0andu3=§u2—l

1 =§x0—l

=-1;

n=1u

=2andu,

n=3u

=-1anduy,

1 = §u3—1

1 = ix—l—l

wehave:

3 =5

Threfore, the first four terms are 2, 0, —%, —% .

(551

Gt 1l

Write down the first four terms of the sequences 1

(@

u,

©

un=5+;1—l,n€Z+

=

n(n+1)’n

—

&

Z

+

n

u,

)

o, =n+2",nEL

=

c 3+3 =, nE€Z

.

®)

Write down the first four terms of the following sequences, where n EZ*. (a)

U, =

4n?

= 22 n

@

u, = %

( o , we see that it is possible to use algebraic manipulation to help in the 8

Sequences and Limits - CHAPTER

1

investigation of the behaviour of the sequence. We will explore such algebraic manipulations in more detail as we encounter sequences for which such an approach is appropriate. It is also worth noting the similarity between the sequence u, = 2nn+ 1 and the function flx)

= 2;: i . In the same way that u, — % as n — o we have that f(x) — % as x —> .

While we could make use of a graph of the sequence, this time we proceed as in the previous example and use an algebraic approach. 4n? u, = —7’—12n*+5

Next,as n

C . (dividing numerator and denominator by n?).

= —— 2 5 tn

5 > o, = -0

n?

.

Thatis,as n =,y

"

4 sothat ——

5 24+ = n?

4 = ——

240

= 2

4n? L 4n? — 2, Or we could write lim ——— 2n?24+5 n—w0

n

and so we have thatas n —> o, u, —>3-0

=

3.

That is, the sequence converges.

(c)

With the sequence u, = (—21")" we first observe that if n is even then (-1)* = 1, however, if n is odd then (-1)" = —1.

That is, we can rewrite the sequence as u, =

Next,as n — o, 27 — o«

and so,

L2)‘1

| -—n

if n is even ifnisodd

ifniseven,as n—> o, y, — 0 (from above) and

ifnisodd,as

n—

%, u, —> 0 (from below).

In either case, as n — o, u, —> 0 and so the sequence converges.

14

Sequences and Limits - CHAPTER | We display the results using the graphics calculator: Hl:ll:_i P1;t2 Flots rMin=

IR 10 R Ay uCxrMin) BN

2

P

2

2 3 y g

WEn )= vinMind=

5 2

Y inR)=

Using both a visual display and

a table of values it is clearly the

case that the sequence

. ¥=.06451613

converges to 0.

iH:E.iBi‘iBSH For the sequence u, n = q

Flati

=14

|ute 00391 - 00 8.8E-4

-‘:_E‘.FE_.".' e §

T e, —_—_

(d)

n_

B

%eH.191468Y

-c

ZE -12¢ AEZE - (313

G156 007

=1 g ig

|ucted

Flotz Flot3

el unlling e EEQ?;”)”

CULRIERSINLE

winMiny=

A

¥=i

sinn we make immediate use of the ographics calculator: p

n__luin

|| gl | |||k 7 | .6Es8g 3

n=1

n_|udwd

O

[

i2e

fomem| 55051

=14

From the table of values we observe that while the values of the sequence are bounded between two values, being —1 and +1, there does not appear to be a value to which the sequence converges. In fact, we observe that the graph of the sequence never settles on a single unique limit. The graph alongside shows the first three hundred values. In this case we say that the sequence diverges. The important point here is that for a sequence to diverge it does not necessarily mean that it

tends to ‘+o0’. If there is no unique value to which the sequence tends, the operative word being ‘unique’, then the sequence diverges.

15

MATHEMATICS

..

- HL (Option): Series and Differential Equations

8n

=

Lo

As n— o,

42n+3)-12

=

Letu, = 5273 12

=

In+3

12

4~

In+3

.

—>1—2=0.'.un—94—0=4.

2n+73

o

Because u,, approaches a fixed, unique value, the sequence converges.

ii. (b)

Fromi., the limit is 4. That is, limu,

= 4.

n—w

We must show that for any £ > 0 there exists a number N sufficiently large such that 25: 3

4| < ¢ forevery integer n> N .

Now,

Br

2n+3

—4l0,

_ 12 . _2———n+3so.

12 ®2n+35(3-3) For example,

0.001 n>l pie. if e¢ ==L > 3\0001

= 5998.5 . That is, N = 5999

e

’

:

Therefore, as we have been able to determine »n (as a function of € > 0 ) such that we have shown that the limit of

8n 2n+3

_

8n

~-4| 0

Consider the sequences

n+1 ]

{“(‘i) } _

©

a1

{3“1} o

i il. iii.

Show that the sequence converges. State the limit, L, of this sequence. Use the definition of a limit to prove that the sequence has the limit L. .

®

1"

®

{(5) _i}

1+2-5"7

Consider nsider th the sequence e u,, = i3 ——= 5 (a) (b)

i Show that the sequence converges. ii. State the limit, L, of this sequence. Use the definition of a limit to prove that the sequence has the limit L.

