Monthly Problem Gems [1 ed.] 0367766787, 9780367766788

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Monthly Problem Gems

Monthly Problem Gems

Hongwei Chen

Christopher Newport University

First edition published [2021] by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN © 2021 Hongwei Chen CRC Press is an imprint of Taylor & Francis Group, LLC Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright. com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. Library of Congress Control Number: 2020951721 ISBN: 978-0-367-76678-8] (hbk) ISBN: 978-0-367-76677-1 (pbk) ISBN: 978-1-003-16806-5 (ebk) Typeset in CMR10 font by KnowledgeWorks Global Ltd.

Contents

Preface

vii

1 Limits 1.1 A rational recurrence . . . . . . . . . . . . . . . . . 1.2 Asymptotic behavior for a P´olya type recurrence . . 1.3 A nonlinear recurrence with Fibonacci exponents . . 1.4 Rate of convergence for an integral . . . . . . . . . . 1.5 How closely does this sum approximate the integral? 1.6 An arctangent series . . . . . . . . . . . . . . . . . . 1.7 Geometric mean rates . . . . . . . . . . . . . . . . . 1.8 Limits of the weighted mean ratio . . . . . . . . . . 1.9 Limits of mean recurrences . . . . . . . . . . . . . . 1.10 A disguised half-angle iteration . . . . . . . . . . . . 1.11 Nested radicals and generalized Fibonacci numbers 1.12 A limit involving arctangent . . . . . . . . . . . . . 1.13 Summing to the double factorials . . . . . . . . . . 1.14 A fractional part sum with Euler’s constant . . . . . 1.15 A Putnam/Monthly limit problem . . . . . . . . . .

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1 1 4 9 12 15 19 21 24 27 32 36 40 43 47 50

2 Infinite Series 2.1 Wilf wants us thinking rationally . . . . 2.2 Old wine in a new bottle . . . . . . . . 2.3 Another Euler sum . . . . . . . . . . . 2.4 Reciprocal Catalan number sums . . . . 2.5 Catalan generating function . . . . . . 2.6 A series with log and harmonic numbers 2.7 A nonlinear harmonic sum . . . . . . . 2.8 A series involving Riemann zeta values 2.9 Abel theorem continued . . . . . . . . . 2.10 A convergence test . . . . . . . . . . . . 2.11 A power series with an exponential tail 2.12 An infinite matrix product . . . . . . .

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53 53 57 70 75 79 87 91 95 99 103 106 110

3 Integrations 3.1 A Lobachevsky integral . . . . . . . . . . . . . . . . . . . . . 3.2 Two log gamma integrals . . . . . . . . . . . . . . . . . . . .

115 115 119

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v

vi

Contents 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15

Short gamma products with simple values . . . . . Evaluate an integral by Feynman’s way . . . . . . Three ways to evaluate a log-sine integral . . . . . A log-product integral . . . . . . . . . . . . . . . . A Putnam problem beyond . . . . . . . . . . . . A surface integral with many faces . . . . . . . . . Evaluate a definite integral by the gamma function Digamma via a double integral . . . . . . . . . . . Another double integral . . . . . . . . . . . . . . . An integral with log and arctangent . . . . . . . . Another integral with log and arctangent . . . . . An orthonormal function sequence . . . . . . . . . An definite integral Quickie . . . . . . . . . . . . .

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123 127 131 137 139 143 148 151 155 161 166 176 179

4 Inequalities 4.1 An inequality from Klamkin . . . . . . . . . . . . . . . . . 4.2 Knuth’s exponential inequality . . . . . . . . . . . . . . . . 4.3 Tight bounds for the normal distribution . . . . . . . . . . 4.4 An inequality due to Knopp . . . . . . . . . . . . . . . . . 4.5 A discrete inequality by integral . . . . . . . . . . . . . . . 4.6 An inequality by power series . . . . . . . . . . . . . . . . . 4.7 An inequality by differential equation . . . . . . . . . . . . 4.8 Bounds for a reciprocal of log sum . . . . . . . . . . . . . . 4.9 Log-concavity of a partial binomial sum . . . . . . . . . . . 4.10 A bound of divisor sums related to the Riemann hypothesis

. . . . . . . . . .

181 181 187 193 197 204 207 212 217 222 225

5 Monthly Miniatures 5.1 Value defined by an integral . . . . . . 5.2 Another mean value theorem . . . . . . 5.3 The product of derivatives by Darboux’s 5.4 An integral-derivative inequality . . . . 5.5 Eigenvalues of a (0, 1)-matrix . . . . . . 5.6 A matrix of secants . . . . . . . . . . . 5.7 A gcd-weighted trigonometric sum . . . 5.8 A lcm-sum weighted Dirichlet series . . 5.9 An infinite sum-product identity . . . . 5.10 A polynomial zero identity . . . . . . .

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235 235 239 243 246 250 255 263 267 273 280

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A List of Problems

285

B Glossary

289

Bibliography

303

Index

311

Preface

Problems and their solutions lie at the heart of mathematics. The problem section in The American Mathematical Monthly (AMM) provides one of the most challenging and interesting problem sections among the various journals and online sources currently available. According to the statistics on JSTOR, not only is the AMM the most widely-read mathematical journal in the world, but its problem section is the most often downloaded article. The published problems and solutions have become a treasure trove rife with mathematical gems. This book is an outgrowth of a collection of 62 monthly problems that the author has worked in the last two decades. Each selected problem has a central theme, contains gems of sophisticated ideas connected to important current research, and opens new vistas in the understanding of mathematics. This collection is intended to encourage the reader to move away from routine exercises and toward creative solutions, as well as offering the reader a systematic illustration of how to organize the transition from problem-solving to exploring, investigating, and discovering new results. This book seeks to elucidate the principles underlying these ideas and immerse the readers in the processes of problem-solving and research. For example, our solution to Problem 10949 (a weak version of the Riemann hypothesis, see Chapter 4.10) provides an elementary gateway to the Riemann hypothesis, which has appeal to anyone with rudimentary exposure to number theory. This book also emphasizes historical connections, provides interesting anecdotes, and adds the details of existing results that have been relegated to widely scattered books and journals. To fulfill the above goals, this book follows a very structured format: 1. State the numbered Monthly problem along with its proposer(s); 2. Address the heuristic consideration of the solution — a careful discussion of the challenges and related mathematical ideas needed to solve the problem, focusing on mathematical insight and intuition; 3. Present one or more solutions based on the heuristics; most are either my published solution in the Monthly or a solution I submitted to the Monthly (which is different from the published one), aimed at presenting and developing problem-solving techniques; 4. Reveal possible extensions of the problem to promote further research. The materials vary in their degree of rigor and sophistication vii

viii

Preface to accommodate readers with diverse interests. This often refers the reader to additional related Monthly problems (more than 180), but some extensions may serve as a springboard to further research and lead to publications.

This book is organized into five chapters as follows: Chapter 1. Limits. This chapter contains 15 Monthly problems involving the various ways in which the notion of limit may appear. In particular, it shows how to use the Stolz-Ces` aro theorem to determine the growth rate of various P´ olya-type sequences. Chapter 2. Infinite Series. This chapter covers 12 Monthly problems related to series, including those involving Euler sums and Riemann zeta values, and presents deeper techniques like special integrals, Abel’s summation formula, and Fourier series to test the behavior of the series and to determine their values. Chapter 3. Integrations. This chapter presents a combination of approaches to evaluate 15 irresistible Monthly integral problems. These evaluations provide nice applications of infinite series, gamma and beta functions, parametric differentiation and integration, Laplace transforms, and contour integration. Chapter 4. Inequalities. This chapter collects 10 Monthly problems involving inequalities. Their solutions encounter many familiar characters, among them: Cauchy’s inequality, L’Hˆopital’s monotone rule, majorization and convexity, Hilbert’s identity, Hardy’s inequality, and differential equations. What makes our trip worthwhile is the realization that these old chestnuts still have something new to tell us. At any rate, they all appear here in novel and surprising ways. Add them to your inequality tool box. Chapter 5. Monthly Miniatures. This chapter introduces 10 Monthly problems related to the mean value theorem, eigenvalues and determinants of some special matrices, weighted trigonometric sums and Dirichlet series, infinite sum-product identity, and polynomial zero identities. In addition to developing valuable techniques for solving these problems, it also includes some interesting historical accounts of ideas and contexts. Among them, we will encounter various extensions of the mean value theorem, Chebyshev polynomials, (0, 1)-matrices, the Cauchy matrix, Dodgson condensation, Dirichlet convolution, the Woods-Robbbins identity, and the alternating sign matrix conjecture. This chapter is intended to facilitate a natural transition from problem-solving to independent exploration of new results. Most of the ideas involved in the solutions will be comprehensible to the good undergraduate math student. A rich glossary of important theorems and formulas is included for easy reference. For some definitely challenging problems, I have endeavored to provide the related background so that they

Preface

ix

can be followed and enjoyed without frustration. As problems in each new issue of the AMM arrive, the reader may regard this book as a starter set, acquire a jumping of point to new ideas, and extend one’s personal profile of problems and solutions. I wish to take this opportunity to express my thanks to Dr. Brian Bradie, who has shared many years of enjoyment of problem-solving, and carefully read the earlier draft of the manuscript, and corrected many errors. I am also grateful to my chair, Dr. Christopher Kennedy, and dean, Dr. Nicole Guajardo, for their constant encouragement and support. Dr. Kennedy also read the most of manuscript and offered his help in making it better. Special thanks to Christopher Newport University for granting me a sabbatical to write this manuscript. Thanks to Steven Kennedy, Gail Nelson, and the anonymous reviewers for their many valuable suggestions for improving the book. Thanks also to Robert Ross and the CRC Press for their help in making this book a reality and for all assistance they offered. Finally, thanks to all individuals who contributed problems and solutions, from whom I have learned much. Naturally all possible errors are my own responsibility. Comments, corrections, and suggestions from readers are always welcome: [email protected]. Thank you in advance.

1 Limits

We begin with 15 limit problems with the intent of providing an implementation organizing the natural transition from problem solving toward exploring, investigating, and discovering new results.

1.1

A rational recurrence

Problem 11559 (Proposed by M. Bataille, 118(3), 2011). For positive p and x ∈ (0, 1), define the sequence {xn } by x0 = 1, x1 = x, and, for n ≥ 1, xn+1 =

pxn−1 xn + (1 − p)x2n . (1 + p)xn−1 − pxn

Find positive real numbers α, β such that limn→∞ nα xn = β. Discussion. Although a rational recurrence is simple in its form, it is usually difficult to find an exact formula for the general term. Here we try to reformulate the problem into an equivalent but simpler form. First we observe the ratio yn+1 := xn+1 /xn satisfies a first-order rational recurrence, and that the limit to be proved implies yn+1 → 1. We then recall that the solution to the special case zn zn+1 = 1 + azn has a closed form. Taking into account these facts, we are ready to start. Solution. Let yn+1 := xn+1 /xn for n ≥ 0. It is easily verified that yn+1 =

p + (1 − p)yn . (1 + p) − pyn

This implies that 1 − yn+1 =

1 − yn . 1 + p(1 − yn ) 1

2

Limits

Setting zn+1 = 1/(1 − yn+1 ) yields zn+1 = zn + p. Repeatedly applying this recurrence leads to zn+1 = z1 + np = Thus yn+1 = 1 −

1 1 + np = + np. 1 − a1 1−x 1

zn+1

=

p(1 − x)n + x , p(1 − x)n + 1

and xn

n−1 Y xk+1 = yk+1 xk k=0 k=0 n−1 Y k + a Y  p(1 − x)k + x  n−1 = , = p(1 − x)k + 1 k+b

=

n−1 Y

(1.1)

k=0

k=0

where a = x/p(1 − x), b = 1/p(1 − x). Since Γ(x + 1) = xΓ(x), we have Γ(n + x) = (n − 1 + x)(n − 2 + x) · · · (1 + x)(x) Γ(x), and so

n−1 Y

(k + x) =

k=0

Thus xn =

Γ(n + x) . Γ(x)

Γ(n + a)Γ(b) . Γ(n + b)Γ(a)

(1.2)

Using Stirling’s formula for the gamma function, √ Γ(x) ∼ 2π xx−1/2 e−x , we readily deduce that √ 2π (n + a)n+a−1/2 e−(n+a) Γ(n + a) ∼ √ Γ(n + b) 2π (n + b)n+b−1/2 e−(n+b)
n+a−1/2 −a nn+a−1/2 1 + na e ∼
n+b−1/2 b nn+b−1/2 1 + n e−b ∼

na ea e−a = na−b = n−1/p . nb eb e−b

From (1.2) it follows that lim n1/p xn = lim n1/p

n→∞

n→∞

Γ(n + a)Γ(b) Γ(b) = , Γ(n + b)Γ(a) Γ(a)

(1.3)

A rational recurrence

3

giving α=

1 Γ(b) , β= . p Γ(a) 

Remark. The asymptotic ratio (1.3) can also be obtained by the formula (see [49], 8.328.2, p. 945) Γ(x + a) lim = 1. x→∞ xa Γ(x) In particular, if 0 < a < 1, Wendel [93] established the following elegant inequality: 1−a  Γ(x + a) x ≤ a ≤ 1, for x > 0. x+a x Γ(x) A full expansion of the asymptotic ratio due to Tricomi and Erd´elyi [91] is given by ∞ Γ(z + a) X (−1)n σ Γ(b − a + n) 1 ∼ Bn (a) , b−a+n Γ(z + b) n! Γ(b − a) z n=0 with |arg z| ≤ π/2 − δ, δ > 0, Re(b − a) > 0 and the Bnσ (z) (often called as generalized Bernoulli polynomials) satisfy ∞ X tσ ezt tn = Bnσ (z) , t σ (e − 1) n! n=0

|t| < 2π.

Mellin’s formula ([49], 8.325.2, p. 945), ∞  Y 1− k=1

y z+k

 e

y/k

eγy Γ(z + 1) , Γ(z − y + 1)

 =

(z > y > 0),

(1.4)

where γ is Euler’s constant, provides another approach to solve this problem. First, we have " !# n−1 n−1 Y X y −y/k lim n e = exp lim y ln n − 1/k = e−γy , (1.5) n→∞

n→∞

k=1

k=1

where we have used the well-known limit lim

n X

n→∞

! 1/k − ln n

= γ.

k=1

Next, we rewrite (1.1) as xn =

n−1 Y k=0



k+a k+b

 =

 n−1  a Y b−a 1− . b k+b k=1

4

Limits

Because a nb−a xn = b

nb−a

n−1 Y

! e−(b−a)/k

k=1

n−1 Y k=1



b−a 1− k+b



! e(b−a)/k

,

Using (1.4) and (1.5), again, we find that lim nb−a xn =

n→∞

a −γ(b−a) eγ(b−a) Γ(b + 1) Γ(b) e = . b Γ(a + 1) Γ(a)

We now end this section with three more Monthly problems for additional practice: 1. Problem E3356 (Proposed by W. W. Chao, 98(4), 1991). Suppose a1 = 1 and an+1 = n/an for n = 1, 2, . . . . Evaluate   1 1 1 −1/2 + + ··· + . lim n n→∞ a1 a2 an 2. Problem 11528 (Proposed by A. Sˆınt˘am˘arian, 117(9), 2010). Let p, a, and b be positive integers with a < b. Consider a sequence hxn i defined by nxn+1 = (n + 1/p)xn and an initial condition x1 6= 0. Evaluate xan + xan+1 + · · · + xbn lim . n→∞ nxan Hint: Let yn = (n−1)xn /(1+1/p). Then show that xn = yn+1 −yn . 3. Problem 11659 (Proposed by A. Stadler, 119(7), 2012). Let x be real with 0 < x < 1, and consider the sequence {an } given by a0 = 1, a1 = 1, and, for n > 1, an = Show that

a2n−1 . xan−2 + (1 − x)an−1

∞ X 1 = (−1)k xk(3k−1)/2 . n→∞ an

lim

k=−∞

1.2

Asymptotic behavior for a P´ olya type recurrence

Problem 11995 (Proposed by D. S. Marinescu and M. Monea, 124(7), 2017). Pn−1 Suppose 0 < x0 < π, and for n ≥ 1 define xn = n1 k=0 sin xk . Find √ limn→∞ xn ln n.

Asymptotic behavior for a P´ olya type recurrence

5

Discussion. Before proceeding to this problem, it is interesting to recall P´olya’s classical recurrence ([75], Problem 173, p. 38): Suppose 0 < x0 < π, and for n ≥ 1 define xn = sin(xn−1 ). Then r n xn = 1. lim n→∞ 3 There are a number of ingenious proofs. Based on the Stolz-Ces`aro theorem, Chen gave a short proof as follows: Since sin x < x for all x > 0, it follows that 0 < xn < xn−1 whenever 0 < x0 < π. Thus, xn converges as n → ∞. Let limn→∞ xn = L. Taking the limit in xn = sin(xn−1 ) yields L = sin L, which implies that L = 0. Define an = n, bn =

1 . x2n

Using the Stolz-Ces` aro theorem, we have lim nx2n = lim

n→∞

n→∞

n an+1 − an = lim . 2 n→∞ bn+1 − bn 1/xn

Setting t = xn , we have lim

n→∞

an+1 − an bn+1 − bn

= = = =

lim

n→∞

lim

t→0

lim

t→0

1 sin2 (xn )

1 −

1 sin2 t 2

1 − sin2 (x1n−1 )

1 t2

t sin2 t t2 − sin2 t

3,

thereby proving the expected limit r q n lim xn = lim nx2n /3 = 1. n→∞ n→∞ 3 We can solve this Monthly problem similarly. Solution. By the assumption, we have Pn−1 sin(xk ) + sin xn nxn + sin xn xn+1 = k=0 = . n+1 n+1 This implies that 0 < xn+1 ≤

nxn + xn = xn n+1

6

Limits

whenever 0 < x0 < π. From the monotone convergence theorem, we conclude that xn converges. Let limn→∞ xn = L. Applying the Cauchy theorem yields L = lim

n→∞

n−1 1X sin(xk ) = lim sin(xn ) = sin(L), n→∞ n k=0

which implies that L = 0. Next, using the Taylor series approximation of sin x, we have   nxn + xn − x3n /6 + o(x3n ) x2n o(x2n ) xn+1 = = xn 1 − + . n+1 6(n + 1) n + 1 Note that (1 − x)−2 = 1 + 2x + o(x) for small x. We find that   o(x2n ) x2n 1 1 + = 2 1+ . x2n+1 xn 3(n + 1) n + 1 Using the Stolz-Ces` aro theorem again, we obtain lim x2n ln n

n→∞

=

lim

ln n

n→∞

1 x2n

= lim

n→∞

ln(n + 1) − ln n 1 − x12 x2 n+1

n

ln(n + 1) − ln n = lim = 3. 1 1 n→∞ 3(n+1) + o( n+1 ) Hence,

√ √ lim xn ln n = 3.

n→∞

 Remark. To study the properties of a sequence without an explicit formula, we often tend to determine the asymptotic behavior of the sequence. For example, P´ olya’s classical case asserts that r   1 3 +o √ xn = . n n Modifying the technique used in the above solution, we can improve this asymptotic formula to r √   3 3 3 ln n ln n √ +o √ − . xn = n 10n n n n This implies that n lim n→∞ ln n

r

 1−

n xn 3

 =

3 . 10

Another famous example is the normalized logistic sequence {xn } defined below.

Asymptotic behavior for a P´ olya type recurrence

7

Problem E3034 (Proposed by D. Cox, 91(4), 1984). Let 0 < x0 < 1 and xn+1 = xn (1 − xn ) for n = 0, 1, . . . . Prove that lim

n→∞

n (1 − nxn ) = 1. ln n

This is a refinement of the well-known results lim xn = 0 and

n→∞

lim nxn = 1,

n→∞

the latter of which appeared as Putnam Problem 1966-A3 (see [4], p. 5.) Motivated by the above results, we offer a generalization as an exercise. Let f (x) = x − axα + bx2α−1 + o(x2α−1 ) (as x → 0), where α > 1 and a > 0. For x0 → 0, define xn+1 = f (xn ) for n = 0, 1, . . .. Prove that  1/(α−1) 1 1 xn = 1/(α−1) n [a(α − 1)]   2 b − αa /2 ln n ln n + + o . [a(α − 1)](2α−1)/(α−1) nα/(α−1) nα/(α−1) As a consequence, this yields another solution to the following Monthly problem. Problem 11773 (Proposed by M. Omarjee, 121(4), 2014). Given a positive real number a0 , let ! n X ak for n ≥ 0. an+1 = exp − k=1

P∞ For which values of b does n=0 (an )b converge? Pn−1 Hint: Note an+1 = exp(− k=1 ak )e−an = an e−an . Thus f (x) = xe−x and   1 ln n ln n an = − 2 + o . n 2n n2 Furthermore, for sequences that converge to zero faster, we have the following better estimate: If xn converges to zero and for k > 1 lim nk (xn − xn+1 ) = A,

n→∞

then

A . k−1 Indeed, for  > 0, there is N ∈ N such that for n > N lim nk−1 xn =

n→∞

A −  ≤ nk (xn − xn+1 ) ≤ A + .

8

Limits

By adding the inequalities (A − )

1 1 ≤ xn − xn+1 ≤ (A + ) k , nk n

for n > N, m ≥ 2, we have (A − )

n+m−1 X j=n

n+m−1 X 1 1 ≤ x − x ≤ (A + ) . n n+m jk jk j=n

P∞ Letting m → ∞, then multiplying by nk−1 , and using ζ(k) = j=1 1/j k , we find that     n−1 n−1 X 1 X 1  ≤ nk−1 xn ≤ (A + )nk−1 ζ(k) − . (A − )nk−1 ζ(k) − k k j j j=1 j=1 Now, the claimed limit follows from the rate of convergence   n−1 X 1 = 1 . lim nk−1 ζ(k) − k n→∞ j k−1 j=1 We have seen that the Stolz-Ces`aro theorem plays a central role in determining the growth rate of a sequence. This approach can be extended to other classes of recurrent sequences. Here are five problems for additional practice: 1. Problem Pn E1557
(Proposed by D. R. Hayes, 70(9), 1963). If Sn = (1/n2 ) k=0 ln nk , evaluate S = limn→∞ Sn . 2. Problem 6376 (Proposed by I. J. Schoenberg, 89(1), 1982). Let an (n = 1, 2, . . .) be reals. Show that lim an

n→∞

n X

a2k = 1

implies that

k=1

lim (3n)1/3 an = 1.

n→∞

3. Problem 12079 (Proposed by M. Omarjee, 125(10), 2018). Choose x1 in (0, 1), and let xn+1 =

n 1 X ln(1 + xk ) n

for n ≥ 1.

k=1

Compute limn→∞ xn ln n. 4. Mathematics Magazine Problem 2087 (Proposed by F. Stanescu, 93(1), 2020). Consider the sequence defined by x1 = a > 0 and   x1 + x2 + · · · + xn−1 xn = ln 1 + for n ≥ 2. n−1 Compute limn→∞ xn ln n.

A nonlinear recurrence with Fibonacci exponents

9

5. Let {an } be a sequence of real numbers such that ean + nan = 2 for all n ∈ N. Show that lim n(1 − nan ) = 1

n→∞

1.3

and

lim n[n(1 − nan ) − 1] =

n→∞

1 . 2

A nonlinear recurrence with Fibonacci exponents

Problem 11976 (Proposed by R. Bosch, 124(4), 2017). Given a positive real number s, consider the sequence {un } defined by u1 = 1, u2 = s, and un+2 = un un+1 /n for n ≥ 1. (a) Show that there is a constant C such that limn→∞ un = ∞ when s > C and limn→∞ un = 0 when s < C. (b) Calculate limn→∞ un when s = C. Discussion. To search for patterns in this sequence, we compute the first few terms. Based on the definition of un , taking n = 1, 2, . . . , 6, we obtain u3 = s, u4 =

s2 s3 , u5 = , 2 2·3

s5 s8 s13 , u7 = 3 2 , u8 = 5 3 2 . ·3·4 2 ·3 ·4·5 2 ·3 ·4 ·5·6 In the above forms, we see that u6 =

22

ˆ The exponents of s are 1, 2, 3, 5, 8, 13, respectively, the Fibonacci sequence. ˆ The denominators are products of the form k Fn−k , where Fn is the nth Fibonacci number.

Thus, the pattern of un emerges and can be established by induction. Solution. (a) Let Fn be the nth Fibonacci number with F1 = F2 = 1, and Fn = Fn−1 + Fn−2 for n > 2. It follows by induction that un+2 =

2Fn−1

sFn+1 . · 3Fn−2 · · · (n − 1)F2 · nF1

Let bn = 2Fn−1 · 3Fn−2 · · · (n − 1)F2 · nF1

10

Limits

and φ = (1 +

√

5)/2, the golden ratio. We now show that bn ∼ C Fn+1

as n → ∞,

where C = eL = 3.20096 . . . and ∞ X ln k L := = 1.16345 . . . . φk k=1

To see this, taking the logarithm of bn yields n X

ln bn =

Fn+1−k ln k.

k=1

Recall Binet’s formula

where φ¯ = −1/φ = (1 − ln bn

= =

Therefore,

√


1 Fn = √ φn − φ¯n , 5 5)/2. We compute n X

1 √ 5

! (φ

n+1−k

¯n+1−k

−φ

) ln k

k=1

n n X 1 1 X ¯n+1−k √ φn+1 φ ln k. φ−k ln k − √ 5 5 k=1 k=1

√

∞ n X ln k 5 ln bn 1 X ¯n+1−k = − φ ln k. φn+1 φk φn+1 n=1 k=1

¯ < 1, using Stirling’s formula for n! yields Since |φ| n n X X n+1−k φ¯ ln k ≤ ln k = ln n! = n ln n − n + O(ln n) k=1

k=1

and so 1 φn+1

n X n+1−k φ¯ ln k → 0,

as n → ∞.

k=1

Thus, as n → ∞, we have

√

5 ln bn ∼L φn+1

and

n+1

bn ∼ eL·φ

√ / 5

n+1

= (eL )φ

√ / 5

∼ C Fn+1 .

When ln s 6= L, i.e., s 6= eL = C, we have ln un+2 = Fn+1 ln s − ln bn ∼ Fn+1 (ln s − l) = Fn+1 ln

s . C

A nonlinear recurrence with Fibonacci exponents Therefore

 lim un =

n→∞

11

0 if s < C, ∞ if s > C.

(b) When s = C, we have ln un+2

= Fn+1 l − ln bn ∞ n X 1 ¯n+1 ln k 1 X ¯n+1−k 1 √ √ − φ l + = √ φn+1 φ ln k φk 5 5 5 k=1 k=n+1 ∼

∞ X ln k 1 1 √ φn+1 ≥ √ ln(n + 1), φk 5 5 k=n+1

and so limn→∞ un = +∞.



Remark. Tyler provided a generalization by relaxing the initial condition to u1 = t > 0, then proving that un → ∞ when stφ−1 ≥ C and un → 0 when stφ−1 < C. Let Un = 1/un . The recurrence becomes Un+2 = nUn Un+1 , which then can be viewed as a variant of derangements in combinatorics. A derangement is a permutation of the elements of a set such that no element appears in its original position. In other words, a derangement is a permutation that has no fixed points. If Dn is the number of derangements of a set of size n, then it is well-known that Dn satisfies Dn+1 = n(Dn + Dn−1 )

and

Dn = n!

n X

(−1)k

k=0

1 . k!

−1

In particular, Dn /n! → e = 0.367879 . . . as n → ∞. So for large n, more than one third of all permutations are derangements. In contrast to what we know about derangements, there are many interesting open questions related to the un . For example, 1. Is there a combinatorics interpretation for un ? Does un count something? P∞ k 2. Is it possible to find the exact value of L = k=1 ln ? φk 3. Is there a better asymptotic formula for bn ? For example, is there an α such that  C Fn+1  α bn = 1 + + o(1/n) ? n n 4. More generally, is there a closed form for the generating function of {ln n}? For this last question, using the Laplace transform L(ln t) = −

γ + ln s , s

12

Limits

we have ∞ X

ln n xn = −

n=1

γx − 1−x

1

Z

xe−t ln t dt, (1 − xe−t )2

0

for x ∈ (0, 1).

But, it is currently unknown whether this integral can be represented by elementary functions or not.

1.4

Rate of convergence for an integral

Problem 11941 (Proposed by O. Furdui, 123(9), 2016). Let 1

Z

p n xn + (1 − x)n dx.

L = lim

n→∞

0

(a) Find L. (b) Find lim n2

n→∞

Z

1

 p n xn + (1 − x)n dx − L .

0

Discussion. When an integral sequence is hopelessly incomputable, a potential approach is to find two simpler sequences that “squeeze” it and use the squeeze theorem. For part (b), observe that the substitution x = t/(1+t) transforms the integral Z 1/2 p Z 1 √ n 1 + tn n n n x + (1 − x) dx = dt (1 + t)3 0 0 and

√ n

1 + tn 1 → , (1 + t)3 (1 + t)3

as n → ∞.

Thus once finding the value of L, the desired result follows upon determining the limit under the integral sign. Solution. R1 p (a) We claim L = 3/4. To see this, let In = 0 n xn + (1 − x)n dx. We have Z In

1/2

p n

=

xn + (1 − x)n dx +

0

Z ≥

1

p n

1/2 1/2

Z

1

(1 − x) dx + 0

Z

x dx = 1/2

3 . 4

xn + (1 − x)n dx

Rate of convergence for an integral

13

On the other hand, since x ≤ 1−x for x ∈ [0, 1/2] and 1−x ≤ x for x ∈ [1/2, 1], we have Z 1/2 p Z 1 p n n n n In = x + (1 − x) dx + xn + (1 − x)n dx 0

1/2 1/2

Z ≤

√ n

Z

√ n

1

2 (1 − x) dx +

0

2 x dx =

1/2

3 √ n 2. 4

The squeeze theorem implies that L = limn→∞ In = 3/4 as claimed. R 1/2 R1 (b) The limit is π 2 /48. Notice that 0 (1 − x) = 1/2 x dx = 3/8. We claim ! Z 1/2 p 3 π2 n lim n2 xn + (1 − x)n dx − = , (1.6) n→∞ 8 96 0 and lim n

Z

2

n→∞

1

p n

1/2

3 xn + (1 − x)n dx − 8

! =

π2 , 96

(1.7)

from which the required limit follows. To prove (1.6), we compute ! Z 1/2  p  n 2 n n lim n x + (1 − x) − (1 − x) dx n→∞

=

= = = =

0

lim n2

n→∞

Z

s

1/2

(1 − x)

n

 1+

0

x 1−x

n

! −1

! dx

(let t = x/(1 − x)) Z 1  √
1 n n − 1 dt lim n2 1 + t (let u = tn ) 3 n→∞ 0 (1 + t) Z 1  √
1/n−1 1 n lim n 1 + u − 1 u du 1/n )3 n→∞ 0 (1 + u  Z 1 √
1/n−1 1 n lim ·n 1+u−1 ·u du n→∞ (1 + u1/n )3 0 Z 1 1 ln(1 + u) π2 du = , 8 0 u 96

where we have used the well-known facts: Z 1 ∞ X √
ln(1 + u) (−1)n+1 π2 lim n n 1 + u − 1 = ln(1+u) and du = = . 2 n→∞ u n 12 0 n=1 Equation (1.7) follows from (1.6) upon substituting 1 − x for x.



Remark. The above results can be generalized as follows. We leave the proof to the reader. For part (a), if f and g are nonnegative and integrable on [a, b], then Z b p Z b n lim f n (x) + g n (x) dx = max{f (x), g(x)} dx. n→∞

a

a

14

Limits

For part (b), if f is positive continuous function on [0, 1] with f (0) = 1 and g(x) is continuous on [0, 1], then lim n

n→∞

2

1

Z

p n

f (xn )g(x) dx

1

Z −

 Z g(x) dx = g(1) 0

0

0

1

ln f (x) dx. x

In particular, letting f (x) = 1 + x and g(x) = 1/(1 + x)3 yields the result in part (b). Finally, we offer seven more problems for additional practice. 1. Problem E1245 (Proposed by M. S. Klamkin, 63(10), 1956). If 1

Z bn+1 =

1

Z min(x, an ) dx, an+1 =

max(x, bn ) dx.

0

0

Prove that the sequences {an } and {bn } both converge and find their limits. 2. Problem 4828 (Proposed by M. S. Klamkin, 66(2), 1959). Do the sequences {an }, {bn }, {cn } converge, where Z

1

an+1 =

min(x, bn , cn ) dx, bn+1 0

Z =

1

Z mid(x, cn , an ) dx, cn+1 =

0

1

max(x, an , bn ) dx 0

and mid(a, b, c) = b if a ≥ b ≥ c. If so, can you find their limits? 3. Problem 11225 (Proposed by J. L. D´ıaz-Barrero, 113(5), 2006). Find Z n  x ln(1 + x/n) 1 dx. lim n→∞ n 1+x 0 4. Problem 11611 (Proposed by O. Furdui, 118(10), 2011). Let f be a continuous function from [0, 1] into [0, ∞). Find 1

Z lim n

n→∞

0

∞ X xk k

!2 f (x) dx.

k=n

5. Problem 12120 (Proposed by M. Bataille, 126(6), 2019). For posPk−1
itive integers n and k with n ≥ k, let a(n, k) = j=0 nj 3j . (a) Evaluate lim

n→∞

n 1 X a(n, k) . 4n k k=1

How closely does this sum approximate the integral?

15

(b) Evaluate n X a(n, k) lim n 4 L − n→∞ k

!

n

,

k=1

where L is the limit in part (a). 6. Problem 12153 (Proposed by O. Kouba, 127(1), 2020). For a real number x whose fractional part is not 1/2, let hxi denote the nearest integer to x. For a positive integer n, let ! n X √ 1 √ − 2 n. an = h ki k=1 (a) Prove that the sequence a1 , a2 , . . . is convergent, and find its limit L. √ (b) Prove that the set { n(an − L) : n ≥ 1} is dense subset of [0, 1/4]. 7. Mathematics Magazine Problem 2097 (Proposed by O. Kouba, 93(3), 2020). For a real number x ∈ / 1/2 + Z, denote the nearest integer to x by hxi. For any real number x, denote the largest smaller than or equal to x and the smallest integer larger than or equal to x by bxc and dxe, respectively. For a positive integer n let 2 1 1 an = √ − √ − √ . h n i b nc d n e P∞ (a) Prove that the series n=1 an is convergent and find its sum L. (b) Prove that the set ( √

n

n X

! ak − L

) : n≥1

k=1

is dense in [0, 1].

1.5

How closely does this sum approximate the integral?

Problem 11535 (Proposed by M. Tetiva, 117(9), 2010). Let f be continuously R1 differentiable function on [0, 1]. Let A = f (1) and let B = 0 x−1/2 f (x)dx.

16

Limits

Evaluate Z

1

f (x)dx −

lim n

n→∞

0

n  2 X k k=1

(k − 1)2 − n2 n2

  ! (k − 1)2 f n2

in terms of A and B. Discussion. The statement to be proved suggests we work backward. Indeed, if we split the summation in the proposed limit into   n 1 X 2(k − 1) (k − 1)2 f n n n2

  n 1 X (k − 1)2 , f n n2

and

k=1

k=1

then they are exactly the equidistant Riemann sums of 2xf (x2 ) and f (x2 ) associated with the left-endpoint rule over [0, 1], respectively. On the other hand, by substitution, we have Z 1 Z 1 Z 1 Z 1 B= x−1/2 f (x) dx = 2 f (x2 )dx, 2xf (x2 ) dx = f (x) dx. 0

0

0

0

Thus, the problem reduces to determining the convergence rate for the equidistant Riemann sum associated with the left-endpoint rule. Solution. R1 It is well-known that the error in approximating 0 f (x)dx by the Riemann sums with left-endpoint rule is O(1/n). More precisely, this result can be rephrased as follows: If f (x) is continuously differentiable on [0, 1], then  ! Z 1 n k−1 1 1 X f = (f (1) − f (0)). (1.8) lim n f (x) dx − n→∞ n n 2 0 k=1

For the sake of completeness, we give a proof. Let F be an antiderivative of f . We have Z

1

f (x) dx − 0

  n 1 X k−1 f n n k=1

= = =

 ! k−1 1 f (x) dx − f n n (k−1)/n k=1        n X k k−1 1 k−1 F −F − F0 n n n n n X

k=1 n X

k=1

Z

k/n

1 F 00 (ξk ), 2n2

(using Taylor’s theorem)

How closely does this sum approximate the integral?