17

MATHEMATICS - HL (Option): Series and Differential Equations

1.2.3

MORE

LIMIT DEFINITIONS

We start by considering increasing and decreasing sequences. For example, if we consider the 4 sequence J — n

. . 4 its terms are given by u;, = 4, u, = 2, uy = 3 and so on. Our first n=1

observation is that u; > u, > u3 > ... . That is, the terms of the sequence are decreasing. Similarly, if we consider the sequence {2 — (§>

n-1

Jtsterms are u;

=

1, u,

=

3

5

Uy

=

7

1 , and so

n=1

on. However, this time we have that 1| o\,

) with the expression

n—ow

n

we end up

lim (n + sin(n)) === which is of the form ; and lim (n% +27)

(n * sm(n)) n? +2n

lim

1

n—=x

n—%

this is undefined! To determine this limit we first re-arrange the expression: n+ sin(n)

= ’L*;%z(”) - _# n

n

n

n

(after dividing through by r)

n

14 sin(n) _

n 211

n+ =

Next, lim (1+Si“(”)) = lim (1) + lim (S‘Ln(”—)) =140 =1 n—»w

and

lim

n—

n

n

(n+2—) n

n—

=

o

lim (n)+

n—

o

n—ow

lim

n—

n

o

(2;)

So that the limit now takes on the form 1

o

n

= o (orrather,asn-—>oo,n+g—-—>°0). n

and so,

23

lim

n—so

(H—Slgfin—))

n?+42n

=0.

MIALHEMALICS ~ HL (Option): Series and Differential Equations

©)

u, = Jn?+4-n,n€Z*

(e)

{3’12 }

nl+n

1

,nEZ

o [geed] —

6]

&)

+

,nEZ

_== {3 +2)(2n-5) LREZY

u, = l("; 1 +’31-i) ,nEZ

@E\’}

u, = (l—’l—l)n(%z—l)

®

{2n3_1}n=1,nEZ

4n?

Jn-11"

”

om0 [FH] h

()

u

O

u,

24

m N+

—

w —~ = ~—r

o

s

(a)

+

3

|

=4+(g)n+r§l,n€Z+

o0

u,

&

@2\-;\

Q

Determine the limit of the following sequences.

=

1.

.

,neZt ,REZY

= (i)”n,nEZ+

n=3.

Sequences and Limits - CHAPTER

‘»':;::xenéisss 125 1.

(a)

- MISCELLANEOUS QUESTIONS

Determine the nature of the sequence u, = cos(@)

(b)

"]

+ 1 ,HEZT. n

Find the minimum upper bound and maximum lower bound of this sequence. n_p2.9n

Consider the sequence u, = Fonr 2 . (n+1)-3" (a)

Plot this sequence.

©)

Use (a) and (b) to determine if the sequence is convergent or divergent and find its limit if it exists.

(b)

Prove, using the definition of a limit, that

4.

. .. Use the definition of a limit to show that

5.

Consider the sequence {n(0.9)"}, _,. (a)

(b)

(¢)

(d) (e)

6.

=2.

. lim

=3,

n—o

n

3n2+1 n2—5n

Plot a graph of this sequence.

Determine the values of »n for which the sequence is 1. increasing ii. decreasing.

Determine the minimum upper bound for this sequence.

Deduce that the sequence converges. State the limit of this sequence.

Determine

. @

© 7.

lim 3+24n

n—>w

3n—1\*

)

Ilh—‘;noo (4n + 1)

®)

lim (M)

n—

0

Prove formally that

@

n3

lim

n—o

3n__

n+1

-

1+2+3+...+n

fim o (

n2

n—

)

lim (ndnZ+1-n)

n—ow

3

2

8.

Using the definition of a limit, prove that lim

9.

(a)

Use the definition of a limit to prove that

(b)

Prove, using the definition of a limit, that

——

2

nson+n—1

25

=

lim 1+2-5"

n—w35+3-5%

lim

n—w

= z

(J/r?>+n-n)

3°

=

=

3.

Find an upper and lower bound for this sequence.

N

2.

MATHEMATICS

~ HL (Option): Series and Differential Equations

10.

Determine lim u, where u, = Gn+ 1)(n-2)

1.

Consider the sequence {

n— =

n(n+3)

nE€Z*.

°

o

@ 12.

13.

lal1.

Consider the sequence {u,}, _,,where 2u, o~ u,, —#,

= 0,u; = 1 and u, = -2.

(a)

Using mathematical induction, show that u, = —1 - (—%)

(b)

Hence, determine

n-2

,nEZ".

lim u, .

n—>w

1 Given the real sequence {un}:= | = {(3” + 4n)n }

. n=1

(@) b

(¢) 14.

Provethat {u,} _,=1 isbounded. Prove that (37+47)

ii.

Hence deduce that u, "+, , 3n+l44n+l p>1. n?

nz1.

Determine whether {u,} _, is convergent, giving the reason for your decision.

Consider the sequence {u, }, _ (a)

n+1

i

where u, = [2u,_, u; = J2.

Make use of mathematical induction show that #,>0

and u,, , > u, for n EZ*.

(b)

From part (a), show that u, 0 and x is real. The graph of the function f consists of the set of points (x, f(x)) which is depicted by a continuous curve, whereas the graph of the sequence {un}:= ; consists of the set of points (#x, u,) which is depicted by discrete points:

A 1

From the graphs we see that the sequence {u,,}:o: , Tepresents a discrete sample of its corresponding continuous function. The key observation is that in both instances, the same

behaviour is displayed. This reflects the sense in which sequences are special subsets of their corresponding functions. The benefit of this observation is that for such sequences we can often make use of their functional counterpart. This leads us to the following statement:

It must be noted that the above statement only considers the case where n — o (ie., x — ).

So, how else can this observation help when working with limits of sequences? Let us return to o

Exercise 1.2.3 Q.3. The corresponding real valued function of the sequence { zn 7 } n givenby f(x) =

is n=1

x . ek If we can show that the function f decreases for some x =z x, we can x2+ 0

n

then deduce that the sequence { 7+ 1 } n?+

.

.

will also decrease for some integer value n = ng . To n=1

show that f decreases for x = x; we need to show that f'(x) x,.

So, with f(x) = xzi 1 =/ Then, for x > 1,1 -x2

X

£ n>1,as x" = @ more rapidly than e*, the

X"

’

quotient would tend to 0. It is clear that I’Hospital’s rule can also be applied to a product of the form 0 x o as this product 0

00

can always be converted to the form ~ or =

This product is of the form 0 x —oo . To be able to use L’'Hospital’s rule we need to rearrange the expression so that we may obtain an indeterminate form. We can rewrite

lim

x—>0

xInx

as

lim

x—0

Applying L'Hospital’s rule we have

Therefore,

lim xInx

x—0

li’f

5

1

lim

C

which is of the form

1 X,

——

x—>0 (_l) x2

= 0.

35

=

lim(-x)

x—0

—00 —

o0

.

= 0

MATHEMATICS - HL (Option): Series and Differential Equations

:x—bfi_

'(xsinr—r) . C

X

This limit is of the form O x c . But to use L’Hospital’s rule we need to rewrite the

expression in the appropriate indeterminate form. In this instance we rewrite

lim (xsm:') XxX—>w

. T sin= 0 lim (which is of the form 5 ).

as

o

X—=>

X

X . , .1 . Applying L’Hospital’s rule we obtain lim x—

It '\—.,COS;

x°

7

x

=

X2

. lim

x>

n (T[COS —)

x

X

=T.

Again, not an obvious result.

. . We now solve Example 1.26 by making use of a known limit, namely,

. by observing that the statements

.

.

. . TC lim (xsm —)

x>

. siny lim ——

x—=0

X

=

lim

X

=

. lim

x—=x

() /]

=

1)

. limmx-

. lim

=

1

X

)

Xx—>

.

) sinu

u—=0

U

lim=~=——

=

X

¢

JI)

. (TC sm(——) X

6

X

. sinu = g lim —, u—>0

U

where u = I

X

=T

36

X

=

. 1.We do this

!

sSin

. (T sm(—)

x—>x

. = 7t lim

=

(5

x>

sm(—-)

1 and

. /1 sin{ -

.

- we have that

=

X

.

That is, letting ¥

So,

.

. sinx lim —=

x—=0

=

1.

. | are equivalent.

Sequences and Limits - CHAPTER

"

EXAMPLE 1.27 |

This is of the form 0 x oo, but it can easily be transformed into the form g . This is done by rewriting 5x - 3cotx as 15 cotx . Therefore, using L’Hospital’s rule we have:

()

lim 15 - €% = 15x lim 2% = 15 lim

x—0

l

x—=0/1

(5

x—=0

§

-

(55)¥ (7 —_1)

= 15x lim ( . ¥ =0

\sin2x,

)

15 x x—=0 lim (i)(—x—) \SlnJx Sinx,

15 x [}iino (;;—xfl[hino (fi})} 15x1x1

15

Note that in this problem we applied L’Hospital’s rule once.

This is of the form 0 x oo, but can be easily transformed into the form %o . We do this by rewriting (1 — cosx)Inx

as

hix

(1-cosx) Therefore, lim (1 — cosx)Inx

=

x—=0

lim

hix

x—=0

(1 -cosx)

.,

Applying I'Hospital’s

-

.

rule, we obtain

1

.

x

xh—‘flo

“oinx (1 —cosx)?

37

.

= xh_r.no (

(1 - cosx)? ~sinx

)

MATHEMATICS

- HL (Option): Series and Differential Equations

. 0 , As this is now of the form g Wecan apply L"Hospital’s rule: 2 = lim . 2U 1i 2sinx—sin2 That L is, lim (_Ufl) zcosx)sinx _ p, . 2sinx— sindx x=0

xsinx

x—0

Sinx+ xcosx

x—0

Sinx+ xcosx

As this is still of the form g we once again apply L"Hospital’s rule, which will give us: m

1—0

2sinx — sin2x ——————

SiNX+XxCOSX

=

. lim

x—0

2cosx—2cos2x

O I

2COSX— xSinx

=0

This is a limit of the form g , therefore we are in a position to apply L’Hospital’s rule. With f(x)

=

1 —tanx

and g(x)

= cos2x then f'(x)

= —sec?x and g'(x)

= —-2sin2x.

. —seclx So, lim .1 — —tanx 220X _ iy 232X

2 ® X3 Micos2y T . adsm2x | 22 - .

This is an interesting case as it appears to be 00 , but we can transform it into the form 5

.

-

=

l

= =

= xlnx

kl»—-)

If we take the function y = x¥,then Iny

, where y > 0.

.

—00

apply L’Hospital’s rule, giving:

lim

x—0

HNlle|_.

In this expression we have that as x tends to O, it takes on the form -

=

lim(-x)

x—>0

So it follows that lim Iny = 0, which means that limy x—>0

x—0

=

and so we can

= 0.

1 thatis,

limx*

x—0

=

1.

To help us determine the limit we made use of a special ‘transformation’, i.e., we used Iny instead of using y = x*.This worked because the log function is a strictly increasing function 38

Sequences and Limits - CHAPTER Y} and so the behaviour of y = x* is reflected by the behaviour of Iny . Also, it should be noted that the limit lim

x—>0

x* should more correctly be written as

lim

x—

x~.

0+

It is also important to once again realise that before applying L'Hospital’s rule we need to make ..

.

.

.

.

0

+00

sure that the expression is indeterminate and of the required form, i.e., of the form -~ or =— . +

However, as we have seen in Examples 1.27, 1.28 and 1.30 it is possible to manipulate an . . 0 o, . e . o expression that is not of the form 0" % into one that is of the required torm for L'Hospital's rule to apply.

Determine the following limits.