17

for some ξk ∈ ((k − 1)/n, k/n) and 1 ≤ k ≤ n. Thus  ! n 1 X k−1 f (x)dx − lim n f n→∞ n n 0 k=1 Z n 1 1 X 00 1 1 00 F (x) dx = lim F (ξk ) = 2 n→∞ n 2 0 k=1 Z 1 1 1 0 f (x) dx = (f (1) − f (0)). = 2 0 2 Z

This proves (1.8). Using

1

R1 0

f (x) dx =

R1 0

2xf (x2 ) dx, we have

  ! (k − 1)2 lim n f f (x) dx − n→∞ n2 0 k=1  ! Z 1 n X (k − 1)2 2(k − 1) + 1 2 f = lim n 2xf (x ) dx − n→∞ n2 n2 0 k=1 "  ! Z 1 n 2 X 1 (k − 1) (k − 1) = lim n 2xf (x2 ) dx − 2 f n→∞ n n n2 0 k=1 #   n (k − 1)2 1 X f − . n n2 Z

1

n  2 X k

(k − 1)2 − 2 n n2

k=1

Replacing f (x) by 2xf (x2 ) in (1.8), we find that   ! (k − 1)2 lim n f (x) dx − f n→∞ n2 0 k=1 Z 1 Z 1 1 −1/2 1 1 (2xf (x2 ))|10 − f (x2 ) dx = f (1) − x f (x) dx = A − B. = 2 2 0 2 0 Z

1

n  2 X k

(k − 1)2 − n2 n2

 Remark. We can generalize (1.8) with a similar proof: If f (x) is continuously differentiable on [0, 1] and α ∈ [0, 1], then  ! Z 1 n 1 X k−α 1 lim n f (x) dx − f = (2α − 1)(f (1) − f (0)). (1.9) n→∞ n n 2 0 k=1

For the midpoint Riemann sum, along the same lines, we see the convergence rate is second order: If f is twice continuously differentiable on [0, 1], then  ! Z 1 n 1 X 2k − 1 1 2 f (x) dx − lim n f = (f 0 (1) − f 0 (0)). (1.10) n→∞ n 2n 24 0 k=1

18

Limits

However, there is no general result on the convergence rates of non-equidistant Riemann sums. Observe that the summation in the proposed problem n  2 X k k=1

(k − 1)2 − n2 n2

   (k − 1)2 f n2

is indeed a left-endpoint Riemann sum of equidistant partition 0
0. If k ≥ 2, x > −1, then   x−1 |x| + 1 √ √ √ arctan . ≤ √ (k + x + 1) k + 1 + (k + 2) k + x k k + 1 + (k + 2) k − 1 By the Weierstrass M-test, the series P (x) converges uniformly, and therefore it is continuous for x > −1. As in (a), we have  ∞  X 1 1 arctan √ P (x) = − arctan √ k+1 k+x k=1 so

1 P (x + 1) = P (x) + arctan √ . 1+x

(1.11)

Thus lim

x→−1+

P (x) =

lim

x→−1+

  3π 1 π . P (x + 1) − arctan √ = P (0) − = − 2 4 1+x 

Remark. The proposers revealed that they discovered the value −3π/4 experimentally: They calculated a high accuracy numerical value, then used this value as input to the Inverse Symbolic Calculator (ISC, available at http: //wayback.cecm.sfu.ca/projects/ISC/ISCmain.html) to identify the exact value. It is interesting to know whether there are more general closed forms for P , say at half-integers. In this case, from (1.11), we only need to figure out the exact value of ! ∞ X 1 1 arctan √ P (1/2) = − arctan p , k+1 k + 1/2 k=1 which seems still an open problem. Here are six Monthly problems for additional practice. 1. Problem 11853 (Proposed by H. Ohtsuka, 122(7), 2015). Find ∞ X n=1

1 . sinh 2n

2. Problem 11930 (Proposed by C. I. V˜alean, 123(8), 2016). Find   ∞ X 1 −1 √ √ sinh . 2n+2 + 2 + 2n+1 + 2 n=1

Geometric mean rates

21

3. Problem 12090 (Proposed by H. Ohtsuka, 126(2), 2019). The PellLucas numbers Qn satisfy Q0 = 2, Q1 = 2, and Qn = 2Qn−1 + Qn−2 for n ≥ 2. Prove     ∞ X 1 1 π2 arctan arctan = . Qn Qn+1 32 n=1 4. Problem 12101 (Proposed by H. Lee, 126(3), 2019). Find the least upper bound of √ √ ∞ X xn+1 − xn p (1 + xn+1 )(1 + xn ) n=1 over all increasing sequences x1 , x2 , . . . of positive real numbers. 5. Problem 12118 (Proposed by H. Ohtsuka, 126(6), 2019). Compute ∞ X n=0

1 F2mn + iFm

where m is an odd integer and Fn is the nth Fibonacci number. 6. Problem 11505 (Proposed by B. Burdick, 117(5), 2010). Define an to be the periodic sequence given by a1 = a3 = 1, a2 = 2, a4 = a6 = −1, a5 = −2, and an = an−6 for n ≥ 7. Let Fn be the Fibonacci sequence with F1 = F2 = 1. Show that ∞ ∞ X ak Fk F2k−1 X (−1)kn π = . 2k − 1 n=0 Fkn+2k−1 Fkn+3k−1 4

k=1

Hint: Show that for |x| < 1,   ∞ X 2x cos θ (−1)k−1 cos(2k − 1)θ 2k−1 x . arctan = 2 2 1−x 2k − 1 k=1

1.7

Geometric mean rates

Problem 11935 (Proposed by D. M. B˘atinetu-Giurgiu and N. Stanciu, 123(8), 2016). Let f be a function from Z+ to R+ such that limn→∞ f (n)/n = a, where a > 0. Find v  v  un+1 u n Y uY u n t lim  n+1 f (k) − t f (k)  . n→∞

k=1

k=1

22

Limits

Discussion. Searching for solutions sometimes can be done by examining the special cases. The solutions to the special cases may shed light on the general solutions. Notice that when f (k) = k the problem becomes the well-known result ([77], p. 9):  p  1 √ n n+1 (n + 1)! − n! = . lim n→∞ e It is fortunate here that the same argument goes through for the general problem. Solution. We show that the limit is a/e. To see this, we first establish three facts: n F1 limn→∞ √ Qn n

k=1

f (k)

s

n p Qn n

k=1

= ae . To prove this, we rewrite

f (k)

=

n

v u n uY k n! n n n Qn √ = t . · √ · n n f (k) f (k) n! n! k=1 k=1

Stirling’s formula implies that n lim √ = e. n n!

n→∞

The Cauchy theorem yields lim

1 2 n ln f (1) + ln f (2) + · · · + ln f (n)

n

n→∞

= lim ln n→∞

n 1 = ln . f (n) a

Exponentiating both sides gives v u n uY n lim t

n→∞

k=1

k 1 = . f (k) a

These limits confirm F1 immediately.  √Q n n+1 n+1 k=1 f (k) √ F2 limn→∞ = e. This follows from Q n n k=1

 

qQ

n

n+1 k=1 f (k)  p Qn n k=1 f (k)

n+1

s =

f (k)

n+1

f n (n + 1) · (n + 1)n

s =

n(n+1)

n+1 qQ n+1 n+1 k=1

f n (n + 1) Qn k=1 f (k) r ·

f (k)

n+1

!n

s =

n+1

f n (n + 1) Qn k=1 f (k)

f (n + 1) e → a · · a0 = e. n+1 a

Geometric mean rates qQ pQn n+1 F3 Let an = n+1 k=1 f (k) − n k=1 f (k). Then

lim

n→∞

Qn n !√ k=1 f (k)/an

an

1+ p Qn n

k=1

23

= e.

f (k)

Recalling the limit definition of e, it suffices to show that an lim p Qn n

n→∞

k=1

f (k)

= 0.

This can be done by qQ n+1

n+1

an p Qn n

k=1 f (k)

=

k=1

f (k)

− 1 ∼ e1/n − 1 → 0,

p Qn n

k=1 f (k)

where we have used F2. Since  qQ

n n+1 f (k) k=1  p Qn n k=1 f (k)

n+1



!n

an

=

1+ p Qn n

k=1

f (k) Qn nan / n !√ k=1 f (k)/an

 an

=  1+ p Qn n

k=1

f (k)

√ Qn n

k=1

f (k)

,



taking logarithm gives Qn n  qQ n !√ n+1 n+1 k=1 f (k)/an f (k) an k=1  = a n pQ n p ln 1+ . ln  p Q Q n n n n n n k=1 f (k) k=1 f (k) k=1 f (k) Finally, letting n → ∞ and using F1–F3, we find that lim an =

n→∞

a e

as claimed. Remark. Here are some similar Monthly problems for your practice. 1. Problem 11771 (Proposed by D. M. B˘atinetu-Giurgiu, 121(4), 2014). Find ! ! p p π n+1 (n + 1)! n √ lim (2n − 1)!! tan −1 . n→∞ 4 n n!



24

Limits 2. Problem 11338 (Proposed by O. Furdui, 115(1), Q 2008). Let Γ n denote the classical gamma function, and let G(n) = k=1 Γ(1/k). Find   lim G(n + 1)1/(n+1) − G(n)1/n . n→∞

3. Problem Pn11724 (Proposed by A. Cusumano, 120(7), 2013). Let f (n) = P k=1 k k and let n g(n) = k=1 f (k). Find   g(n + 2) g(n + 1) lim − . n→∞ g(n + 1) g(n) 4. Problem 11808 (Proposed by D. M. B˘atinetu-Giurgiu, 121(10), 2014). Compute lim n2

n→∞

Z

(n!)−1/n

Γ(nx) dx. ((n+1)!)−1/(n+1)

Hint: Use the mean value theorem for the integrals. 5. Problem 11875 (Proposed by D. M. B˘atinetu-Giurgiu and N. Stanciu, 122(10), 2015). Let fn = (1 + 1/n)n ((2n − 1)!!Ln )1/n . Find lim (fn+1 − fn ).

n→∞

Here Ln denotes the nth Lucas number, given by L0 = 2, L1 = 1, and Ln = Ln−1 + Ln−2 for n ≥ 2.

1.8

Limits of the weighted mean ratio

Problem 11811 (Proposed by V. Mikayelyan, 122(1), 2015). Let {a} and {b} be infinite sequences of positive numbers. Let {x} be the infinite sequence given for n ≥ 1 by xn = 

ab11 ab22 · · · abnn b1 +···+bn .

a1 b1 +a2 b2 +···+an bn b1 +b2 +···+bn

(a) Prove that limn→∞ xn exists. (b) Find the set of all c that can occur as this limit, for suitably chosen {a} and {b}.

Limits of the weighted mean ratio

25

Discussion. As usual, we search for ideas from a special case: Let bn = 1 for all n ≥ 1. Then √ n n a1 a2 · · · an a1 a2 · · · an xn = a1 +a2 +···+an
n = a1 +a2 +···+an . n

n

The AM-GM inequality implies that xn ≤ 1 for all n ≥ 1. Moreover, if an = n we have n! n! xn = 1+2+···+n
n = n+1
n , n

2

which is monotonically decreasing. This suggests that xn is decreasing in general. Consider
n n an+1 a1 +a2 +···+a xn+1 n =  n+1 . xn a1 +···+an +an+1 n+1

Notice that xn is decreasing if and only if xn+1 /xn ≤ 1, which is equivalent to  n  n+1 a1 + a2 + · · · + an a1 + · · · + an + an+1 an+1 ≤ . n n+1 Taking logarithms on both sides, then dividing by (n + 1) we obtain     n a1 + a2 + · · · + an a1 + · · · + an + an+1 1 ln an+1 + ln ≤ ln . n+1 n+1 n n+1 This holds by applying Jensen’s inequality for ln x: α ln x + (1 − α) ln y ≤ ln(αx + (1 − α)y) with

a1 + a2 + · · · + an 1 , x = an+1 and y = . n+1 n Hence, by the monotone convergence theorem, we see that limn→∞ xn exists and the limit is between 0 and 1. Now we just need to extend the above arguments to the weighted means. α=

Solution. (a) Let Bn = b1 + b2 + · · · + bn and Sn = a1 b1 + a2 b2 + · · · + an bn . Since xn ≥ 0 for all n ≥ 1, by the monotone convergence theorem, it suffices to show that xn is decreasing. To this end, we have x1 = 1 and for n ≥ 1 b

a n+1 (Sn /Bn )Bn xn+1 = n+1 . xn (Sn+1 /Bn+1 )Bn+1 It is clear that xn+1 /xn ≤ 1 if and only if b

n+1 an+1 (Sn /Bn )Bn ≤ (Sn+1 /Bn+1 )Bn+1 .

26

Limits

Taking logarithms on both sides yields bn+1 ln an+1 + Bn ln(Sn /Bn ) ≤ Bn+1 ln(Sn+1 /Bn+1 ), or

bn+1 Bn ln an+1 + ln(Sn /Bn ) ≤ ln(Sn+1 /Bn+1 ). Bn+1 Bn+1 This inequality follows from Jensen’s inequality α ln x + (1 − α) ln y ≤ ln(αx + (1 − α)y) with α = bn+1 /Bn+1 , x = an+1 , and y = Sn /Bn . (b) By the weighted AM-GM inequality, we see that limn→∞ xn ∈ [0, 1]. Now we show that [0, 1] is exactly the set of limit points of {xn }; i.e., if c ∈ [0, 1], then there exist sequences {a} and {b} such that xn converges to c. If c = 0, letting an = n, bn = 1, from Stirling’s formula, we have  n  n √ 2 n n! xn = n+1
n ∼ 2πn → 0. e n+1 2

Next, let an = 1 + t/n, bn = 1 with t ≥ 0. We find that

Qn Qn t t k=1 1 + k k=1 1 + k xn =
n =
(n/tHn )·tHn 1 + tHnn 1 + tHnn 1 1 → = γt , teγt Γ(t) e Γ(t + 1) where we have used the gamma product formula:   ∞  Y 1 t = teγt 1+ e−t/n . Γ(t) k k=1

By the intermediate value theorem, the continuous function 1/eγt Γ(t + 1) attains every value in (0, 1] as t ∈ [0, ∞). This completes the proof.  Remark. In the published solution [61], Kouba gave an elementary proof on Part (b) as follows: if c ∈ (0, 1), let bn = 1 for all n and √ √ 1− 1−c 1 1+ 1−c √ √ , a2 = , an = √ for n ≥ 3. a1 = c c c Direct computation shows that xn = c for all n ≥ 2 and so xn → c. Here is one more problem for your practice. Let {an } be an arithmetic sequence of positive numbers. Show the limit set of √ n a1 a2 · · · an lim a1 +a2 +···+an n→∞

is either 1 or 2/e.

n

Limits of mean recurrences

1.9

27

Limits of mean recurrences

Problem 12057 (Proposed by P. K´orus, 125(7), 2018). (a) Calculate the limit of the sequence defined by a1 = 1, a2 = 2, and a2k+1 =

a2k−1 + a2k √ and a2k+2 = a2k a2k+1 2

for positive integers k. (b) Calculate the limit of the sequence defined by b1 = 1, b2 = 2, and b2k+1 =

b2k−1 + b2k 2b2k b2k+1 and b2k+2 = 2 b2k + b2k+1

for positive integers k. Discussion. Normally, we expect to find formulas for a2k and a2k+1 in terms of a1 and a2 . However, direct calculations lead to nested radical sequences. Because they are patternless, there is no way to condense them into a closed form. The key insights needed are to use trigonometric substitution in (a) and to apply the homogeneous property of the means in (b). Solution. (a) By induction, based on the definition of the sequences, it is easy to show that a2k−1 ≤ a2k+1 ≤ a2k+2 ≤ a2k , for k = 1, 2, . . . . The monotone convergence theorem, together with the recurrences, implies that the two sequences a2k+1 and √ a2k converge to a common limit M (a1 , a2 ). We now show that M (1, 2) = 3 3/π. Indeed, we can determine M (a1 , a2 ) for any positive initial values a1 and a2 . We distinguish three cases. (i) If a1 = a2 , the given recursions imply that an = a1 for all n ∈ N. Thus, M (a1 , a1 ) = a1 . (ii) If a1 < a2 , we construct a right triangle with adjacent side a1 and hypotenuse a2 . Let θ ∈ (0, π/2) be the angle such that cos θ = a1 /a2 and let a be the opposite side. Then a1 = a cot θ, a2 = a csc θ; q θ = cos−1 (a1 /a2 ), a = a22 − a21 .

(1.12)

We claim a2k−1 =

a a cot(θ/2k−1 ), a2k = k−1 csc(θ/2k−1 ), 2k−1 2

for k = 1, 2, . . . . (1.13)

28

Limits

The base case k = 1 follows from (1.12). Assume (1.13) holds for some k. By the trigonometric identities, for 0 < α < π/2, r cot(2α) + csc(2α) cot α 1 1 csc(2α) · = cot α, = csc α, 2 2 2 2 we find that a2k+1 = a2k =

a 2k−1 a 2k−1

cot(θ/2k−1 ) + csc(θ/2k−1 ) a = k cot(θ/2k ), 2 2 q a k−1 k−1 cot(θ/2 ) · csc(θ/2 ) = k csc(θ/2k ). 2

By induction (1.13) holds for all k ≥ 1. As x → 0, cot x ∼ csc x ∼ x−1 . Therefore a a lim a2k−1 = lim k−1 cot(θ/2k−1 ) = ; k→∞ k→∞ 2 θ a a k−1 lim a2k = lim k−1 csc(θ/2 )= . k→∞ k→∞ 2 θ Thus p a22 − a21 a M (a1 , a2 ) = = . −1 θ cos (a1 /a2 ) √ In particular, if a1 = 1, a2 = 2, we obtain M (1, 2) = 3 3/π as claimed. (iii) If a1 > a2 , using the identities r 1 1 coth(2α) + csch(2α) coth α = coth α, = cschα, csch(2α) · 2 2 2 2 similarly, we find that p a21 − a22 M (a1 , a2 ) = . −1 cosh (a1 /a2 ) (b) Similarly, by induction we can show that b2k−1 ≤ b2k+1 ≤ b2k+2 ≤ b2k ,

for k = 1, 2, . . . .

Thus the two sequences b2k+1 and b2k also converge to a common limit M (b1 , b2 ). By the definition of the sequence, we have M (αb1 , αb2 ) = αM (b1 , b2 ) for any positive α, b1 , b2 . Thus, without loss of generality, it suffices to consider the case where b1 = 1 + x, b2 = 1 with x > −1. Applying the definition of the sequence yields b3 = 2−1 (2 + x), b4 = 2 b6 = 22

2+x (2 + x)(23 + x) , b5 = 2−2 , 2 2 +x 22 + x

(2 + x)(23 + x) . (22 + x)(24 + x)

Limits of mean recurrences

29

In general, by induction, we readily conclude that b2n+1

=

2−n

n Y

(22k−1 + x)/

k=1

b2n+2

=

2n

n Y 

n−1 Y

(22k + x), n = 1, 2, . . . ,

k=1

 (22k−1 + x)/(22k + x) , n = 1, 2, . . . .

k=1

So M (1 + x, 1) = lim 2n n→∞

∞ n Y Y 1 + x/22k−1 22k−1 + x . = 22k + x 1 + x/22k

k=1

k=1

In particular, letting x = −1/2 yields M (1, 2) = 2M (1 − 1/2, 1) = 2

   ∞  Y 1 1 (1/4; 1/4)∞ 1− k / 1− = , k 4 2·4 (1/2; 1/4)∞

k=1

where (a; q)∞ is the q-Pochhammer symbol, so called a q-series (see [8], Chapter 10).  Remark. The early history of using mean recursions can be traced back to the calculation of accurate values of π. Archimedes’ process (250 BCE) constitutes the first rigorous algorithm for π. This process can √ be stated √ as the following harmonic-geometric mean recursion. Set a1 = 3 3, b1 = 3 3/2. Then define an+1 =

p 2an bn , bn+1 = an+1 bn , an + bn

for n ≥ 1.

Geometrically, an and bn denote the semi-perimeters of circumscribed and inscribed regular (n + 2)-gons of the unit circle. For a regular 96-gon, these recursions give an intriguing estimate for π: 3.1408 ' 3

10 1 < π < 3 ' 3.1429. 71 7

No wonder Archimedes is viewed as the first numerical analyst. Another famous mean recursion is the Gauss arithmetic-geometric mean (AGM) ([20], p. 5-7), which is defined by the two-term recurrence: an+1 =

p an + bn , bn+1 = an bn , 2

for n ≥ 0,

Gauss observed that the common limit M satisfies M (ta, tb) = tM (a, b), for any t > 0, (homogeneous);   a+b √ M (a, b) = M , ab , (invariant), 2

30

Limits

and found that M (a, b) =

2 π

π/2

Z

!−1

dθ p a2 cos2 θ + b2 sin2 θ

0

.

This led to his discovery of the theory of elliptic functions. Back to the proposed problem (b). Using the Jacobi triple product identity ([8], p. 497, [25], p. 50), we have ∞ X

xk(k+1)

=

k=0

= = = =

∞ 1 X k(k+1) x 2

1 2

k=−∞ ∞ h Y

(1 − x2n )(1 + x2n )(1 + x2(n−1) )

i

n=1

∞ Y   (1 − x2n )(1 + x2n )2 n=1 ∞ Y n=1 ∞ Y n=1

(1 − x4n )2 1 − x2n 1 − x4n . 1 − x4n−2

Setting x = 2−1/2 , we find another alternative representation of M (1, 2): M (1, 2) = 2M (1/2, 1) =

∞ Y k=1

∞ X 1 1 − 1/22k = . 2k−1 n(n+1)/2 1 − 1/2 2 n=0

Historically, the q-series is associated with the Ramanujan theta function ([8], p. 501-505): ∞ X f (a, b) := an(n+1)/2 bn(n−1)/2 . n=−∞

For example, M (1, 2) is the direct consequence (with q = 1/2) of the formula f (q, q 3 ) =

∞ X

q n(n+1)/2 =

n=0

(q 2 ; q 2 )∞ . (q; q 2 )∞

As in (a), trigonometric substitution is particularly useful in solving recurrence relations. Here we present one more example. Find the limits of the sequences defined by an+1 =

p

an bn and bn+1 =

2an+1 bn , an+1 + bn

for n ≥ 0.