(@

lim (x——+ anzx) \X—SIn2x

®)

lim ("'.‘ ”)

©

lim 12X

©

lim (%—_1)

©

r—x \ SInX

lim (*‘“2“)

-

CoSx

19

=0

=

1.

2.

Determine the following limits.

(8 3.

x>

®

2x)

lim x—0 \X + sinx

X

@

sinx -1 -

-

hm:_t (

CcOSXx

2

. )

®)

a

_Yll_{l })+ ! ln(

14

1

-

* x)

lim (.\'—sinx) 3

ey

1

.

©

_\-1_1311

. jlmg (tanx+ secx)

. 1 _jlml (l-

(b)

1

1)

(©)

. ] im

2

7.

2%

X+ Inx

Inx—(x-1)

A7 -/

x-1

Determine the following limits, if they exist. (a)

6.

lim

x—o

Determine the following limits. .

5.

®)

Determine the following limits.

@ 4.

lim (%)

. . . . What is wrong in the calculation lim

x>0

COSX

. —sinx = lim

x?

x>0

2x

x—1

. —COSX = lim x>0

Determine the following limits, if they exist.

@

lim Le*

x—=>x

X

(®)

lim &

x—=w

¥

2

) 39

lim (sinx)(Inx)

x—0*

Inx (_\_2 )

2

1 —=? 2

MATHEMATICS

Evaluate the following limits, if they exist. (a) (d)

(&)

(a) )

lim

X — sinx x2e*

(b)

lim

(cosecx - i)

(e)

( cotx) cot2x

(h)

x>0

x>0

m x>0

i

Determine

lim

x?Inx

®

i, (%5]

im (e

)

lim

(cos 2x— cosx)

x>0

sin?x

Sin‘x

0

o

(

ii.

lim

x—=1

)

(559)

Consider the continuous function f with a continuous first derivative such that

o)

Given that lim

X—=7n

Determine

X

lim

= 2, calculate the value f'(st).

xsi®+, [Hint: Let z = x5"¥

-0

1)

)

1.

Show that (1 + x)/x = e(

12.

Determine

13.

Determine the following limits.

lim

lim

x—1*

14.

Determine

15.

(a)

X

and take Inz to transform it to the form ;J .

.Hence. prove that lim (1 +x)!/* = e. x -0+

x(a!/*-1).

X —> 00

1

(a)

©)

X—>

lim

f(@) =0.

10-

1—-cosx —

x—=0

x—>8

(x“ —7x3+8x2 - 2) x3+5x-6

. lim

x—0

x¥-1!

(b)

©

lim (sinx)*

x=0

lim (cosx)l/x

(a)

(b)

x—0

lim

x—0

&N x—1

(b)

Given that lim

S

16.

Evaluate

(a)

17.

Evaluate

(@)

lim

x—=>0

Iim

x—0

L1 x2-1

e , where f is a continuous differentiable function, find

x—»0e*—1

JF'(0) given that f(0)

(x+ 1)

lim

Giventhat f{(1) = 1 and f'(1) = e, calculate lim

!

8.

- HL (Option): Series and Differential Equations

0.

sinox

(b)

sinBx

cosax

— x2

cosfx

40

lim

sinx®

x—0 (sinx)¢

(b)

lim

x—>B

,where o €Z* and § €7Z*. sinZx — sin2f

XZ_p?

1.4.1

WHAT

ARE

IMPROPER

INTEGRALS?

Throughout your course you have, up until now, only encountered proper integrals, that is, b

integrals of the form f f(x)dx where the interval of integration, [a, b], is a finite interval in a

which the function, f(x), is a continuous, bounded function.

In essence, when dealing with improper integrals there are two types of ‘impropriety’ that occur:

Type 1.

The interval of integration may be oo

1

©

1

©

1

e.g.,fx—zdx,f—fidx,fx2+1dx, 1

1

infinite. ©

0

1

.

-2

1

o0

fx2+1dx,

f}—cdx,fcosxdx.

—00

—Co

0

Geometric interpretations (where f(x) > 0) are shown below.

Ay = f(x)

y = f(x) ff(x)dx

ff(x)dx

41

MATHEMATICS —~ HL (Option): Series and Differential Equations Type 2.

The integrand may be unbounded on the interval of integration. ]

2

1

T

2 1 e.g., {x—zdx , _fz;cdx,{lnxdx, —famdx,{tanxdx 1

1

2

The reason here is that the integrand is undefined for particular values of x in the interval of integration. For example, in the expression f Jl—cdx ,the integrand, f(x) -2

=

S 1—

2

is undefined at x = 0.

19

2 Similarly, in the expression f tanxdx , the integrand, f(x) = tanx is undefined at x = T5

0

1.4.2

EVALUATING

The improper integral J‘ f(x)dx

IMPROPER =

INTEGRALS

lim J f(x)dx is the one which will be of most use to us when

we come to deal with infinite series later in this book. However, we now proceed with a number

of examples to see how these integrals can be evaluated.

£)

T%dx: X

lim (fLax = n—»ol tim [—1]X = lim [-1+1] =1 | X2 n

n—>osd

1

n—>®

That is, the integral converges to 1.

(b)

Tidx = n—wdl fim (1dx = n—» lim o [Inx]’ = lim (lnn) X n—»

Now, as n — o, Inn — o and so the integral diverges, i.e., the integral doesn’t exist.

42

Sequences and Limits - CHAPTER

1

E);gAMpLE 1.32

(a)

f: T +x2dx = nler:0 oI5 s2d)c =

lim [arctanx ]

n—ow

=

lim

n—w

0

arctann

T

2 .

b

®)

.

.

n

That is, the integral converges to 5

® ] dx f-ool+x2x

=

0 1 dx f—w1+x2x

+

However, by symmetry, fo

© ] fol+x2

1

—o] + x2

Therefore,

o]

4 x2

dx

=

+ = =

)

dx = =

[

1

01 +x2

de = T 2

g .That is, the integral converges to m.

[EXAMPLE 1,33

(a)

x

fo cosxdx

=

.

1

lim [ cosxdx

n—>wJ0

=

.

-

lim [smx

n—

o

}

n

Q

=

-

lim

n— oc

.

sinn.

The integral diverges because sinn oscillates between —1 and 1 as n —> . »

] (b)

]

—

f 1 ,\/}dx

=

nanZo 1 ,\/} dx

=

i

n erio [2 J}

:l

=

.

-2]. ]

nan:O [2 ’\/;l

The integral diverges because as n —> © , [2/n—2]—> o .

We first note that the graph of f(x)= —2 is discontinuous at x = 0 and so we need to x express the integral as follows: 1 flxzd

01 11 f—lx_zdx+fo,r_2dx

= CErg\_

1;§dx+cl_1.rr(}+

43

11 xzdx

MATHEMATICS We first consider

- HL (Option): Series and Differential Equations

-

-

lim f

c—0-

1 ~dx

C

-1

x=

the integral diverges.

-

Similarly,

lim

1

lzdx =

Therefore, the integral f

lim

c—=»0-L

lim [—1]

ca>0VJeX

integral diverges.

=

c—0tL

!

[—1} X

=

X

c

=

-1

lim

c—>0—

[— %‘ + 1} , which does not exist and so

lim [— 1+ 1] , which also doesn’t exist as the c

c—>0*

! —I—,dx diverges and so doesn’t exist. ~1x?

The integral has two “improprieties”, one at x = 0 (where the function is discontinuous)

and one at the upper limit (). We make use of Examples 1.31 and 1.34 to help us out. ..

= 1

|

»1

We start by writing fo x—zdx = fo x—zdx +f1 x—zdx . Now, fm —ladx = 1 x

1,1.e., the integral converges to 1 (Example 13 1).

I However, the integral fo —lidx was shown to diverge (Example 1.34). X . . . # 1 . e Therefore, the integral fo —dx diverges and so doesn’t exist. X

The function f(x)

|

= ———

is an improper integral.

Lo

———dx

=

f“.\/l—x

J1—x

.

———dx

N—>l'f0 AMl=x

=

.

lim

N—1-

=

[—2,,/1 -X

.

.

l

1

0 J1l=x

]

N 0

lim [-2J1-N+2] N=>I-

=2

So, the improper integral f

!

1

f 0J1-x

N1

lim

.

is discontinuous at x = 1, Therefore the integral [ ——dx

dx converges to 2.

Sequences and Limits - CHAPTER

1.4.3

COMPARISON

TEST

FOR

IMPROPER

1

INTEGRALS

When it is difficult to evaluate an improper integral due the nature of the integrand, for example J‘ e**dx , and all that is required is to determine if the integral converges or diverges, we can 1 compare the improper integral to another improper integral that is known to either converge or

diverge and then make an appropriate decision. This is the essence of the comparison test for improper integrals:

We first notice that there is no immediate known result for f e~Y’dx . So, we make use of a second function that can be readily compared to f(x)

= e* ,forx=1.

Using the function g(x) = e¢* we have that for x=1, g(x) = f(x) . Therefore, as f(x)= 0, if we can show

that fT g(x)dx converges, then by the comparison test o«

we can conclude that f e~**dx also converges. ] = fTe"‘dx

Now, ng(x)dx

=

lim f’:e“xdx

nh—

®

lim [—e‘x

=

n—w

=

}"

1

lim [-e"+1]

=1 =]

=]

Therefore, as f e~Ydx converges, so too does f e~dx. 1 1

Notice then, that the first stage in the comparison process is to guess a function that behaves like

the integrand, which is done by looking at the behaviour of the integrand for large values of x.

45

MATHEMATICS

- HL (Option): Series and Differential Equations

We first search for a function that will ‘behave’ like —31— for large values of x. x> +9 Ask—)w,—~—1—,sothatforx21

J3+9

Next,

i3

f——dx = f x32dx=

lim

—1— §',, . So our problem is now:

Is it any easier to prove the divergence of the comparison series? w

We see that §', = §, = 5+ S0 we need only analyze terms for which n > 2.

C o1y

S4—1+§+(Z+Z)-—1+2X2

-

B

If we consider the sum that will now include the next group of terms, i.e. = + - , we have

1

Then, the sum including the next group of terms is

Sy=

1

/1

1

1.

1.1 —

1

15+ (z+3) + (557 5%8) =

o=

-

-4+

=

— 4

-

-}

=1

14373 3

1

-

Therefore, we can generalize the sum Sy

=

1+kx%,wherék>2.

As k can be as large as we want, it follows that lim §', = n— =%

.

As we already had concluded that S, > §', , it is natural to assert that lim §,= h—>

.

o

Which proves that the harmonic series diverges. This important observation means that we need another condition to prove that a series converges. That is, showing that lim u, = 0 is simply not enough! n—

o

The only conclusions we have at this point is that:

0

At this point it is useful to warn the reader that although the harmonic series

E 1 , a series with n

n=1

.\

.

.

.,

only positive terms, is divergent, the ‘cousin series

o0

1.

E (-1)"= is convergent. n

n=1

Note that the latter series is made up of positive and negative terms that alternate (hence the term

alternating series), according to » being even or odd. We shall expand on alternating series later 0

on and we shall show how

E (-1 )"rll converges. n=1

55

MATHEMATICS

- HL (Option): Series and Differential Equations

2.1.5 AN IMPORTANT MATHEMATICAL THE NUMBER ¢.

CURIOSITY:

Leonhard Euler, the 18th century Swiss mathematician, showed how the number e (base of the

natural logarithm) could be calculated out as the limit of a sequence, namely

lim (1 + }z)

n—

o

.