Along the same lines in (a), first, let 0 < a0 < b0 . Construct a right triangle

Limits of mean recurrences

31

√ √ with the adjacent side a0 and the hypotenuse b0 . Let θ ∈ (0, π/2) be the angle such that cos2 θ = a0 /b0 and let r a0 a= b0 = b0 cot θ. b0 − a0 Then b0 = a tan θ and a1

=

b1

= =

√ a0 b0 − a0 b0 · √ = a sin θ; b0 − a0 b0 2a1 b0 2a2 tan θ sin θ = a1 + b0 a sin θ + a tan θ sin θ = 2a tan(θ/2). 2a 1 + cos θ

p

r

a0 b0 =

Based on these results and the calculations of a few next iterations, we conjecture that
an = 2n−1 a sin θ/2n−1 , bn = 2n a tan (θ/2n ) , (for n ≥ 1). (1.14) We have already proved that (1.14) holds for n = 1. Assume (1.14) holds for some k. Then q p ak+1 = ak bk = 22k−1 a2 sin (θ/2k−1 ) tan (θ/2k ) q 22k a2 sin (θ/2k ) cos (θ/2k ) tan (θ/2k ) =
= 2k a sin θ/2k ;

22k+1 a2 sin θ/2k tan θ/2k 2ak+1 bk bk+1 = = k ak+1 + bk 2 a sin (θ/2k ) + 2k a tan (θ/2k )
k+1
2 a sin θ/2k = = 2k+1 a tan θ/2k+1 . k 1 + cos (θ/2 ) By induction (1.14) holds for all n ∈ N. As x → 0, sin x ∼ tan x ∼ x. We find that
lim an = lim 2n−1 a sin θ/2n−1 = aθ;

n→∞

n→∞

lim bn = lim 2n a tan (θ/2n ) = aθ.

n→∞

n→∞

Hence, r M (a0 , b0 ) = aθ =

p a0 b0 · cos−1 ( a0 /b0 ). b0 − a0

In particular, if a0 = 2, b0 = 4, we have M (2, 4) = 4 ·

π = π. 4

32

Limits

Thus (1.14) presents another algorithm for π. For example, we have 3.141591 ' a10 < π < b10 ' 3.141593. Second, if 0 < b0 < a0 , we have an

=


2n−1 a sinh θ/2n−1 ;

bn

=

2n a tanh (θ/2n ) ,

where a0 = cosh2 θ, a = b0 Thus

r

a0 b0 . a0 − b0

r

p a0 b0 · cosh−1 ( a0 /b0 ). a0 − b0 We leave the details to the reader. We end this section with two additional practice problems. M (a0 , b0 ) =

1. In part (b), for any positive initial values b1 and b2 , prove that M (b1 , b2 ) =

(α; 1/4)∞ , (β; 1/4)∞

where α = 1/2 − b1 /(2b2 ), β = 1 − b1 /b2 . 2. Let a be a positive constant. It is well-known that, for any initial term x1 > 0, the sequence {xn } defined for n ≥ 1 by   a 1 xn + xn+1 = 2 xn √ always converges to a. Consider the sequence yn defined for n ≥ 1 by   1 a yn+1 = yn − . 2 yn For any positive integer p > 1, show that one can choose the initial term y1 such that {yn } is periodic with the period p.

1.10

A disguised half-angle iteration

Problem 10973 (Proposed by L. D. Servi, 109(9), 2002). With Rk (s) defined as below, prove that limk→∞ Rk (2)/Rk (3) = 3/2. k square roots z }| { v v u s u u r u q u √ t t Rk (s) = 2 − 2 + 2 + · · · + 2 + s .

A disguised half-angle iteration

33

Discussion.√ Let f (x) = 2 + x. Define f 1 = f, f k = f ◦ f k−1 for all k ≥ 2. Observe that k square roots zv }| { s u r u q √ t f k (x) = 2 + 2 + 2 + · · · + 2 + x . To find a closed form for f k , the key insight comes from the proof of the Vi`ete formula: v s r s r u r u 1 1 1 1 t1 1 1 1 2 · + · + + ··· = . 2 2 2 2 2 2 2 2 π Setting t = π/2 in the well-known limit lim cos

n→∞

t t t sin t cos 2 · · · cos n = , 2 2 2 t

we find that

π π t 2 cos 3 · · · cos n+1 = . 2 n→∞ 2 2 2 π The Vi`ete formula now follows from r r π θ 1 1 1 cos = and cos = + cos θ. 4 2 2 2 2 lim cos

Once we rewrite the last half angle formula as √

θ 2 + 2 cos θ = 2 cos , 2

similarly, we generate f k as the required nested radicals. Solution. √ Let f (x) = 2 + x and x = 2 cos α for α ∈ (0, π/2). Since √ f (2 cos α) = 2 + 2 cos α = 2 cos(α/2), it follows that f 2 (2 cos α) = f (2 cos(α/2)) = 2 cos(α/22 ). By induction, for all k ≥ 1, we find that f k (x) = f k (2 cos α) = 2 cos(α/2k ) = 2 cos[cos−1 (x/2)/2k ]. Thus Rk (4 cos2 α) =

q

2 − f k−2 (2 cos α) =

q


2 − 2 cos(α/2k−2 ) = 2 sin α/2k−1 .

34

Limits

In general, for any s, t ∈ (0, 4), we have √ √ Rk (s) 2 sin[21−k cos−1 ( s/2)] cos−1 ( s/2) √ √ lim = lim = . k→∞ Rk (t) k→∞ 2 sin[21−k cos−1 ( t/2)] cos−1 ( t/2) In particular,

√ Rk (2) cos−1 ( 2/2) π/4 3 √ lim = = = . −1 k→∞ Rk (3) π/6 2 cos ( 3/2)  k

Remark. There is another way to find f by using the Chebyshev polynomials. If x ∈ [−2, 2], the inverse function of f is f −1 (x) = x2 − 2 for x ∈ [0, 2]. Recall that the Chebyshev polynomial of the first kind of degree 2 is T2 (x) = 2x2 − 1. Then f −1 (x) = 2T2 (x/2). Since T2k (x) = T2k (x)

and

Tn (x) = cos(n cos−1 x),

it implies that f −k (x) = 2T2k (x/2) = 2T2k (x/2) = 2 cos[2k cos−1 (x/2)]. Once again we find that f k (x) = 2 cos[cos−1 (x/2)/2k ]. In general, consider a class of periodic continued radicals of the form s r q √ (1.15) a0 2 + a1 2 + a2 2 + a3 2 + · · · , where an+k = ak for some positive integer n and ak = ±1 for k = 0, 1, 2, . . . , n−1. By induction, we can show that these radicals given by (1.15) have limits two times the fixed points of the Chebyshev polynomials T2n (x). Explicitly, we have v s u r u q √ t a0 2 + a1 2 + a2 2 + a3 2 + · · · + an−1 2 h a0 a1 a0 a1 · · · an−1  π i = 2 sin a0 + + ··· + . 2 2n−1 4 Since the sequence a0 +

a0 a1 a0 a1 · · · an−1 + ··· + 2 2n−1

A disguised half-angle iteration

35

converges absolutely, let its limit be α. Hence, the original radical converges to 2 sin(απ/4). This immediately yields a solution to the following question. Problem 12129 (Proposed by H. Ohtsuka, 126(7), 2019). Compute v s u r u q √ t 2 + 2 + 2 + · · · + 2 − 2 + · · ·, where the sequence of signs consists of n − 1 plus signs followed by a minus sign and repeats with period n. Having found the limits of (1.15), the next natural question is to determine the limit of the radical s r q √ a0 b + a1 b + a2 b + a3 b + · · · for values of the variable b that make the radical (and the limit) well defined. However, a direct application of the above method fails and so far a convenient variation has been elusive. Therefore, the limit of the last radical in the general case remains an open problem although it is known at least in one case. Putnam Problem 1953 A-6 ([47], p. 39). Show that the sequence s r r q q q √ √ √ √ 7, 7 − 7, 7 − 7 + 7, 7 − 7 + 7 − 7, . . . converges, and evaluate the limit. Recently, Fern´ andez-S´ anchez and Trutschnig in an elegant paper [39] show how various aforementioned results on the nested square roots can be derived easily via a topological conjugacy linking the following tent map and the logistic map. See Figure1.1.

FIGURE 1.1 The tent (left) and logistic (right) maps

36

Limits

Moreover, applying this clever approach, they find some new and striking results for the radicals which are not periodic. Here is another Monthly problem which may use trigonometric substitution to compute the limits. Problem E 2835 (Proposed by M. Golomb, 87(6), 1980). Let −1 < a0 < 1, and define p recursively an = (1 + an−1 )/2, n > 0. Find limn→∞ 4n (1−an ), B := limn→∞ a1 a2 · · · an , and lim 4n (B − a1 a2 · · · an ). n→∞

A much more challenging limit problem involving radicals is as follows: Problem 11367 (Proposed by A. Cusumano, 115(5), 2008). Let x1 = p √ √ 1 + 2, x2 = 1 + 2 1 + 3, and in general, let xn+1 be the number obtained by replacing the innermost expression (1 + (n + 1)) in the nested square root p formula for xn with 1 + (n + 1) 1 + (n + 2). Show that lim

n→∞

xn − xn−1 = 2. xn+1 − xn

As an continuation of this Monthly problem, the reader may study the conditions on x and an for which s r q p 1 + x 1 + (x + 1) 1 + · · · + (x + n − 1) 1 + (x + n)an converges and then determine the limit.

1.11

Nested radicals and generalized Fibonacci numbers

Problem 12063 (Proposed by H. Ohtsuka, 125(8), 2018). Let p and q be real numbers with p > 0 and q > −p2 /4. Let U0 = 0, U1 = 1, and Un+2 = pUn+1 + qUn for n ≥ 0. Calculate v v u s u u r u q u t t 2 2 2 lim U1 + U2 + U4 + · · · + U22n−1 . n→∞

Discussion. For each n ∈ N, the challenge is determining how to generate the nested radicals systematically. To gain insight, we analyze a simpler case: v v u s u u r u q u tF 2 + tF 2 + F 2 + · · · + F 2 , 1 2 4 2n−1

Nested radicals and generalized Fibonacci numbers

37

where Fn is the nth Fibonacci numbers. Recall Catalan’s identity Fi2 − Fi+k Fi−k = (−1)i−k Fk2 . Setting i = 2n + 2, k = 2n in this identity yields F22n +2 − F2n+1 +2 = F22n . Thus F2n +2 =

q

F22n + F2n+1 +2 .

(1.16)

Repeatedly applying (1.16) leads to r q 2

q = F20 +2 = F220 + F21 +2 = F12 + F221 + F22 +2 v v u v u u s u u r u u q u u t t 2 2 2 t = · · · = F1 + F2 + F4 + · · · + F22n−2 + F22n−1 + F2n +2 .

That is exactly what we expect. Taking an = F2n +2 , (1.16) becomes an = p F22n + an+1 . This suggests we find a proper sequence an which satisfies the iteration q an = U22n + an+1 . Solution. p The limit is (p + 4 + p2 + 4q )/2. To see this, let p p p + p2 + 4q p − p2 + 4q α= , β= . 2 2 Similar to Binet’s formula for Fibonacci numbers, we have Un = p

1 p2 + 4q

(αn − β n ), for n ≥ 0.

Let Vn = αn + β n for n ≥ 0. Direct calculations give U2n = Un Vn ,

(1.17)

Vn2 + (p2 + 4q)Un2 = 2V2n , and

(1.18)

V2n = Vn2 − 2(−q)n .

(1.19)

Define an =

V2n +

p

4 + p2 + 4q U2n , for n ≥ 0. 2

38

Limits

Applying (1.17) and (1.18) yields 4a2n

=

V22n + (4 + p2 + 4q)U22n + 2

p

4 + p2 + 4q U2n V2n p + 2 4 + p2 + 4q U2n V2n

4U22n + V22n + (p2 + 4q)U22n p = 4U22n + 2V2n+1 + 2 4 + p2 + 4q U2n+1

=

=

4U22n + 4an+1 .

Thus, an =

q U22n + an+1 .

Repeatedly using this identity gives r q q 2 a0 = U1 + a1 = U12 + U22 + a2 = · · · v s u r u q t 2 2 U1 + U2 + · · · + U22n−1 + an . = Since an > 0, it follows v v u s u u r u q u t t a > U2 + U2 + U2 + · · · + U2 0

1

2

4

2n−1 .

(1.20)

On the other hand, given any  ∈ (0, a0 ), let t = 1 − /a0 . Then v s u r u q t 2 2 a0 −  = ta0 = t U1 + U2 + · · · + U22n−1 + an

=

v s u r u q t2 2 2 2 2 t U1 + t U2 + · · · + t2n (U22n−1 + an )


0 (as n → ∞). 2 2 U2n−1 + an 1 + an /U2n−1 1+L

Since 0 < t < 1, there exists an N ∈ N such that n

t2
N ). U22n−1 + an

Therefore, when n > N , we have v v u s u u r u q u t 2 t 2 2 a −  < U + U + U + · · · + U2 0

1

2

2n−1 .

4

(1.21)

Since  is arbitrary, from (1.20) and (1.21), by the squeeze theorem, we find v v u s u u r p u q u p + 4 + p2 + 4q t t 2 2 2 2 , lim U1 + U2 + U4 + · · · + U2n−1 = a0 = n→∞ 2 as claimed.



Remark. Similarly, for any k > 0, we have v v u s u u r u q u t t 2 2 2 lim kU + kU + kU + · · · + kU 2 n→∞

1

2

4

2n−1

=

p+

p

4k + p2 + 4q . 2

Nested radicals were a favorite topic of Ramanujan. One of his famous results s r q √ 1 + 2 1 + 3 1 + 4 1 + · · · = 3, appeared as Putnam Problem 1966 A-6 ([4], p. 5). There are three more problems for additional practice. 1. (A Convergence Test for the Nested Radicals) Let {an } be a positive sequence and s r q √ Rn := a1 + a2 + a3 + · · · + an . Assume the following limit exists: L = lim sup n→∞

ln(ln an ) . n

40

Limits ˆIf L < 2, then Rn converges. ˆIf L > 2, then Rn diverges. ˆIf L = 2, the test is inconclusive.

2. This problem involves the rate of convergence: Let s r q √ an = 1 + 2 + 3 + · · · + n for n ∈ N.√Prove that limn→∞ an = a exists and limn→∞ n · √ n a − an = e/2. 3. Problem 11967 (Proposed by H. Ohtsuka, 124(3), 2017). Let Fn n be the nth Fermat number 22 + 1. Find v s u r u q p t lim 6F1 + 6F2 + 6F3 + · · · + 6Fn . n→∞

1.12

A limit involving arctangent

Problem 11592 (Proposed by M. Ivan, 118(8), 2011). Find ! n X 1 lim − ln n + arctan . n→∞ k k=1

Discussion. Pn 1 Since there is no closed form for k=1 arctan k , based on the well-known limit ! n X 1 lim − ln n = γ, (1.22) n→∞ k k=1 Pn we can replace the arctan sum by k=1 k1 . Another observation is Z 0

1

x2

k 1 dx = arctan = Im(ln(1 + i/k)) = Arg(1 + i/k), 2 +k k

√ where i = −1. This makes the problem more manageable in the complex domain. Here we give two solutions. One uses the gamma function while the other uses the digamma function.

A limit involving arctangent

41

Solution I. Let L denote the desired limit. From (1.22) we have L = γ − lim

n→∞

∞  X 1

k

k=1

− arctan

1 k

 .

Recall the product representation of the gamma function ∞  Y 1 z  −z/k e . = zeγz 1+ Γ(z) k k=1

Taking the logarithm of both sides, and then setting z = i, we obtain ln Γ(i) = −iγ − ln i +

∞  X i k=1

  i − ln 1 + . k k

From Im(ln(1 + i/k)) = Arg(1 + i/k) = arctan k1 , so equating the imaginary parts yields ∞

π X Arg(Γ(i)) = −γ − + 2



k=1

1 1 − arctan k k

 ,

which implies that L = −π/2 − Arg(Γ(i)) = −Arg(Γ(1 + i)), with the principle branch cut.



Solution II. Recall two properties of the digamma function ψ: P1 ψ(z + n) = 1.2.7).

1 z

+

1 z+1

+ ··· +

1 z+n−1

+ ψ(z) for n ∈ N ([8], p. 13, Theorem

P2 ψ(z) = ln z + O(1/z) for |argz| < π − δ, δ > 0 ([8], p. 22, Corollary 1.4.5). By partial fractions and P1-2, we have n X k=1

k 2 x + k2

=

 n  1 X 1 1 + 2 k + ix k − ix k=1

1 = (ψ(n + 1 + ix) − ψ(1 + ix) + ψ(n + 1 − ix) − ψ(1 − ix)) 2 1 1 = − (ψ(1 + ix) + ψ(1 − ix)) + ln(n2 + x2 ) + O(1/n). 2 2

42

Limits

Thus n X k=1

1

Z

1 arctan k

= 0

n X k=1

k x2 + k 2

! dx

i 1 (ln Γ(1 + ix) − ln Γ(1 − ix)) |10 + ln(n2 + 1) 2 2 +n arctan(1/n) − 1 + O(1/n) i 1 = (ln Γ(1 + i) − ln Γ(1 − i)) + ln(n2 + 1) 2 2 +n arctan(1/n) − 1 + O(1/n). =

This implies lim

n→∞

− ln n +

n X k=1

1 arctan k

! =

i Γ(1 + i) ln = −Arg(Γ(1 + i)). 2 Γ(1 + i) 

Remark. Beckwith noted this problem is a special case of Formula 6.1.27 in [1]:   ∞  X y y Arg(Γ(x + iy)) = yψ(x) + − arctan . x+k x+k k=0

Replacing arctan with arctanh  in the proposed problem, then telescoping  1 1+x based on arctanhx = 2 ln 1−x , we have a surprisingly simple result lim

n→∞

− ln n +

n X k=1

1 arctanh k

! =−

1 ln 2. 2

Here are five more problems for additional practice. 1. Find lim

n→∞

n √ Y n k=1

e1−1/k
k . 1 + k1

2. Problem 11494 (Proposed by O. Furdui, 117(3), 2010). Let A be the Glaisher-Kinkelin constant, given by A = lim n

−n2 /2−n/2−1/2 n2 /4

n→∞

e

n Y

k k = 1.2824 . . . .

k=1

Prove that ∞  Y n=1

n! √ 2πn(n/e)n

(−1)n−1 =

A3 . 27/12 π 1/4

Summing to the double factorials

43

3. Problem 11612 (Proposed by P. Bracken, 118(10), 2011). Evaluate in closed from n ∞  Y n+z+1 e(2z−2n+1)/(2n) . n + z n=1 4. Problem 12029 (Proposed by H. Ohtsuka, 125(3), 2018). For a > 0, evaluate  n  Y k lim a+ . n→∞ n k=1

5. Problem 11677 (Proposed by A. Stadler, 119(10), 2012). Evaluate ∞  Y

1 + 2e−nπ

√ 3

 cosh

n=1

nπ √ 3

 .

Q∞ Hint. Recall the eta function η(z) = eπiz/12 n=1 (1 − e2nπiz ). Express this product in terms of a ratio of eta function values. Stenger [89] presents an interesting solution to this problem via the approach of “Experimental Math.”