In the beginning this was thought to be just a mathematical curiosity, but it then became clear that

‘e’ was a constant of paramount importance. Euler’s starting point was the sequence defined by its general term u, = (1 + l) n expressionof u,, , viz.,

n

. If we use the binomial expansion we can write the full

u, = (g) 17+ (T)(l"‘ ‘)(%) + (;)(1"—2)(}1)2 + ...+ (Z) (%) !

This is more conveniently written as:

"y

= 1+n‘1+n(n—1)__1_+n(n—1)(n—2)__1_+m_*_n(n—1)(n—2)...|:n—(n~1)].i n

n

1+1+i

2!

1

n?

n-1 W

3!

n3

1 '(n—l)(n—2)+“

+§

n?

T

n!

nn

+l‘n(n—1)(n—2)...[n—(n—1)] on! nt-1

and in its final form as:

b (100225

oz tete ke (1D e (Y0

Note that all these brackets are positive, therefore the sequence defined by u, is made up of monotonically decreasing terms. Besides, all these brackets in the above expression are smaller than 1.

Therefore '

u,l2.

The right-hand side of the above inequality, from the second term onwards, is the sum of a .

geomeltric

sequence

.

.1

with common

ratio

i .

1

Thus we have

u, 2n

=3-

1

n—1

2 We can see that, for whatever large n, the value of u,, has a finite upper bound, smaller than 3. In fact

This can be seen to be correct by numerically calculating the value of (1 + %z)

56

n

for any large

Series - CHAPTER

value of n, e.g.if n = 101, (1+fi)

1010

2

~2.7182822.

Besides being the base of the natural logarithm, the number e is the base of the exponential function f(x) = e*, which happens to be the inverse of f(x) = In(x). The exponential function has many applications in mathematics (in real and complex analysis) and in physics (radioactive decay; charge and discharge of a capacitor in a DC circuit; as a general description of a travelling wave; in quantum mechanics it gives the interpretation to the wave function and is part of the Schrédinger formalism of QM , etc...)

2.1.6 THE SUFFICIENT CONDITION CONVERGENCE

FOR SERIES

We already said that the necessary condition for series convergence, namely

lim u,

=

n— o«

0, is not

enough to ensure convergence. We now present the other half of the problem, the sufficient condition. When both conditions are taken together, they will provide the certainty that a series converges. As we show next, there are several criteria that can be used to test for series convergence. The

choice of the particular test to be used depends on the specific problem at hand, i.e. on the series we are testing. There is no general recipe as each case is its own, however we shall indicate at the

end some general features that a series must have in order to ascertain the relevance of one or the

other criteria. Ultimately, experience will be the best guide as to the choice of test in each case.

TEST 1 - THE

LIMIT COMPARISON

TEST

This test applies only to series that have positive terms. Let us consider two series that have positive terms S, =uj+u,+us+...+u,+...and

S, = ¥ +uy tu's+ o+

n

U+

If the terms of the first series are smaller than or equal to those of the second series for all values of n,ie., u, u; and &', = ¥ u';.Then,as u, cw

As the terms are positive, we can say that S, < $' and because S, = $', then §, 00("l+1)2+1

.

n-+1 One can see that, as n — oo, this limit tends to the ratio of the monomials of highest

. n? o degree,ie. — — 1, thus D’Alembert’s criterion does not allow us to tell whether the n

series tested converges or diverges. We may suspect the series to converge, as the limit of the general term tends to 0 as n — o . That is, lim

n—x“

>

+

= 0. But for the time being this is only a conjecture as it constitutes only the

necessary condition. In order to convince ourselves that this series indeed converges, we note that

T_, 11 13,1 _3 2

27

2’572

10010

10

10077

So, let us rewrite it as the sum of pairs of terms, viz.,

3t

= (73 60 (o) (o im0+

)G B-G-3-

1 — small number

Thus, after cancellation by pairs, we can see that the sum of the series tends to a fixed, finite value, which is slightly smaller than 1. Hence we can say that the series converges.

61

MATHEMATICS - HL (Option): Series and Differential Equations

. 1 The general term can be written as u, = — , therefore n!

. lim i, =

.1 lim —

n—>x

n—

ol

= 0.

We suspect that the series may converge as the limit of its general term is 0, but we have to

make sure by applying D’ Alembert’s criterion. 1

Asu,=—and n!

o

1im(”_+1_)-

un

"n—x

.

= 1 , the D’ Alembert’s criterion tells us that: (n+1)!

_

u

lim ~2+! = n—

e

u,,,

!

t

=

lim %

_L

n——»oc(”l'f‘l)!

n!

=

. n! lim ——

n——»oc(l'l'f‘

1)

= lim — n-—>

x(ll

+

-n!

1 )

=0 Therefore, as

u

.

lim ol

n—ax

(which is smaller than 1), we can now conclude that according to

U,

D’Alembert’s criterion the series converges .

L ‘Jlnvesti gate the behaviour of the series: E

.

In this case, the limit of the nth term u,, is lim n—

i+

= 0, which indicates that the

series might converge. Next we have

«, =

Zn nZ+1

and u,,,

= __n+2] (n+1)2+1

= =5 n n2+2n+2

.

So, using D’ Alembert’s criterion we have: n+1

lim

n—o

U,

0

=

fim n+2n+2

n—co

n

n?+1

_

i

(n+1)(n2+1)

n—>won(n?+2n+2)

_

Again, we cannot tell whether the series converges or diverges from this result. However we observe that

62

Series - CHAPTER 2

1

N2

/1

1y

3

12\

4

2

74

~=(1-z);==(z—-—=];===+=); = ={-=+-—=];etc..

3

(

2)’5

(2

10)’10

(10+10)’17

Therefore, we can rewrite the series as

(

e N

2)+(2

/11 e

(=12 )+

(-22

10)+(10+10)+
’e°

(- 274

10+170)+
1 and is not a fixed finite number, thus the series diverges.

LS

: Invésti:gatej-:fhé b,éiiaiziour\_of the

We can already tell the behaviour of this series from the limit of the general term. . 2n . As

.

lim =— = oo (i.e., two to the power of n grows much faster than ») then the series

n—-x

diverges as the limit doesn’t equal zero. .

,

,

.

.

o

20+ 1

On the other hand if we use D’Alembert’s criterion, we have u,, = — and u,, | = i1 n

on+

Therefore,

.

u

lim —*!

n—o

u”

=

.

lim

n—

1

n+1 %

.

=

2"

2n+l

lim

n—

o

.

2"

g n+l

n

=

Then, as

.

lim

n—

2n xn

TR £

5

S

. lim

2n

s

L.

n—ooh

R

T

+

1

.

.

= 2> 1, from D’Alembert’s criterion, we also see that the series diverges.

U

Investigate. the behaviour of the series‘:-'l + 2 +é to ~ "2 3 4 .

n

The limit of the general term is lim

1nH—>x

n

[ =

.

n+1

.

.

1 , which is nonzero. Therefore the series

diverges. s

s

Had we used D’ Alembert’s criterion we have u, = .

lim

n—w

22t!= U,

i

n—o

n+1 n

22 N

.

(

n

n+1)?

ll—)oo"l(n+2)

and u,

n+1

| = 35O n

that

.

n+1

Meaning that we cannot deduce the behaviour of the series from this result. However, we note 63

MATHEMATICS - HL (Option): Series and Differential Equations th

-

e

1

n

ntl_

ey

(n+1)2-n(n+2)

=

1

=

(n+2)(n+1)

h

.

(n+2)(n+1)>0teserles

diverges.

The limit of the general term is

lim [r?r%l_)]

n—o

= 0, thus suggesting that the series

might converge.

If we apply D’ Alembert’s criterion, we have u,, = n(n—l-i-l—) and u, ., = m Therefore:

fim .

n E;nor:

u,Y1+l ll"

o gy .

1

n—l—l;noc

+1)(n+2) et D +2) 1

o .

n(n+1) alntl)

=

——

;1£nao(n+

1)(n+2)

n(n+1) lim

n l—l;nw(n + 2)

=1 This shows that we cannot deduce the behaviour of the series from D’ Alembert’s criterion.

However. we can rewrite the general term of the series using the method of partial fractions. That is, using the fact that

_v

11

nn+1)

n

n+l’

. . 1 1 1 1 1 The series can be rewritten as (1 - 5) + (i - 3) + (3 - Zl) +

Cancelling by pairs, we have the partial sum

(3=

+{~-n

53)

n+l

+ ..

S, = l——l— , and thus n+1l

lim$, = lim (1 -—1—)

n—

o

n—

o«

n+1

=1

That is, the sum tends to a fixed finite limit, 1 , which means that the series converges.

In some of the preceeding examples we made use of two common methods (of rewriting terms) when studying the behaviour of series, namely — 1. 2.

Telescoping series Partial fractions

We consider each of these in turn:

64

.

Series - CHAPTER Telescoping series E (% - —

=

+

+

S

S——

I

S[ -

+

+

+

Al—

S——

|

]

+

W=

S—

= +

|

|—

+

S——

[ —

|

+

+

[

ok | pomd

—_

1) we end up with a series that looks like

G Gyt

n=1

|

4+ | —

(Y

e

=

|

S

—A

=

When the terms of series are of the form

W[

1.

IS0k = (o] e he Do (b v (B = lim[l—

()]

1 ]

n—o

n+1

=1

-

o/l

1

giving the result that E](;l_n+l)

=1

n=

The great thing about telescoping series is that not only can we determine that the series converges, but it has the advantage that it provides the limiting value of the series.

2.

Partial fractions

Closely following on from telescoping series, there are expressions in which we first need to

rewrite the terms as the sum of partial fractions. o

For example, consider the series

partial fractions. That is,

El n(n+l

ne

1 nn+1)

1

)

, we would first express

n(Tl-l-l—)

as a sum of

1 n n+1l’

We obtain the partial fractions by solving for A and B and assuming that the fraction can be ‘broken up’ as: 1

A

= —

n(n+1)

B

h

n+n+1,fromwerewege

Therefore, equating the coefficients we have: And so,

Eln(Tl-l-—l—) n=

=

El[}l_fi]

=

t

1

n(n+1)

=E

A(n+ 1)+ Bn

n(n+1)

=1

=A(r+1)+Bn

lim % l:ln(lnx) } , x = 2lim —>

[In(lna)-In(In2)]

0 o

Therefore, as f: %vdx

0,

E 2 & nln(n)

diverges

ges.

N Let us define the function f(x) = L . Consider the integral f —l—a’x . xP

pr#l,fl

A

x—pdx

_[xetl =

N

[:p_fi]l

. N1 Then, if p = l,f1 ;dx

=

1 xP

1

—_;(N

Cpet 4

—'1)

N = fl }dx = In(N).

If we now make N — o, we can study the convergence of the series.

1.

fp>1, fT :},-’dx

—1—1 the integral is finite and thus the series converges. p_.