1.13

Summing to the double factorials

Problem 11821 (Proposed by F. Holland and C. Koester, 122(2), 2015). Let p be a positive integer. Prove that   Y p n 1 X 2p n lim n p (n − 2k) = (2j − 1). n→∞ 2 n k j=1 k=0

Discussion. We can proceed with this problem in two different ways. Let Sp (n) =

n X k=0

  n (n − 2k) . k 2p

First, to get a feel for this sequence, we compute the first few terms by Math-

44

Limits

ematica: n X

  n = 2n n, k k=0   n X n S2 (n) = (n − 2k)4 = 2n (3n2 − 2n), k k=0   n X 6 n S3 (n) = (n − 2k) = 2n (15n3 − 30n2 + 16n), k k=0   n X n S4 (n) = (n − 2k)8 = 2n (105n4 − 410n3 + 588n2 − 272n). k S1 (n) =

(n − 2k)2

k=0

n The emerging pattern suggests that Sp (n) = 2Q Pp (n), where Pp (n) is a polyp p nomial in n with the leading coefficient n j=1 (2j − 1), from which the proposed limit follows immediately. The second approach is to show that 2−n Sp (n) is a (2p)th derivative of a well-known function.

Solution I. Let Sp (n) =

n X

(n − 2k)2p

k=0

  n k

for n, p > 0.

We claim that Sp (n) satisfies the recurrence Sp+1 (n) = n2 Sp (n) − 4n(n − 1)Sp (n − 2).

(1.23)

Indeed, we have 4n(n − 1)Sp (n − 2)

=

=

=

n−2 X

  n−2 k k=0   n−1 X n−2 4n(n−1)(n−2k)2p (replacing k + 1 by k) k−1 k=1   n−1 X 2p n 4k(n − k)(n − 2k) k 4n(n − 1)(n − 2 − 2k)2p

k=1

(using binomial coefficient formula). Thus (1.23) follows from n2 Sp (n) − 4n(n − 1)Sp (n − 2)

= =

n X k=0 n X k=0

(n2 − 4k(n − k))(n − 2k)2p (n − 2k)2(p+1)

  n k

  n = Sp+1 (n). k

Summing to the double factorials

45

Next, let Mp (n) = 2−n Sp (n). So M0 (n) = S0 (n) = 1. We show by induction on p that Mp (n) is a polynomial of degree p in n with the leading coefficient Q p desired result follows immediately. j=1 (2j − 1). From this theQ p We use the notation that j=1 (2j−1) = (2p−1)!!. Let cp be the coefficient p−1 of n in Mp (n). From (1.23), the inductive computation for p ≥ 0 is Mp+1 (n)

= n2 Mp (n) − n(n − 1)Mp (n − 2) = (2p − 1)!!np+2 + cp np+1 − (n2 − n)[(2p − 1)!!(n − 2)p +cp (n − 2)p−1 ] + O(np ) = cp np+1 − cp np+1 + n(2p − 1)!!(n − 2)p +n2 (2p)(2p − 1)!!np−1 + O(np ) = (2p + 1)(2p − 1)!!np+1 + O(np ). 

Solution II. Let Mp (n) = 2−n Sp (n). Applying the binomial theorem, we have 2p   n   1 X 2p 2p−m X n Mp (n) = n n · (−2k)m . k 2 m=0 m k=0

n

n

Let [x ]f (x) denote the coefficient of x in the power series of f (x). Then  2p−m   m n   X x x n enx = n2p−m and (1 + e−2x )n = (−2k)m . (2p − m)! m! k k=0

By the Leibniz rule for derivatives, we deduce that   m 2p    x 1 X 2p x2p−m nx e · (1 + e−2x )n Mp (n) = n 2 m=0 m (2p − m)! m!  2p   2p   x n 1 x x e + e−x nx −2x n = e (1 + e ) = 2n (2p)! (2p)! 2 = D2p (coshn (x))(0), where D2p indicates the (2p)th derivative. We now use the following Fa`a di Bruno’s formula [55] to compute the derivative: k 2p  m X Y (2p)! D g(x) m D2p (f ◦ g)(x) = (Dk f )(g(x)) · , k1 ! · · · k2p ! m! m=1 where k = k1 + k2 + · · · + k2p and the sum is over all partitions of 2p that satisfy k1 + 2k2 + . . . + (2p)k2p = 2p. Let f (x) = xn , g(x) = cosh x. Note that Dk (xn )(cosh(0)) = n(n − 1) · · · (n − k + 1)  1, if m is even, Dm (cosh(x))(0) = 0, if m is odd.

and

46

Limits

This implies that k1P= k3 = · · · = k2p−1 = 0. Thus k = p constraint becomes m=1 mk2m = p. Furthermore, D2p (coshn (x))(0) =

Pp

m=1

k2m and the

k−1 Y (2p)! (n − m). k 2m m=1 (k2m !)[(2m)!] m=0

X

Qp

Note that the right hand side is a p-th degree polynomial in n. The degree is attained only at k2 = p, k4 = · · · = k2p = 0. Thus we finally obtain D2p (coshn (x))(0) (2p)!! Mp (n) = lim = = (2p − 1)!!. p p n→∞ n→∞ n n p! (2!)p lim

 Remark. It is possible to find an explicit representation of Sp (n) in terms of the Stirling numbers of the second kind S(i, j) ([1], p. 822). To see this, recall the generating function of S(i, j): xi =

i X

S(i, j)(x)j ,

j=0

where (x)j = x(x − 1) · · · (x − j + 1), called the falling factorial. This implies that n   i X n i X n−j k = 2 S(i, j)(n)j . k j=1 k=0

Hence, we have !  2p   X 2p n i 2p−i (−2k) n Sp (n) = i k i=0 k=0 !   2p n   X X n i i 2p 2p−i = (−2) n k i k i=0 k=0     2p i X X 2p 2p−i  n = 2n (−1)i 2i−j S(i, j)(n)j  i i=0 j=1 n X

Moreover, Mp (n) = 2−n Sp (n) =

2p X

(−1)i

i=0



   i 2p X i−j 2 S(i, j)n2p−i (n)j  . i j=1

Fa` a di Bruno’s formula generalizes the chain rule to higher derivatives. Let |A| be the cardinality of the set A. The formula has a “combinatorial” form X Y Dn (f ◦ g)(x) = D|π| f (g(x)) · · · D|B| (x), π∈Π

B∈π

A fractional part sum with Euler’s constant

47

where π runs through the set Π of all partitions of the set {1, 2, . . . , n}, B ∈ π means the variable B runs through the list of all of the blocks of the partition π. There is a nice explanation via an explicit example in https://en.wikipedia.org/wiki/Fa_di_Bruno\%27s_formula, which also contains a memorizable scheme for 1 ≤ n ≤ 4.

1.14

A fractional part sum with Euler’s constant

Problem 11206 (Proposed by M. Ivan and A. Lupas, 113(2), 2006). Find lim

n→∞

n 1 X n n o2 , n k k=1

where {x} = x − bxc denotes the fractional part of x. Discussion. Note that

2 n n  1 X n n o2 X 1 1 = · n k k/n n k=1

k=1

R1

2

is a Riemann sum for 0 {1/x} dx. Since {1/x}2 is bounded and continuous on (0, 1) except at {1/n : n ∈ N}, it is Riemann integrable. In this way, we R1 are led to evaluate the integral 0 {1/x}2 dx. Solution. We show that Z 1  2 n 1 X n n o2 1 lim = dx = ln(2π) − γ − 1. n→∞ n k x 0 k=1



 1 For x ∈ n+1 , n1 ,     1 1 1 1 = − = − n. x x x x Then Z 0

1

 2 1 dx = x =

∞ Z X

1/n

n=1 1/(n+1) ∞  X



1 − 2n ln

n=1

1 −n x

2 dx

n+1 n + n n+1

 .

48

Limits

Let the N -partial sum be SN . Rewrite 1 + N X n=1

n ln

n+1 n

=

N X

n n+1

=2−

1 n+1

and note that

(n ln(n + 1) − (n − 1) ln n − ln n)

n=1

= N ln(N + 1) −

N X

ln n = N ln(N + 1) − ln(N !).

n=1

We obtain that SN = 2N − 2N ln(N + 1) + 2 ln(N !) − HN +1 + 1, Pn where Hn = k=1 1/k is the kth harmonic number. By Stirling’s formula, we have   1 1 ln N − N + ln(2π) + O(1/N ). ln(N !) = N + 2 2 This, in addition to the fact that HN = ln N + γ + O(1/N ), where γ is Euler’s constant, implies that   N SN = 2N ln + (ln N − HN ) + ln(2π) + 1 + O(1/N ) N +1 = ln(2π) − γ − 1 + O(1/N ). Letting N → ∞ yields the claimed limit.



Remark. One well-known fractional part integral is Z 1  1 dx = 1 − γ. x 0 One may wonder: Are there other similar formulas? Is it possible to extend these equalities from one dimension to multiple dimensions? Furdui’s Book [43] provides a beautiful collection of these types of problems and solutions. Their solutions cover a host of mathematical topics including integrals, infinite series, exotic constants, and special functions. R1 It is interesting to see that we can explicitly represent 0 {1/x}k dx in terms of the Riemann zeta function for any positive integer k. To this end, by the substitution u = 1/x, we have Z 1  k Z ∞ 1 1 k dx = {u} du 2 x u 0 1 ∞ Z n+1 X (u − n)k = du (let y = u − n) u2 n=1 n ∞ Z 1 X yk = dy (n + y)2 n=1 0 ! Z 1 ∞ X 1 k = y dy. (y + n)2 0 n=1

A fractional part sum with Euler’s constant In view of

1 = (y + n)2

Z

49

∞

te−(y+n)t dt,

0

we have ∞ X n=1

1 = (y + n)2

and so Z

∞ X

∞

Z 0

! −(y+n)t

te

Z

∞

dt = 0

n=1

te−yt dt, et − 1

 k Z 1  Z ∞ t 1 k −yt dx = y e dy dt. x et − 1 0 0

1

0

Since Z

1

y k e−yt dy = k!e−t

0

we finally obtain Z 0

1

∞ X i=1

ti−1 , (k + i)!

 k ∞ X i! 1 dx = k! (ζ(i + 1) − 1) x (k + i)! i=1

as expected. Here are three more Monthly problems for additional practice: 1. Problem 11637 (Proposed by O. Furdui, 119(4), 2012). Let m ≥ 1 be a nonnegative integer. Prove that Z 1  m m 1 X 1 xm dx = 1 − ζ(k + 1) x m+1 0 k=1

where ζ is the Riemann zeta function. 2. Problem 12031 (Proposed by O. Furdui, 125(3), 2018). (a)Prove 1

1

 x dxdy = 1 − γ, 1 − xy 0 0 where {a} denotes the fractional part of a, and γ is Euler’s constant. (b)Let k be a nonnegative integer. Prove k Z 1  k Z 1Z 1 x 1 dxdy = dx. 1 − xy x 0 0 0 Z

Z



3. Problem 12181 (Proposed by G. Apostolopoulos, 127(5), 2020). Prove  Z  ∞ X 1 1 1 √ dx = γ, k k 0 x k=2

where γ is Euler’s constant.

50

1.15

Limits

A Putnam/Monthly limit problem

Problem 11837 (Proposed by I. Pinelis, 124(1), 2017). Let a0 = 1, and for n ≥ 0 let an+1 = an + e−an . Let bn = an − ln n. For n ≥ 0, show that 0 < bn+1 < bn ; also show that limn→∞ bn = 0. Discussion. This problem is a modification of Putnam Problem 2012 B-4. The original problem asked whether bn has a finite limit as n → ∞. Two published solutions indeed show that bn converges to 0 as n → ∞. Please refer [57] and the solution by Kedlaya and Ng in http://kskedlaya.org/putnam-archive/2012s.pdf Here we give another proof based on the estimates of two Riemann sums. Solution. Note that an is strictly increasing by its definition. Partition [a0 , an ] with subintervals [ak−1 , ak ], k = 1, 2, . . . , n. Since ex is increasing, from ak −ak−1 = e−ak−1 , the Riemann sum with the left-endpoint rule gives Z an n Z ak X ean − e = ex dx = ex dx a0

>

n X

k=1

ak−1

eak−1 (ak − ak−1 ) =

k=1

n X

1 = n.

(1.24)

k=1

It follows that an > ln(e + n) > ln(n + 1) > ln n for all n ∈ N. Since Z n+1 dx 1 < = ln(n + 1) − ln n, 0 < an+1 − an = e−an < n+1 x n this shows that 0 < bn+1 < bn . Thus bn is decreasing, and so bn has a finite limit as n → ∞. We now show that bn → 0 as n → ∞. It suffices to show that ean /n converges to 1 as n → ∞. By (1.24), we have ean e >1+ . n n

(1.25)

On the other hand, the Riemann sum with the right-end point yields Z an n Z ak X an x ex dx e −e = e dx = a0


(k − 1) + e > k from (1.24). Thus, e−ak−1 < 1/k and Z n n X ean − e ≤ e1/k ≤ e + e1/x dx. k=1

1

A Putnam/Monthly limit problem Hence,

ean 2e 1 ≤ + n n n

51 Z

n

e1/x dx.

(1.26)

1

Notice that, for example, using the Stolz-Ces`aro theorem, Z n 1 lim e1/x dx = 1. n→∞ n 1 Now, ean /n → 1 as n → ∞ follows from (1.25), (1.26), and the squeeze theorem.  Remark. One reviewer of this book offered an alternative solution based on the following inequality √ ( n + e)2 . 0 < an − ln n < ln n The proof is a nice application of the mean value theorem. As an exercise, the interested reader may derive this inequality. In general, this problem is an example of the principle that one can often predict the asymptotic behavior of a recursive sequence by studying solutions of a sufficiently similar-looking differential equation. In this proposed problem, as Kedlaya and Ng suggested in http://kskedlaya.org/putnam-archive/ 2012s.pdf: We can start with the equation an+1 − an = e−an , then replace an with function y(x) and replace the difference an+1 − an with the derivative y 0 (x) to obtain the differential equation y 0 = e−y , which has the solution y = ln x.

2 Infinite Series

We select 12 infinite series problems that have appeared in the Monthly and try to present solutions in a cohesive and engaging way. Through various solutions and proofs, we illustrate how problem-solving evolve over time — from the specific to the general, from the simplified scenario to the theoretical framework, from the concrete to the abstract. You may find some problems ultimately reach results related to current research. Along the way, we hope you can learn how to solve hard problems and the motivation behind them, and expand the breadth and depth of your mathematical knowledge.

2.1

Wilf wants us thinking rationally

Problem 11068 (Proposed by H. Wilf, 111(3), 2004). For a rational number x that equals a/b in lowest terms, let f (x) = ab. (a) Show that 1 5 = , f 2 (x) 2

X x∈Q+

where the sum extends over all positive rationals. (b) More generally, exhibit an infinite sequence of distinct rational exponents P s such that x∈Q+ f −s (x) is rational. Discussion. Note that every positive rational number x = a/b can be viewed as a ordered pair (a, b). Let d = gcd(a, b). Then there exists a unique ordered pair (p, q) such that gcd(p, q) = 1, a = dp, b = dq. By the definition, we have ab = d2 (pq)

and

f (a/b) = pq.

It is easy to verify that (a, b) 7→ (p, q, d) is bijective. Collecting together those P∞ 1 a, b in a,b=1 (ab) 2 which have the same gcd(a, b) yields   ! ∞ ∞ X X X 1 1 1   = · . (ab)2 f 2 (x) d4 + a,b=1

x∈Q

d=1

53

54

Infinite Series

Thus we see the rationality of the required series depends on the ratio of the Riemann zeta function values. Solution. (a) Based on the above discussion, for any s > 1, we have ! ! ∞ ∞ ∞ X X X 1 1 1 · = s s a b (ab)s a=1 b=1 a,b=1   ∞ ∞ X X 1   = d2s (pq)s d=1 p,q=1, gcd(p,q)=1   ∞ ∞ X 1 X 1  = · d2s (pq)s d=1

=

p,q=1, gcd(p,q)=1

∞ X X 1 1 · . 2s s (x) d f + d=1

x∈Q

In terms of the Riemann zeta function ζ(s) = X x∈Q+

P∞

k=1

1/k s , it follows that

ζ 2 (s) 1 = . f s (x) ζ(2s)

(2.1)

In particular, since ζ(2) = π 2 /6 and ζ(4) = π 4 /90, From (2.1) we find that X x∈Q+

1 (π 2 /6)2 5 = = . f 2 (x) (π 4 /90) 2

Here are a few more exponents such that the corresponding series have rational values: X 1 7 X 715 X 7297 1 1 = , = , = . 4 (x) 6 (x) 8 (x) f 6 f 691 f 7234 + + + x∈Q

x∈Q

x∈Q

(b) In view of the above results for exponents s = 4, 6, 8, it is natural to think of even exponents. Recall Euler’s famous result on the Riemann zeta function at even positive integers: ζ(2n) = (−1)n−1

22n−1 B2n π 2n , (2n)!

(2.2)

where Bk is the kth Bernoulli number. Let s = 2n. By (2.1) and (2.2), we find that   2 X ζ 2 (2n) 1 4n B2n 1 = = , f 2n (x) ζ(4n) 2 2n |B4n | + x∈Q

Wilf wants us thinking rationally

55

which is rational because Bk is rational for all k ∈ N.



Remark. Since there is no closed form similar to (2.2) for ζ(2n + 1), it is unknown whether there exists any other rational number s > 1 such that ζ 2 (s)/ζ(2s) is rational. The question whether ζ(2n + 1) is irrational has been asked since the time of Euler. For almost 200 years there had been no progress until 1978 when Ap´ery [9] proved that ζ(3) is irrational. Despite considerable effort we still know very little about the irrationality of ζ(2n + 1) for n ≥ 2. Ap´ery, who used the series for ζ(3) below in his proof of irrationality of ζ(3), suggested representing ζ(n) via the series involving the central binomial coefficients. For example, ζ(2) ζ(3) ζ(4)

=

3

∞ X

1

n=1 ∞ X

n2


,

2n n

(−1)n−1
, n3 2n n

=

5 2

=

∞ 36 X 1
. 4 17 n=1 n 2n n

n=1

But this kind of Ap´ery-like analogous formula fails at ζ(5) = 2 ·

∞ ∞ n−1 X (−1)n−1 5 X (−1)n−1 X 1

. − 2 n=1 n3 2n k2 n5 2n n n n=1 k=1

Thus it would seem we can’t find ζ(2n + 1) in terms of any numbers whose names we have already known. They are actually “new” numbers! Another path to follow is to replace the function f (x). For example, let f (x) = ab(a + b) if gcd(a, b) = 1. Then   ∞ ∞ ∞ X X X 1 1   = ab(a + b) d3 pq(p + q) a,b=1 d=1 p,q=1, gcd(p,q)=1   ∞ ∞ X 1 X 1  = · d3 pq(p + q) d=1

= ζ(3) ·

p,q=1, gcd(p,q)=1

X x∈Q+

1 . f (x)

56

Infinite Series

On the other hand, we have ∞ X a,b=1

1 ab(a + b)

Z 1 ∞ X 1 xa+b−1 dx ab 0 a,b=1 ! ∞ ! Z 1 X ∞ X xb dx xa = a b x 0 a=1 b=1 Z 1 2 Z ∞ 2 −u ln (1 − x) u e du (using x = 1 − e−u ) = dx = x 1 − e−u 0 0 ! Z ∞ ∞ ∞ X X 2 2 −ku = u e du = = 2ζ(3). k3 0 =

k=1

k=1

Here the interchange of integration and summation is justified by the positivity of the summands. Thus we obtain X 1 = 2, f (x) + x∈Q

a rational number! It is interesting to see that f (x) = ab and f (x) = ab(a + b) both are just special cases of the following Witten zeta function [94] X 1 W(r, s, t) := , ar bs (a + b)t a,b=1

which is a function associated to a root system that encodes the degrees of the irreducible representations of the corresponding Lie group. Witten’s work has launched extensive research to obtain exact values of W(r, s, t) in terms of the Riemann zeta function. Some typical evaluations include  n  4 X 4n − 2k − 1 ζ(2k)ζ(6n − 2k), (2.3) W(2n, 2n, 2n) = 3 2n − 1 k=0  n  X 4n − 2k + 1 W(2n + 1, 2n + 1, 2n + 1) = −4 ζ(2k)ζ(6n − 2k + 3). 2n k=0

(2.4) Let f (x) = ar bs (a + b)t if gcd(a, b) = 1. Then X W(r, s, t) = ζ(r + s + t) · x∈Q+

1 . f (x)

1 is rational when r = s = t = 2n. From (2.2) and (2.3) we see that x∈Q+ f (x) But, note that Z 1 2 ln x ln2 (1 − x) W(1, 1, 3) = = −2ζ(2)ζ(3) + 4ζ(5), x 0

P

Old wine in a new bottle

57

we don’t know whether ∞ X p,q=1, gcd(p,q)=1

ζ(2)ζ(3) 1 =4−2 pq(p + q)3 ζ(5)

is rational. Here we skim the terms in Riemann zeta function to produce a rational sum. By Riemann’s theorem on conditionally convergent series, we can rearrange a conditional convergent series that sum to any rational number (indeed, any number!) we wish. For example, with the alternating harmonic series, we have 1 1 1 1 1 1 1 1 1 1− − − − + − − − − + · · · = 0. 2 4 6 8 3 10 12 14 16 In general, as early as 1953, in the following Monthly problem, Klamkin asked to characterize the rational sum by rearrangement. We leave it to the reader for additional practice. Problem 4552 (Proposed by M. S. Klamkin, 60(7), 1953). What derangement of terms of ∞ X (−1)n+1 n n=1 will produce a sum which is rational?