2.

Kp11< 5 + %

.

+... 4+ m

+ ...

Apply D’ Alembert’s criterion to study the convergence of the series: (a) (b)

1 3x2 1

! 4x22 8

i—!+3_!

1 + ! + +—1—+ 5x23 6x24 T (n+2)2n 3 27+64 n E

§+.-.+zn+—1)!+.-.

©

Lpastomoo@e-nt 2Tttt T

@

% 22x|4+2x34!x6+"'+2x4xn.'..x2n

(e)

3+§x%+g

®

1+§—?+§—?+...+z—7+...

®

F+() () (i)

(h)

1+§8-!+%+...+(2Ln—_31)—!+...

%++giii;x3—,}_—l+

Use the integral test to study the series: (a)

1%2-+$+...+1117n+...

b)

1+%+é+...+nlz+

©

%+A—}_g+%+...+

@

é+;2—2+%+..

2:+1+...

g

o«

Determine if the series

E

7

converges or diverges and determine its value if possible.

n=1

69

IVIALHEMATLICS — HL (Option): Series and Differential Equations 9.

Determine if the series

}: 2(%)

converges and determine its value if it does.

n=1

10.

(@)

(b)

i

On the same set of axes sketch the graphs of h(x) = ./x and g(x) = Inx.

.. il.

. Find

. Inx lim —.

xX—>

™

X

Determine if the series

2

lz_zk converges or diverges.

k=1

M.

(a)

Using the limit comparsion test, show that the series

2 —2——,where 1 n=1

a>0,b>0,c>0

an*+bn+c

converges.

@

(b) (c)

Show that

2

& n?+3n+2

converges by making use of limit comparsion test.

i.

Express 1

ii.

Use the integral test to show that

.

Show that

n2+3n+2

in its partial fractions, A

n+l

+ B

,A BER.

n+2

12.

(a)

L1 50 =

2

n+20

n=1

g

o«

ii. b)

Ne

=]

2

2

& n2+3n+2

converges.

1c.

k=11

1

Hence, deduce that "21 57230

diverges.

Make use of the integral test to confirm your result of (a) ii. 1

13.

Using the limit comparison test, decide if 2 converges or diverges. a1 AR+ 2n

14.

Use the integral test to decide which of the following series converge. x

(@)

=]

2

Efi o

(b)

n=

15.

(@

«©

2

2

m

©

n=2

IT(x)

= ln(xj_ 1)

X

, show that lim T(x) = -1. X—

Hence decide if the series

2

ln(fi—i)

n

converges or diverges.

n=1

16.

Use the ratio test to determine the convergence or divergence of

@

L

3Z02

1

=14+=—1. +n 1.

—

!

1

=, 2Infact,AU

—

>=

+

)

e

< foxdx < A, , but more importantly, as the number of partitions increases

-

1 and A, 1 5 n

We have shown that

Ay

+

1

, it must be the case that foxdx _

1

n

=

D=

Asn—— o,

1

2 (k—n—l) _lrl sfoxdx < E (5)’1’ and that as n — m,foxdx — % . k=1

72

Series - CHAPTER 2 The reason for having spent time going through a detailed example of how the integral is

evaluated via the use of an infinite series is to highlight the relationship that exists between

integration and series and to provide a little more justification for the method used in Examples 2.10,2.11 and 2.12.

The difference between the example we have just worked through and the limit of a series is that with a series, the partitions [or ‘widths’] are always 1 as opposed to the partitions [or widths] of -, as in the example given. So, when referring to a series, we can simplify our result to: n

of course, if

¢ a

x)dx

diver, g €s, SO too will the series — which

was the P urpose of making& use of

improper integrals when using the integral test!

o

From our result above, we have izl (21+—17)2 =~ f:o mdx Now, let u =

2x+3:>d—u

oLl=[a

=

2.'.-1-du

= dx.

2

1, 2du 1o = zfu s i

1 Andf(2x+3)2dx

dx

.

du

—1u‘1+c 2

1

T2(2x+3)" € o

1

.

n

1

. There [ o—— heref fore[,, Te v apdx 3™ == lim 2 w3 ydx = = I

1

n—w

_l

lim

2n— o

1

10

1 So,0, we have have thattha i§1(2i+3)2 So0 ——— ~ —G1 .

73

[ T2(2 2 1 and so

n:lnp

E

1

converges,

n=1n3

meaning that we can make use of the above result to determine R, . Thatis,from 0

(b)


—,ifa=1;

(©

1

12.

(b)

-1

13.

(©

convergent

14.

(¢

2

M

1

(©)

2

2

(a

0

M

0

()

b

=

(b)

~%

(c)

1

6.

(b)

0

(©)

0

—

Exercises

@a

©

—%

@

0

(@

M.

a

Exercises 1.

(@

3.

(a

7.

i. (©)

1

.

1,35

¢

lb=

n=9 0

81

ii.

minub=1;maxlb=-1

(c)

converges to 0.

n=29

(d)

—%,ifa=—1

(©

diverges

9(0.9)° 10.

(e)

0

3

and n is odd; «, if a = -1 and» is even

1.3 3

|1

(a)

doesn’t exist.

(a)

undefined

@@

O

(b)

D

6.

ub=

(b)

©@

:

4

169

@

0

®

0

©

2

o0

see soln. manual

O

(f

doesn’texist

MATHEMATICS

- HL (Option): Series and Differential Equations

®@

2

®

5

O

-

9.

@@

i

—%

i

e

B

2

10.

1

M.

e

1B.

@

e

b

1

©

1

14

(@

1

b

1

5.

@

2

B

e

16.

(@