2.2

Old wine in a new bottle

Problem 4305 (Proposed by H. F. Sandham, 55(7), 1948). Prove that 

1 + 1/2 1+ 2

2  2  2 1 + 1/2 + 1/3 1 + 1/2 + 1/3 + 1/4 17π 4 + + +· · · = . 3 4 360

Discussion. This problem was initially published as the Monthly Advanced Problem in 1948. Let Hn be the nth harmonic number. The stated identity can be rewritten in the form nowadays they are called Euler sums: S :=

∞ X Hn2 17π 4 17 = = ζ(4). 2 n 360 4 n=1

(2.5)

This identity apparently remained unnoticed until 1993 when Au-Yeung, an undergraduate at the University of Waterloo, numerically rediscovered (2.5). Shortly thereafter it was rigorously proven true by Borwein and Borwein in [16]. This empirical result launched a fruitful search for Euler sums through

58

Infinite Series

a profusion of methods: Combinatorial, analytic, and algebraic. Here we give four proofs to (2.5). The first algebraic proof is a modification of the original 1950 published solution by Kneser [58]. He was the solo solver beside the Proposer. Applying the Riemann zeta function and the approach we used in the previous section makes the original proof more compact. The second solution is based on logarithmic integrals and Abel’s summation formula. The third solution uses Parseval’s identity in Fourier analysis. In contrast to Borweins’ original complex-variable proof in [16], here we present a real-variable proof. Bearing in mind Hadamard’s dictum ([52], p. 123) “The shortest and best way between two truths of the real domain often passes through the imaginary one.” We give the final proof by using complex residue theory. Before proceeding the proofs, we give a brief historical account of Euler sums. In response to a letter from Goldbach in 1742, for integers m ≥ 1 and n ≥ 2, Euler [37] studied sums of the form S(m, n) :=

∞ X k=1

1 (k + 1)n



1 1 1 + + ··· + 2 k

m =

∞ X k=1

Hkm (k + 1)n

and established the following beautiful formula S(1, n) =

n−2 n 1 X ζ(n + 1) − ζ(k + 1)ζ(n − k), 2 2 k=1

or equivalently, ∞ n−2 X 1 X Hk n+2 ζ(n + 1) − ζ(k + 1)ζ(n − k). = kn 2 2

k=1

(2.6)

k=1

However, over the centuries progress on evaluating S(m, n) for m ≥ 2 (often called nonlinear Euler sums) has been minimal. For example, using the method of partial fractions, Nielsen (1906) found the formula for S(m, n) when m = n [71]; Georghiou and Philippou (1983) established [45] ∞ (2) X Hk k 2n+1

= ζ(2)ζ(2n + 1) −

k=1

+2

n+1 X

(n + 2)(2n + 1) ζ(2n + 3) 2

(j − 1)ζ(2j − 1)ζ(2n + 4 − 2j),

n ≥ 1,

j=2 (2)

where Hk = 1 + 212 + · · · + k12 . Until 1993, especially after the publication of the very insightful solution to this Monthly problem [16], Borweins and their co-researchers have generated a revival of interest in Euler sums. Since then, a large class of Euler sums and their variations have been explicitly evaluated in terms of the Riemann zeta function values. The derivations make a very pleasant journey through

Old wine in a new bottle

59

diverse topics in classical analysis include generating functions, special integrals, and special functions. For example, Borwein and Bradley [21] collected 32 intriguing proofs of S(1, 2) = ζ(3). For reasonably detailed accounts of the recent developments of Euler sums, please refer the Chapter 3 in BorweinBailey-Grigensohn’s book [19] . Solution I — The initial proof by Kneser. First, by partial fractions, telescoping yields ∞ X i=1

∞

X 1 1 = i(i + n) n i=1



1 1 − i i+n

 =

Hn . n

Then S

=

=

=

=

 1   = ij(i + n)(j + n) n=1 i,j=1 n=1 i=1   ∞ ∞ X X X 1 1   (let j = i + k) +2 2 (i + n)2 i ij(i + n)(j + n) n=1 i 1≤i k. We now conclude this section with some problems for additional practice and independent study. 1. Fourier series. Prove that
P∞ sin n P∞ sin n 2 (a) = n=1 n=1 n = n P∞ sin2 n (π−1)2 (b) . n=1 n4 = 6

π−1 2 .

2. Problem 4946 (Proposed by P M. S. Klamkin, 68(1), 1961). Let ∞ Sn = 1 + 1/2 + · · · + 1/n. Sum n=1 Sn /n!.

68

Infinite Series 3. Problem 11885 (Proposed by C. I. V˘alean, 123(1), 2016). Prove that ∞ X ∞ X ∞ X

3 5 1 = ζ(3) − ζ(4). (m + n)4 + ((m + n)(m + p))2 2 4

p=1 n=1 m=1

4. Ramanujan’s constant. Ramanujan [12] claimed that, without a proof, G(1) :=

where hk =

∞ ∞ ∞ 1 X hk π X (−1)k π X 1 √ = − , 3 3 8 k 4 (4k + 1) 3 3 k=0 (2k + 1)3 k=1 k=0

Pk

i=1

1/(2i − 1). Can you prove or disprove it?

5. Evaluate log-sine and log-cosine integrals in closed form. Find the values of (m, n, x) to calculate the following integrals in closed form. Z x Lc(m, n, x) = θm lnn (2 cos(θ/2)) dθ, 0 Z x Ls(m, n, x) = θm lnn (2 sin(θ/2)) dθ. 0

Some known results for your tests: Z

π/2

θ ln(2 sin(θ/2)) dθ = 0

Z

1 35 ζ(3) − πG, 32 2

π/3

θ ln2 (2 sin(θ/2)) dθ =

0

17 π, 6480

where G denotes the Catalan constant defined by G = P∞ (−1)k k=0 (2k+1)2 . P∞ hm k 6. Independent study on S(m, n) = k=1 kn , where hk = Pk i=1 1/(2i − 1). (a) Show that ∞ X k=1

(−1)k+1

hk 7 = πG − ζ(3), k2 4

where G is the Catalan constant. (b) When n is even, show that n/2−1 1 X n−2j+1 2n+1 − 1 S(1, n) = ζ(n+1)− (2 −1)ζ(2j)ζ(n−2j+1). 4 2 j=1

Old wine in a new bottle

69

In particular, ∞ X 31 hk 7 = ζ(5) − ζ(2)ζ(3). k4 4 2

k=1

(c) For n ≥ 2, let (−1)n−1 2n (n − 1)!

J(n) :=

1

Z 0

lnn−1 (1 − x) √ dx. x 1−x

Show that ∞

J(n) =

X 1 1 (ζ(n) + η(n)) = , 2 (2k − 1)n k=1

where η is the Dirichlet eta function, an alternating zeta function. i.e., η(x) :=

∞ X (−1)k+1 = (1 − 21−x )ζ(x). kx

k=1

(d) Show that, for n ≥ 1, ∞ X k=1

=

∞

X hk hk + (2k − 1)2n+1 (2k)2n+1 k=1

n−1 X 22i+1 − 1 1 J(2n+2)+J(2n+1) ln 2− ζ(2i+1)ζ(2n−2i+1). 2 22n+2 i=1

(e) Let G(1) be Ramanujan’s constant defined in Problem 4. Show that   Z 1 1 ln x 2 1 + x ln G(1) = − dx 8 0 x 1−x ∞ (2) Hk 3 1 X = ζ(4) − (−1)n+1 . 16 4 (n + 1)2 k=1

(f) Show that ∞ X

(2)

(−1)n+1

k=1

Hk 65 7 = ζ(4) − ζ(3) ln 2 + ζ(2) ln2 2 2 (n + 1) 16 2

1 4 ln 2 − 4 Li4 (1/2), 6 P∞ where Lin (x) = k=1 xn /k n denotes the polylogarithm function. −

70

Infinite Series 7. Independent study on hyperharmonic numbers. (k) Let Hn be the hyperharmonic numbers defined by (2.13). (a) Recall the Hurwitz zeta function defined by ζ(m, n) =

∞ X i=0

1 . (n + i)m

For m ≥ k + 1, show that ∞ ∞ (k) X X Hn = Hn(k−1) ζ(m, n). m n n=1 n=1 H(k) n n n=1 n x

P ∞ Hn n Pk−1  1 P∞ (k) n − = x + n=1 n j=1 j n=0 Hn x 
P ∞ n+j−1 1 xn . n=0 n j2
1
Pk−1 P∞ (c) H(k, m) = H(1, m)+ i=1 1i H(k, m − 1) − i12 n=0 n+i−1 n nm−1 . 3 5 (d) H(2, 3) =
−ζ(2)15 + 2ζ(3) + 4 ζ(4); H(3, 4) = − 4 ζ(2) − 1 + ζ(2) ζ(3) + 8 ζ(4) + 3ζ(5). 4 (b)

2.3

P∞

Another Euler sum

Problem 11810 (Proposed by O. Furdui, 122(1), 2015). Let Hn = and let ζ be the Riemann zeta function. Find ! n ∞ X X 1 . Hn ζ(3) − k3 n=1

Pn

k=1

1/k,

k=1

Discussion. P n Let bn = ζ(3) − k=1 1/k 3 be the nth tail (remainder) in the series expansion of ζ(3). Applying Abel’s summation formula with an = Hn yields ! ! n n n−1 k X X X X Hn bn = H k bn − Hi (bk+1 − bk ). n=1

k=1

k=1

i=1

Pk The identity i=1 Hi = (k + 1)Hk − k makes the limit process manageable. An alternative approach is based on the fact that Z 1 2 xk−1 ln2 x dx = 3 , k 0

Another Euler sum

71

from which we have 1 bn = 2

Z 0

1

xn ln2 x dx. 1−x

This transforms the problem into the computation of a definite integral. Solution I. We claim the answer is 2ζ(3) − ζ(2). To prove this, we begin with n X

Hk =

k=1

n n n X k n X 1 XX 1 X n + 1 − i = = i i i i=1 i=1 i=1

k=1

k=i

= (n + 1)Hn − n = (n + 1)(Hn+1 − 1). (2.14) Pn Pn Let an = Hn , bn = ζ(3) − k=1 1/k 3 and An = k=1 Hk . Applying Abel’s summation formula and (2.14), we find that n X

Hn bn = An bn −

n=1

n−1 X

Ak (bk+1 − bk ) = An bn +

n−1 X k=1

k=1

Hk+1 − 1 . (k + 1)2

(2.15)

Using (2.14) andPHn = O(ln n), we have An = (n + 1)(Hn+1 − 1) = O(n ln n). ∞ Note that bn = k=n+1 1/k 3 . We have 1 = 2(n + 1)2

Z

∞

n+1

1 dx ≤ bn ≤ x3

Z

∞

n

1 1 dx = 2 , x3 2n

and so bn = O(1/n2 ). Letting n → ∞ in (2.15) we obtain ∞ X

Hn bn =

n=1

∞ ∞ X Hk+1 − 1 X Hk = − ζ(2) = 2ζ(3) − ζ(2) (k + 1)2 k2

k=1

k=1

as claimed, where we have used the well-known identity of Euler (see (2.6)): ∞ X Hk = 2ζ(3). k2

k=1

For completeness, we give an elementary proof of (2.16). Since ∞ ∞ ∞ k−1 X X Hk Hk−1 + 1/k X X 1 = = + ζ(3), k2 k2 ik 2 i=1

k=1

k=1

k=2

it suffices to show that ∞ k−1 X X 1 A := = ζ(3). ik 2 i=1 k=2

(2.16)

72

Infinite Series

In fact, we have ∞ X

1 (let k = i + j) i(i + j)2 i,j=1   ∞ X 1 1 1 i . − − i2 j i+j (i + j)2 i,j=1

A =

=

Notice that, for every i ∈ N, telescoping yields  X ∞  i X 1 1 1 − = . j i + j j j=1 j=1 Thus A =

∞ X i=1

 i ∞ X X 1 1 1   − i2 j=1 j j=1 i(i + j)2 

=

i ∞ X ∞ k−1 X X X 1 1 − 2 ji ik 2 i=1 j=1 i=1

=

∞ X 1 = ζ(3). i3 i=1

k=2

 Solution II. Since Z

1

xk−1 ln2 x dx =

0

2 , (for k = 1, 2, . . .), k3

we have n ∞ X X 1 ζ(3) − = k3 k=1

k=n+1

1 1 = 3 k 2

Z ∞ X k=n+1

0

1

x

k−1

1 ln x dx = 2 2

Z 0

1

xn ln2 x dx. 1−x

Thus ∞ X n=1

Hn

n X 1 ζ(3) − k3

! =

k=1

= =

Z 1 n 2 ∞ 1 X x ln x Hn dx 2 n=1 1−x 0 ! Z ∞ 1 1 X ln2 x n Hn x dx 2 0 1−x n=1 Z 1 1 ln(1 − x) ln2 x − dx. 2 0 (1 − x)2

Here interchanging the order of the summation and integration is asserted by

Another Euler sum

73

P∞ the positivity of the general terms, and − ln(1−x) = n=1 Hn xn is used in 1−x the last equation. Moreover, using integration by parts, we have Z 1 Z 1 Z 1 ln2 x ln x ln(1 − x) ln(1 − x) ln2 x dx = dx − 2 dx. (2.17) 2 2 (1 − x) (1 − x) x(1 − x) 0 0 0 P∞ Recall that 1/(1 − x)2 = n=1 nxn−1 . We find the first integral on the righthand side of (2.17) Z 1 Z 1 ∞ ∞ X X ln2 x 2 n−1 2 dx = = 2ζ(2). n x ln x dx = 2 n2 0 (1 − x) 0 n=1 n=1 To compute the second integral on the right-hand side of (2.17), in view of the facts: Z 1 Z 1 1 1 1 ln x ln(1 − x) ln x ln(1 − x) = + , dx = dx, x(1 − x) x 1−x 0 x 1−x 0 and ln(1 − x) = −

∞ X xn , n n=1

we have Z 0

1

ln x ln(1 − x) dx = 2 x(1 − x)

1

ln x ln(1 − x) dx x 0 Z 1 ∞ X 1 = −2 xn−1 ln x dx = 2 ζ(3). n 0 n=1 Z

In summary, we finally find that ! ∞ n X X 1 1 = − (2ζ(2) − 4ζ(3)) = 2ζ(3) − ζ(2). Hn ζ(3) − 3 k 2 n=1 k=1

 Remark. It is interesting to see that the identity (2.16) and its generalizations have appeared in the Monthly multiple times. For example, Advanced Pn Problem 4431 (Proposed by M. S. Klamkin, 58(2), 1951). If Sn = k=1 1/k, prove: ∞ ∞ ∞ X X X Sn 1 Sn = 2 = 2 . 2 3 n n (n + 1)2 n=1 n=1 n=1

Advanced Pn Problem 4564 (Proposed by M. S. Klamkin, 62(2), 1955). If Sn = k=1 1/k, prove: ∞ X Sn π4 = . n3 72 n=1

74

Infinite Series

Problem 10635 by N. R. Farnum, 105(1), 1998). Show that the P(Proposed ∞ value ofP ζ(3) = k=1 k −3 , also called Ap´eP ry’s constant, can be expressed as ∞ n ζ(3) = n=1 rn /n, where rn = (π 2 /6) − k=1 k −2 is the nth remainder of the series expansion of ζ(2). Problem 10754 (Proposed P by P. Bracken, 106(8), 1999). Let ζ(s) = P∞ ∞ −s −s , and let ρ(s, n) = . Show that for positive integers k=1 k k=n+1 k s ≥ 2, ∞ s−2 X ρ(s, k) s 1 X = ζ(s + 1) − ζ(s − k)ζ(k + 1). (2.18) k 2 2 k=1

k=1

You may notice that (2.18) is actually a rearranged form of Euler’s formula (2.6) because ∞ ∞ ∞ i ∞ X X X 1 1 1 ρ(s, k) X X = = S(1, s). = s k k(i + 1)s (i + 1) k i=1 k=1 i=k

k=1

Recall (2.16): 2ζ(3) = tail of 2ζ(3) as

P∞

k=1

k=1

Hk /k 2 . Motivated by (2.18), we define the nth

∞ n X X Hk = tn = 2ζ(3) − k2 k=1

k=n+1

Hk . k2

We challenge the reader to prove   ∞ n X X 1 H1 tk H2 Hk π4 = 2ζ(3) − 2 − 2 − · · · − 2 = 3ζ(4) = . k k 1 2 k 30

k=1

k=1

Applying the approach in Solution I, we can easily prove the following identity: Problem 12102 (Proposed by O. Furdui and A. Sˆınt˘am˘arian, 126(3), 2019). Prove ! ∞ n X X 1 1 2 Hn ζ(2) − − = 2 − ζ(2) − 2ζ(3), k2 n n=1 k=1

where Hn is the nth harmonic number. Moreover, along the same lines in Solution II, we can generalize Problem 10635 above as follows: For k ∈ N, ∞ X ∞ X n1 =1 n2 =1

···

∞ X nk =1

1 n1 n2 · · · nk

∞ X p=n1 +n2 +···+nk +1

1 = k!ζ(k + 2). p2

We now end this section with another extension of the problem for your exercise: Let   ∞ n X X 1 T (k) := Hn ζ(k) − . k j n=1 j=1

Reciprocal Catalan number sums

75

For k ≥ 3, show that T (k) =

k−3 k+1 1 X ζ(k) − ζ(k − 1) − ζ(k − 1 − j)ζ(j + 1), 2 2 j=1

where the last sum is empty if k = 3.

2.4

Reciprocal Catalan number sums

Problem 11765 (Proposed by D. Beckwith,
121(3), 2014). Let Cn be the 2n 1 nth Catalan number, given by Cn = n+1 n . Show that P∞ 2n 3 (a) n=0 Cn = 5 + 2 π, √ P∞ 3n (b) n=0 Cn = 22 + 8 3 π. Discussion. Since the Mathematica command Sum[2^n/CatalanNumber[n], {n, 0, Infinity}] // Expand gives the exact stated answer 5 + 3π/2, it is more ambitious to find a closed form of the generating function for the reciprocal of the Catalan numbers: ∞ X 1 n x . C n n=0

Here we demonstrate two approaches. The first is based on an integral representation which is established by the Euler beta function. The second is based on manipulation of the well-known power series of arcsin2 (x/2), which Lehmer [65] extensively used to find a large class of interesting series; interesting in the sense that the series has a closed form in terms of well-known constants. Solution I. Recall that the Euler beta function is defined by Z 1 B(α, β) = tα−1 (1 − t)β−1 dt. 0

Using B(α, β) =

Γ(α)Γ(β) , Γ(α + β)

we have 1 (n + 1)(2n)! = = n(n + 1)B(n, n + 1) = n(n + 1) Cn n!n!

Z 0

1

tn−1 (1 − t)n dt.

76

Infinite Series

Thus ∞ X 1 n x C n n=0

= =

∞ X

1+ 1+

n=1 Z 1 X ∞ 0

Z =

n(n + 1)xn

Z

1

tn−1 (1 − t)n dt

0

n(n + 1)xn tn−1 (1 − t)n dt

n=1 1

2x(1 − t) dt, (1 − xt(1 − t))3

1+ 0

where the interchange of summation and integration is justified by the positivity of the general terms. In the last equation we have used the identity ∞ X

n(n + 1)z n =

n=1

2z . (1 − z)3

Calculating the integral by partial fractions, we find that √ √ √ ∞ X 1 n (8 + x) 4 − x + 12 x arcsin( x/2) x =2 . Cn (4 − x)5/2 n=0

(2.19)

By the ratio test, the series in (2.19) converges for |x| < 4. Thus, setting x = 2 and 3 in (2.19), respectively yields the desired formulas for (a) and (b).  Solution II. Begin with the power series [65] f (x) := arcsin2 (x/2) = which converges for |x| < 2. Let Dx = (

∞ 1 1 X
x2n , 2 2 n=1 n 2n n

d dx .

Direct calculation yields !#) ∞ X 1 1
x2n Dx (x3 Dx (xDx f )) = Dx x3 Dx xDx 2 n=1 n2 2n n " !# ∞ X 1
2n = Dx x3 Dx 2n x n n n=1 ! ∞ X 1
2n+2 = 2Dx 2n x "

n=1

= =

n

∞ X n + 1 2n
4x 2n x

4x

n=1 ∞ X n=1

n

1 2n x . Cn

Reciprocal Catalan number sums Thus

77

∞ X 1 2n 1 x =1+ Dx (x3 Dx (xDx ))[arcsin2 (x/2)]. C 4x n n=0

This leads to √ ∞ X (8 + x2 ) 4 − x2 + 12x arcsin(x/2)) 1 2n x =2 . Cn (4 − x2 )5/2 n=0 Replacing x2 by x yields (2.19) immediately. Thus, setting x = yields the desired formulas for (a) and (b) again.

√ √ 2 and 3 

Remark. The Catalan numbers are probably one of the most ubiquitous sequence of numbers in mathematics. There are many counting problems in combinatorics whose solution is given by the Catalan numbers. The monograph by Stanley [86] compiles much fascinating information on the Catalan numbers which includes 66 different combinatorial interpretations of the Catalan numbers. √ Since arcsin t = arctan(t/ 1 − t2 ), we have an alternative formula from (2.19) p √ √ ∞ X (8 + x) 4 − x + 12 x arctan( x/(4 − x)) 1 n . x =2 Cn (4 − x)5/2 n=0 Lehmer [65] used ∞ ∞ X X (2x)2n (2x)2n 2x arcsin x
= = √ 2n n(n + 1)Cn n n 1 − x2 n=1 n=1

to produce many interesting series including ∞ X 2n
2n n=1 ∞ X n=1

=

n

3n
2n n

=

∞ X n 2n π
+ 1, 2n = π + 3, 2 n n=1 √ √ ∞ X n 3n 4π 3 20π 3
+ 3, + 18. 2n = 3 3 n n=1

The formulas (a) and (b) now also follow easily from these results above. Similarly, we can find additional interesting series based on (2.19). For example, we have the alternating version for (a) and (b): √ √ ! ∞ X (−1)n 14 24 5 1+ 5 = − ln , Cn 25 125 2 n=0 √ ∞ X √ (−2)n 1 3 = − ln(2 + 3), Cn 3 9 n=0 √ √ ! ∞ X (−3)n 10 36 21 5 + 21 = − ln . Cn 49 1029 2 n=0

78

Infinite Series

Finally, we end this section with some problems for additional practice. 1. Show that Cn =

1 π

2

Z

x2n

0

Z ∞ p 2 x2 4 − x2 dx = dx. 2 π 0 (x + 1/4)n+2

2. Let y(x) be the generating function of the reciprocal of the Catalan numbers. Show that y(x) satisfies x(x − 4)y 0 + 2(x + 1)y = 2. Then solve this differential equation to obtain (2.19). 3. Prove that the exponential generating function of the reciprocal Catalan numbers is given by √  ∞ X √ 1 xn x 1 x = 1 + + ez/4 πx(x + 6)erf , C n! 4 8 2 n n=0 where erf is the error function defined by Z x 2 2 e−t dt. erf(x) = √ π 0 4. Problem 11509 (Proposed by W. Stanford, 117(6), 2010). Let m be a positive integer. Prove that m2X −m+1 k=m

m2 −m+1 k−m 2
k mk


=

R1 Hint: Use 0 xm−1 (1 − x)m−1 dx 2 x)m−1 x−(m−1) dx.

1 m =


.

2m−1 m

R1 0

2

xm+(m−1)

−1

(1 −

5. Problem 11716 (Proposed by O. Knill, 120(6), 2013). Let α = √ ( 5−1)/2. Let pn and qn be the numerator and denominator of the nth continued fraction convergent to α. Thus, pn = Fn−1 and qn = Fn , where Fn is nth Fibonacci number defined by the recurrence F0 = 0, F1 = 1, and Fn+1 = Fn + Fn−1 for n > 1. Show that √

  X ∞ pn (−1)(n+1)(k+1) Ck 5 α− = , qn qn2k+2 5k k=0

where Ck denotes the kth Catalan number, given by Ck =

2k 1 k+1 k




.

Catalan generating function

2.5

79

Catalan generating function

Problem
11832 (Proposed by Donald Knuth, 122(4), 2015). Let C(z) = P∞ 2n z n n=0 n n+1 (thus C(z) is the generating function of the Catalan numbers). Prove that  ∞  X 2n zn ln2 (C(z)) = (H2n−1 − Hn ) . n n n=1 Here, Hk =

Pk

j=1

1/j; that is, Hk is the kth harmonic number.

Discussion. From the well-known generating function of the central binomial coefficients  ∞  X 2n n 1 z =√ , n 1 − 4z n=0 integrating both sides yields the generating function of the Catalan numbers √ 1 − 1 − 4z 2 √ = . C(z) = 2z 1 + 1 − 4z Logarithmic differentiation implies that   1 1 C 0 (z) √ = −1 , C(z) 2z 1 − 4z which enables us to find a power series expansion of ln C(z). After some manipulations, the problem remains to show that     n−1  2n − 2k 1 2n 1 X 2k = (H2n−1 − Hn ). 2 k n−k k n k=1

For this purpose, here we present two proofs — one by the method of generating function, another by the Wilf-Zeilberger algorithm [74] with telescoping sums. Solution I. Note that C 0 (z) 1 = C(z) 2z



1 √ −1 1 − 4z



 ∞  1 X 2n n−1 = z . 2 n=1 n

Integrating both sides yields  ∞  1 X 2n z n ln C(z) = . 2 n=1 n n

(2.20)

80

Infinite Series

By the Cauchy product, we find that ∞ 1 X ln C(z) = 4 n=1 2

Since

n−1 X k=1



2k k

1 1 = k(n − k) n

!   2(n − k) 1 zn. n−k k(n − k) 

1 1 + k n−k

 ,

by symmetry, we have ∞ 1 X ln C(z) = 2 n=1 2

n−1 X k=1

1 k



2k k



! n 2(n − k) z . n−k n

Comparing this with the proposed identity, we need to show      n−1 2(n − k) 2n 1 X 1 2k = (H2n−1 − Hn ). 2 k k n−k n k=1

Adding

1 2n


2n n

both sides yields      n 1 X 1 2k 2(n − k) 2n = (H2n − Hn ). 2 k k n−k n

(2.21)

k=1

To prove (2.21), we invoke Formula 7.43 in the Table 351 ([50], p. 351) that  ∞  X m+n 1 1 (Hm+n − Hm )z n . ln = n (1 − z)m+1 1 − z n=0 Let m = n. From

∞

X 1 = n+1 (1 − z)

k=0



 n+k k z , n

n

matching the coefficients of z on both sides yields     n X 1 2n − k 2n = (H2n − Hn ). k n n

k=1

Therefore it suffices to show that      n n X 1 X 1 2k 2(n − k) 1 2n − k = . k n 2 k k n−k k=1

k=1

(2.22)

Catalan generating function

81

We now prove (2.22) by using the method of generating functions. We compute  !  ! ∞ n ∞  ∞ X X X 1 2n − k 1 X 2n − k n n z = z k n k n n=1 k=1 n=k k=1 !  ∞ ∞  X 1 X 2m + k m+k = z k m=0 m + k k=1

=

= = =

(let n = m + k)  ! ∞ ∞  X z k X 2m + k m z m k m=0 k=1 √ k !  ∞ X zk 1 1 − 1 − 4z √ k 2z 1 − 4z k=1 √   1 1 + 1 − 4z −√ ln 2 1 − 4z 1 √ ln C(z), 1 − 4z

where we have used Formula 5.72 ([50], p. 203) that √   k ∞  X 2m + k m 1 1 − 1 − 4z z =√ . m 2z 1 − 4z m=0 1 ln C(z), we extract the coefficient By (2.20), in the Cauchy product of √1−4z n of z as    n 1 X 1 2k 2(n − k) 2 k k n−k k=1

and prove (2.22) as desired.



Solution II. We prove     n−1  1 X 2k 2(n − k) 1 2n = (H2n−1 − Hn ) 2 k n−k k n k=1

by using the Wilf-Zeilberger algorithm. First, we rewrite the above identity as n−1 X F (n, k) = 2(H2n−1 − Hn ), (2.23) k=1

where F (n, k) =

   −1 1 2k 2(n − k) 2n , k k n−k n

82

Infinite Series

which is hypergeometric in both arguments. Invoking the Wilf-Zeilberger algorithm — EKHAD (available at http://sites.math.rutgers.edu/ ~zeilberg/tokhniot/EKHAD), we find that F (n + 1, k) − F (n, k) = G(n, k + 1) − G(n, k), where G(n, k) = −

k 2 (2n − 2k + 1)F (n, k) . (n + 1)(2n + 1)(n + 1 − k)

Thus n−1 X

F (n, k)

=

k=1

n−1 X

F (n − 1, k) +

k=1

n−1 X

(G(n − 1, k + 1) − G(n − 1, k))

k=1

2(H2n−3 − Hn−1 ) + F (n − 1, n − 1) + G(n − 1, n) −G(n − 1, 1) (2n − 3)G(n − 1, 1) 1 +0+ = 2(H2n−3 − Hn−1 ) + n−1 n(2n − 1)(n − 1) 1 1 + = 2(H2n−3 − Hn−1 ) + n − 1 n(2n − 1) = 2(H2n−1 − Hn ). =

 Remark. As Solution II indicates, by using the Wilf-Zeilberger algorithm, we can now follow a mechanical procedure to evaluate sums of binomial coefficients and discover the answer quite systematically. We refer the interested reader to [70] or [74] for details. In fact, these procedures enable us to find many similar formulas that arise frequently in practice. Here we illustrate how to combine Abel’s summation formula with the Wilf-Zeilberger algorithm to derive the following identity on harmonic numbers:    n n X X n+k (−1)k−1 (2) n−k n (−1) Hk = 2 . k k k2 k=1

k=0

Let S(n) :=

n X k=0

(−1)n−k

      n n+k n n+k (2) Hk , F (n, k) = (−1)n−k . k k k k

The Wilf-Zeilberger algorithm gives F (n + 1, k) − F (n, k) = G(n, k + 1) − G(n, k), where



2(−1)n−k k 2 nk n+k k G(n, k) = . (n − k + 1)(n + 1)

Catalan generating function

83

Thus S(n + 1) − S(n) =

X

(2)

(G(n, k + 1) − G(n, k)) Hk .

k

Applying Abel’s summation formula on the right-hand side, we have X −G(n, k + 1) (k + 1)2 k X = (T (k + 1) − T (k))

S(n + 1) − S(n)

=

k≥0

= where T (k) =

2

(−1)n , (n + 1)2 n k+1 + 1)3

2(−1)n−k−1 (k + 1)2 (n − k)(n




n+k+1 k+1


.

Since S(0) = 0, we find that S(n) = S(0) + 2

n n X X (−1)k−1 (−1)k−1 = 2 . k2 k2

k=1

k=1

Another powerful approach to study sums involving harmonic numbers is to express them in terms of differentiation of binomial coefficients. Let H0 (x) = 0

and

Hn (x) =

n X k=1

1 for n = 1, 2, . . . . x+k

For a differentiable function f (x), denote Dx f (x) =

d f (x) dx

and

D0 f (x) =

d f (x)|x=0 . dx

If m ≤ n, applying logarithmic differentiation yields     m   x+n x+n X 1 x+n Dx = = (Hn (x) − Hn−m (x)) . 1+x+n−i m m m i=1 In particular, we have  D0

x+n m

 =

  n (Hn − Hn−m ) . m

Chu and Donno [32] applied these derivative operators to hypergeometric series, and derived many striking summation identities involving harmonic numbers like   n  2 X n 2n Hk = (2Hn − H2n ). k n k=0

84

Infinite Series

Indeed, the featured solution to Problem 11832 due to James Smith also employed this technique to the identity      n X x + 2k 2(n − k) x + 2n x = x + 2k k n−k n k=0

to obtain (2.21). Here we compile a list of problems for your practice in using generating functions: 1. Problem 11343 (Proposed by D. Backwith, 115(2), 2008). Show that when n is a positive integer, X n2k  X  n 2k  = 3n−2k . k k 2k k k≥0

k≥0

You may generalize this identity to X n 2k  X  n 2k + m = 3n−(2k+m) . k k+m 2k + m k k≥0

k≥0

2. Problem 11356 (Proposed by M. Poghosyan, 115(4), 2008). Prove that for any positive integer n,
n n 2 X 24n (n!)4 k
2n = (2n)!(2n + 1)! . (2k + 1) 2k k=0

3. Problem 11406 (Proposed by A. A. Dzhumadil’daeva, 116(1), 2009). Let n!! denote the product of all positive integers not greater than n and congruent to n mod 2, and 0!! = (−1)!! = 1. For positive integer n, find n   X n (2i − 1)!![(2(n − i) − 1]!! i i=0 in closed form. 4. Problem 11757 (Proposed by I. Gessel, 121(2), 2014). Let [xa y b ] be the coefficient of xa y b in the Taylor series expansion of f . Show that 1 [xn y n ] = 9n . (1 − 3x)(1 − y − 3x + 3x2 ) 5. Problem 11862 (Proposed by D. A. Cox and U. Thieu, 122(8), 2015). For positive integers n and k, evaluate k X i=0

(−1)i

   k kn − in . i k+1

Catalan generating function

85

6. Problem 11897 (Proposed by P. P. D´alyay, 123(3), 2016). Prove for n ≥ 0,

  2k 2j+2 X 2n + 2 k j+1 =2 . k+1 n k+j=n,k,j≥0

A list of problems for your practice in using the Wilf-Zeilberger algorithm: 1. Problem 11274 (Proposed by D. Knuth, 114(2), 2007). Prove that for nonnegative integers m and n, n X k=0

2k



2m − k m+n



= 4m −

 n  X 2m + 1 . m+j j=1

2. Problem 11940 (Proposed by H. Ohtsuka, 123(9), 2016). Let Tn = n(n + 1)/2
and C(n, k) = (n − 2k) nk . For n ≥ 1, prove n−1 X

C(Tk , k)C(Tk+1 , k) =

k=0

   Tn+1 n3 − 2n2 + 4n Tn . n+2 n n

3. Problem 12049 (Proposed by Z. K. Silagadze, 125(6), 2018). For all nonnegative integers m and n with m ≤ n, prove    n X (−1)k+m n + k 2k 1 . = 2k + 1 2n + 1 n−k k−m

k=m

For n, p ∈ N, n ≥ p, prove the following identities        n X n n+k k n+p n (−1)n−k Hk = (−1)n (2Hn − 4. k k p p p k=0 Hp ).    n  2   X n k 2n − p n 5. Hk = (2Hn − H2n−p ). k p n p k=0  2    n X 2n (−1)n 2n n 6. (−1)k Hk = (Hn + H2n ). k 2 n p k=0

A list of problems for your practice in using derivative operators: 1. For all n ∈ N, prove that   2 n  2  X n 2n + k 2n (H2n+k − Hk ) = 2 (H2n − Hn ). k k n

k=0

86

Infinite Series 2. Let (x)n = x(x + 1) · · · (x + n − 1) for n ∈ N. Prove that m X

(−1)k

k=0

   (n − x)m m x + k Hn+k − Hk (x) = Hm+n .
n+k k k (n + 1)m k

3. For all n ∈ N, prove that ∞ X k=1

∞ X k   X 1 n! k (−1)m = 2. (Hk+n − Hn ) = 2 (k)n+1 (m + n + 1) n m m=0 k=0

4. Problem 11026 (Proposed by J. Sondow, 110(7), 2003). Let Hn denote the nth harmonic number. Let H0 = 0. Prove that for positive integers n and k with k ≤ n, n k−1 XX

   k−1 X n2 n n = (Hn−i − Hi ). i j i i=0

(−1)i+j−1

i=0 j=k

5. Problem 11164 (Proposed by D. Barrero, 112(6), 2005). Show that if n is a positive integer then n X

k+1

(−1)

k=1

  n k

X 1≤i≤j≤k

1 1 = 2. ij n

(2.24)

Comments. Tauraso (https://www.mat.uniroma2.it/~tauraso/AMM/ AMM11164.pdf) generalized (2.24) to n X k=1

k+1

(−1)

  n k

X 1≤i1 ≤i2 ≤···≤il ≤k

l Y 1 1 = l. i n j=1 j

Subsequently, introducing an extra variable x, the above identity can be extended further: n X k=0

(−1)k

  −1 n x+k k k

X

l Y

1≤i1 ≤i2 ≤···≤il ≤k j=1

1 x = . x + ij (x + n)l+1

The interested reader is encouraged to prove this identity and pursue new results in this direction.

A series with log and harmonic numbers

2.6

87

A series with log and harmonic numbers

Problem 11499 (Proposed by O.P Kouba, 117(2), 2010). Let Hn be the nth n harmonic number, given by Hn = k=1 1/k. Let Sk =

∞ X

(−1)n−1 (ln k − (Hkn − Hn )).

n=1

Prove that for k ≥ 2,   [k/2] 1 π X (2l − 1)π k−1 (k + 1 − 2l) cot ln 2 + ln k − 2 . Sk = 2k 2 2k 2k l=1

Discusion. We attempt to evaluate this series by integration. How can we obtain it? The key step is to represent ln k − (Hkn − Hn ) in terms of an integral. Let Pk (x) = 1 + x + · · · + xk−1 . Then Z 1 0 Pk (x) dx = ln Pk (x)|10 = ln k. 0 Pk (x) Combining this result with Z 1 Z 1 1 − xm dx = (1 + x + · · · + xm−1 ) dx = Hm , 1−x 0 0 we arrive at the desired integral formula (2.25) below. Solution. Let Pk (x) = 1 + x + · · · + xk−1 . First, we prove that Z 1 0 Pk (x) kn x dx. ln k − (Hkn − Hn ) = 0 Pk (x)

(2.25)

Since (1−x)Pk (x) = 1−xk , differentiating gives (1−x)Pk0 (x) = Pk (x)−kxk−1 . Thus Pk0 (x) (1 − xkn ) Pk (x)

= = =

1 − xkn 1 − xk 1 − xkn (Pk (x) − kxk−1 ) 1 − xk (1 + x + · · · + xk−1 − kxk−1 ) (1 − x)Pk0 (x)

(1 + xk + x2k + · · · + x(n−1)k ) =

kn X i=1

xi−1 − k

n X i=1

xki−1 .

88

Infinite Series

Integrating this expression yields (2.25) as desired. Next, we have Z 1 0 ∞ X Pk (x) kn Sk = (−1)n−1 x dx P k (x) 0 n=1 Z 1 0 ∞ Pk (x) X (−1)n−1 xkn dx = 0 Pk (x) n=1 Z 1 0 Pk (x) xk dx. = k 0 Pk (x) 1 + x Since the series above converges absolutely for 0 ≤ x < 1, the dominated convergence theorem justifies to exchange the order of summation and integration. Appealing to 2xk = 1 + xk − (1 − x)Pk (x) and Z 1 k−1 Z x dx 1 1 dt 1 = = ln 2, k 1 + x k 1 + t k 0 0 we find that Sk

=

1 2

=

1 2

=

1 2

= =

1

(1 + xk )Pk0 (x) − (1 − x)Pk0 (x)Pk (x) dx Pk (x)(1 + xk ) 0  Z 1 0 Z 1 (1 − x)Pk0 (x) Pk (x) dx − dx 1 + xk 0 0 Pk (x) Z 1 0  Z 1 Pk (x) − kxk−1 Pk (x) dx − dx 1 + xk 0 0 Pk (x) Z 1 0  Z 1 k−1 Z 1 Pk (x) x dx Pk (x) dx + k − dx k 1 + xk 0 Pk (x) 0 0 1+x Z

1 2 1 (ln k + ln 2 − Tk ) , 2

where Z

1

Pk (x) dx. 1 + xk We now compute Tk by partial fractions. Let Tk :=

0

X Aj 1 + x + · · · + xk−1 = , 1 + xk x − ξj where the sum is over the kth roots of −1. Then, by L’Hˆopital’s rule, for 1 ≤ j ≤ k, (1 − xk )(x − ξj ) 2 ξj Aj = lim = . x→ξi (1 − x)(1 + xk ) k ξj − 1 Except for ξ = −1 when k is odd, the summands in the partial fraction decomposition are in conjugate pairs. If ξ = eiθ , then ξ 1 ξ¯ 1 1+x + ¯ = . ¯ ξ−1x−ξ ξ−1x−ξ (x − cos θ)2 + sin2 θ

A series with log and harmonic numbers

89

Moreover, we have Z 1 1+x 2 dx 2 0 (x − cos θ) + sin θ   1  1 1 + cos θ x − cos θ = ln(x2 − 2x cos θ + 1) + tan−1 2 sin θ sin θ     0  1 θ 1 − cos θ cos θ = ln(2 − 2 cos θ) + cot tan−1 + tan−1 2 2 sin θ sin θ     θ θ π−θ = ln 2 sin + cot . 2 2 2 Let θ = (2j − 1)π/k. We get   [k/2]   2 X (2j − 1)π (2j − 1)π (k − (2j − 1))π Tk = ln 2 sin cot + k j=1 2k 2k 2k =

  [k/2] [k/2] (2j − 1)π π X (2j − 1)π 2 X ln 2 sin + 2 (k − (2j − 1)) cot . k j=1 2k k j=1 2k

If k is odd, then j = (k + 1)/2 in the first sum corresponds to ξ = −1. In view of Z 1  π 2 1 1/2 dx = ln 2 sin , k 0 1+x k 2 Qk and the well-known fact that j=1 sin (2j−1)π = 1/2k−1 , the first sum be2k comes   k (2j − 1)π 1 1 X ln 2 sin = ln 2. k j=1 2k k In summary, we find that Tk =

[k/2] 1 π X (2j − 1)π ln 2 + 2 . (k + 1 − 2j) cot k k j=1 2k

Substituting Tk into Sk yields the proposed identity. Remark. There is another representation for Tk :   k−1 1 π X lπ Tk = ln 2 + csc . k 2k k l=1

To this end, notice that Z 1 Z 1 Pk (x) − xk−1 1 + x + · · · + xk−2 dx = dx k 1+x 1 + xk 0 0 k−1 Z 1 X 1 xl−1 + xk−l−1 = dx. 2 1 + xk 0 l=1



(2.26)

90

Infinite Series

The substitution x = 1/t yields Z

1

xk−l−1 dx = 1 + xk

0

Thus, we have Z 1 0

xl−1 + xk−l−1 dx = 1 + xk

Z 0

∞

Z 1

∞

tl−1 dt. 1 + tk

xl−1 π dx = csc k 1+x k



lπ k



and so 1

Z 0

  k−1 π X Pk (x) − xk−1 lπ dx = , csc 1 + xk 2k k l=1

from which (2.26) follows. These two different ways to calculate Tk indeed present a solution to the following Problem 11873 (Proposed by E. Ionascu, 122(10), 2015). Show that for n ∈ N with n ≥ 2, n  X j=1

2j − 1 1− n



  n−1 X (2j − 1)π jπ cot = csc . 2n n j=1

A nice generalization is   n n−1 X X (2j − 1)π jπ sin((n − 2j + 1)y) cot = csc y + , sin ny 2n n j=1 j=1 for any y 6= mπ/n for some m ∈ N. Can you provide a proof? Let ψ be the digamma function. Applying ψ(x + n) =

1 1 1 + + ··· + + ψ(x), x x+1 x−n−1

we also find that Sk = −

  k−1 1 X 1 j k−1 ψ + − γ, 2k j=1 2 2k 2k

where γ is Euler’s constant. The proposed identity can be derived by formulas ψ(x) − ψ(1 − x) = −π cot(πx) and ψ(p/q) = −γ − where 0 < p < q.

  [q/2] X π pπ 2npπ nπ cot − ln q + 2 cos ln 2 sin 2 q q q n=1

A nonlinear harmonic sum

91

One similar problem appeared in the SIAM Problems and Solutions Online (no longer active, for archival purposes only now): For k ≥ 2, find a closed form for ∞ X ln k − (Hkn − Hn ) Sk = . n n=1 By using the Multisection formula and polylogarithm functions, Rousseau obtained   k−1 1 2 1 X 2 jπ (k − 1)(k + 2) ζ(2) − ln k − ln 2 sin . Sk = 4k 2 2 j=1 k See https://archive.siam.org/journals/problems/downloadfiles/06007s.pdf Indeed, by (2.25), we obtain Z 1 0 Z 1 0 Pk (x) 1 Pk (x) Sk = − ln(1 − xk ) dx = − ln2 k − ln(1 − x) dx. 2 0 Pk (x) 0 Pk (x) Now, it remains to show that Z 0

1

  k−1 jπ (k − 1)(k + 2) 1 X 2 Pk0 (x) ln 2 sin ln(1 − x) dx = ζ(2) − . Pk (x) 4k 2 j=1 k

I leave the proof to the reader. As additional practice, the reader may try 1. Determine a closed form for Tk =

∞ X n=1

(−1)n−1

Hkn n

for k ≥ 1.

2. Problem 12189 (Proposed by H. Katsuura, 127(6), 2020). Evaluate Pk Z 1 (k + 1)xk − m=0 xmk dx, xk(k+1) − 1 0 where k is a positive Pkinteger. Hint: Let Pk (x) = i=0 xi . Prove that Z 1 Z 1 0 (k + 1)xk − Pk (x) Pk (x) dx = dx = ln(k + 1). k+1 − 1 x P k (x) 0 0

2.7

A nonlinear harmonic sum

Problem 12060 (Proposed byP O. Furdui and A. Sˆınt˘am˘arian, 125(7), 2018). ∞ Let ζ(3) be Ap´ery’s constant n=1 1/n3 , and let Hn be the nth harmonic

92

Infinite Series

number 1 + 1/2 + · · · + 1/n. Prove ∞ X Hn Hn+1 5 π2 = − − ζ(3). 3 n −n 2 24 n=2

Discussion. In contrast to the series in Sections 2.3 and 2.5, this proposed problem involves products of two harmonic numbers. We now try to reformulate the nonlinear harmonic sum into linear form by manipulation of the summations. Solution. Let the proposed series be S. Since   1 1 1 1 = − , n3 − n 2 n(n − 1) n(n + 1) we have S

=

1 2

∞ ∞ X Hn Hn+1 X Hn Hn+1 − n(n − 1) n=2 n(n + 1) n=2

=

1 2

∞ ∞ X Hn+1 Hn+2 X Hn Hn+1 − n(n + 1) n(n + 1) n=1 n=2

!

!

(shifting the index in the first sum) ∞ 1 X Hn+1 (Hn+2 − Hn ) 3 = + 2 n=1 n(n + 1) 8 =

∞ ∞ 1 X Hn+1 Hn+1 3 1 X + . + 2 n=1 n(n + 1)2 2 n=1 n(n + 1)(n + 2) 8

To proceed forward, we need the following two auxiliary results: (a)

∞ X n=1

(b)

∞ X n=1

Hn+1 = 2. n(n + 1) Hn+1 = 2ζ(3) − 1. (n + 1)2

In view of

Hn + 1/(n + 1) Hn+1 Hn+1 = − , n(n + 1) n n+1

(a) follows from ∞ X n=1

 X ∞ ∞  X Hn+1 Hn Hn+1 1 = − + = 2. n(n + 1) n=1 n n+1 n(n + 1) n=1

(2.27)

A nonlinear harmonic sum

93

(b) is the direct consequence of the well-known Euler formula (2.16) ∞ X Hn = 2ζ(3). n2 n=1

Now we are ready to prove the claimed result. Using partial fractions 1 1 1 = , − n(n + 1)2 n(n + 1) (n + 1)2 then applying (a) and (b), we find that ∞ X n=1

∞ ∞ X X Hn+1 Hn+1 Hn+1 = − = 3 − 2ζ(3). n(n + 1)2 n(n + 1) (n + 1)2 n=1 n=1

(2.28)

Similarly, using partial fractions 1 1 = n(n + 1)(n + 2) 2



1 1 − n(n + 1) (n + 1)(n + 2)

 ,

we have ∞ X n=1

Hn+1 n(n + 1)(n + 2)

=

= = = =

 ∞  Hn+1 1 X Hn+1 − 2 n=1 n(n + 1) (n + 1)(n + 2) ! ∞ 1 1 X Hn+1 − Hn + 2 n=1 n(n + 1) 2 ∞ 1 1 X 1 + 4 2 n=1 n(n + 1)2  ∞  1 1 1 X 1 + − 4 2 n=1 n(n + 1) (n + 1)2   1 1 π2 5 π2 + 2− = − . 4 2 6 4 12

Plugging this result and (2.28) into (2.27) yields S=

3 5 π2 3 5 π2 − ζ(3) + − + = − − ζ(3) 2 8 24 8 2 24

as desired. Remark. We may attack this problem directly by using Z 1 Hn xn−1 ln(1 − x) dx = − n 0



(2.29)

94

Infinite Series

and by transforming the series into an integral. We single out one calculation as follows: ∞ X Hn Hn+1 I := = ζ(2) + 2ζ(3). n(n + 1) n=2 To see this, by (2.29), we compute I

∞ Z X

=

1

y n−1 ln(1 − y) dy

0

n=1

Z

1

Z

x 0

0

Z

1

= 0

Z = 0

1

1

xn ln(1 − x) dx

0 ∞ X

1

=

Z

! (xy)n−1

ln(1 − x) ln(1 − y) dxdy

n=1 1

x ln(1 − x) ln(1 − y) dxdy 1 − xy 0  Z 1 ln(1 − y) dy dx. x ln(1 − x) 1 − xy 0

Z

For the inner integration, the substitution 1 − xy = u yields   Z 1 Z ln(1 − y) 1 1 ln(1/x) ln(x − 1 + u) dy = + du 1 − xy x 1−x u u 0 Z 1 1 1 ln(x − 1 + u) = ln x ln(1 − x) + du. x x 1−x u Another substitution (1 − x)u → u gives 1

Z 0

1 ln(1 − y) dy = 1 − xy x



 1 2 − ln (1 − x) − Li2 (x) , 2

where Li2 (x) is the dilogarithm function. Thus, we have I=−

1 2

Z

1

ln3 (1 − x) dx −

0

Z

1

ln(1 − x)Li2 (x) dx = ζ(2) + 2ζ(3). 0

The following six problems will provide additional practice: 1. Prove

∞ X n=2

7 π2 3 Hn2 = + − ζ(3). 3 n −n 4 12 2

2. Problem 11302 (Proposed by H. Alzer, 114(6), 2007). Find ∞ X k=2

(2k + 1)Hk2 . (k − 1)k(k + 1)(k + 2)

A series involving Riemann zeta values

95

3. Problem 11633 by A. Sofa, 119(3), 2012). For real a, Pn (Proposed a let Hn (a) = j=1 1/j . Show that for integers a, b and n with a ≥ 1, b ≥ 0, and n ≥ 1, n n X X Hk (1)Hk+b−1 (a) Hk (1)2 + Hk (2) + 2 a (k + b) k k=1

k=1

= Hn+b (a)(Hn (1)2 + Hn (2)). 4. Problem 11802 (Proposed P by I. Mez¨o, 121(8), 2014). Let Hn,2 = Pn n −2 and let Dn = n! k=0 (−1)k /k! (i.e., the derangement k=1 k number of n). Prove ∞ ∞ X (−1)n Hn,2 Dn π2 X = − . n! 6e n=0 n!(n + 1)2 n=1

5. Problem 11921 (Proposed by C. I. V˘alean, 123(6), 2016). Prove ln2 2

∞ X k=1

∞ ∞ X X Hk Hk Hk + ln 2 + k+1 2 k (k + 1)2 (k + 1) 2 (k + 1)3 2k k=1

k=1

1 = (ζ(4) + ln4 2). 4 Pk Here Hk = j=1 1/j and ζ denotes the Riemann zeta function. (2)

6. Nonlinear Sums. Let Hn = 1 + 1/22 + · · · + 1/n2 . Show that ln2 2

∞ X k=1

(2)

∞

(2)

X Hk Hk + (k + 1)2k+1 (k + 1)2 2k+1 k=1

1 1 1 ζ(4) + ζ(2) ln2 2 − ln4 2. = 16 4 8

2.8

A series involving Riemann zeta values

Problem 11400 (Proposed byPP. Bracken, 115(10), 2008). Let ζ be the Rie∞ mann zeta function. Evaluate n=1 ζ(2n)/n(n + 1) in closed form. Discussion. P∞ One way to proceed is to use the power series of n=1 xn /n(n + 1) and to convert the problem into an infinite product. Another way to proceed is to start with the generating function of ζ(2n)/n and to transform the series into some well-known integrals. Based on the discussion, we now show the required closed form equals ln(2π) − 1/2 in two distinct ways.

96

Infinite Series

Solution I. Denote the proposed series by S. Note that S=

∞ X n=1

∞ ∞ X ∞ X X 1 1 1 = , 2n n(n + 1) k n(n + 1)k 2n n=1 k=1

k=1

where interchanging the order of summation is justified by the positivity of summands. Recall the power series ∞ X n=1

xn n(n + 1)

=

x + ln(1 − x) − x ln(1 − x) . x

We obtain S

=

=

  ∞  X 1 1 + (k 2 − 1) ln 1 − 2 k k=2 ! 2 k −1  2 ∞ Y k −1 . 1 + ln e k2 1+

k=2

To compute the infinite product above, let PN =

N Y

 e

k=2

k2 − 1 k2

k2 −1 .

We have 22 −1  32 −1  (N −1)2 −1 2·4 (N − 2) · N ) 1·3 · · · 22 32 (N − 1)2  N 2 −1 (N − 1) · (N + 1) × . N2 

PN =

Observe that, for each 2 ≤ k ≤ N − 1, k appears in the numerator with exponents [(k − 1)2 − 1] + [(k + 1)2 − 1] = 2k 2 and in the denominator with exponent 2(k 2 − 1). After cancelation, only a factor of k 2 remains in the numerator. The factor N appears in the numerator with exponent N 2 −2N and in the denominator with exponent 2(N 2 −1). After cancelation, N remains in the denominator with exponent N 2 + 2N − 2. The factor (N + 1) occurs only in the numerator with exponent N 2 − 1. Therefore, PN =

eN −1 (N − 1)!2 (N + 1)N N N 2 +2N −2

2

−1

=

eN −1 (N !)2 (N + 1)N N N 2 +2N

2

−1

.

A series involving Riemann zeta values Rewrite PN

2π = 2 · e

2

97

(N + 1)N −1 eN −1 N N 2 −1

!  ·

eN N ! √ N N 2πN

2 .

Using Stirling’s formula, we have lim

N →∞

eN N ! √ = 1; N N 2πN

also 2


 (N + 1)N −1 2 lim = exp lim = e1/2 . (N − 1) ln(1 + 1/N ) − N + 1 2 N →∞ eN −1 N N −1 N →∞ Hence

∞ Y k=2

 e

k2 − 1 k2

k2 −1

and so

= lim PN = N →∞

 S = 1 + ln

2π e3/2



2π 2π 1/2 ·e = 3/2 , e2 e

1 = ln(2π) − . 2 

Solution II. We begin by determining the generating function of {ζ(2n)/n}. For x ∈ [0, 1), let ∞ X ζ(2n) n G(x) := x . n n=1 We have !  √ 2n ! ∞  √ 2n ∞ ∞ X X X x 1 x G(x) = m n m m=1 m=1 n=1 # " "    √ 2 #! √ ∞ ∞ 2 X Y x x = − ln 1 − = − ln 1− m m m=1 m=1  √  sin(π x) √ = − ln . π x ∞ X 1 = n n=1

Here interchanging the order of summation is justified by the absolutely convergence. In the last equation, we have used Euler’s sine infinite product formula:  ∞  Y sin(πz) z2 = 1− 2 . πz n n=1

98

Infinite Series

Since the series converges uniformly, term by term integration yields √  Z 1 Z 1  sin(π x) √ S = G(x) dx = − ln dx π x 0 0 Z 1 Z 1 √ √ √ ln(π x) dx (let t = π x) = − ln(sin(π x) dx + 0 0 Z π 1 2 t ln(sin t) dt + ln π − . = − 2 π 0 2 By symmetry, we have Z π Z π Z t ln(sin t) dt = (π−t) ln(sin(π−t)) dt = π 0

0

Z ln(sin t) dt−

0

From the well-known integral π

Z

π

t ln(sin t) dt = 0

π 2

Rπ 0

t ln(sin t) dt.

0

ln(sin t) dt = −π ln 2, we have

π

Z

π

ln(sin t) dt = − 0

π2 π π ln 2 = − ln 2. 2 2

Hence, S=−

  2 π2 1 1 · − ln 2 + ln π − = ln(2π) − . π2 2 2 2 

Remark. The generating function G(x) in Solution II also can be derived from ∞ X √ 1 √ ζ(2n)xn = − π x cot(π x), 2 n=0 from which, together with the power series of cotangent, Euler found the exact value of ζ(2n). Rewrite G(x) as   ∞ X ζ(2n) 2n sin(πx) x = − ln . n πx n=1 From here it follows that ∞ X π ζ(2n) = ln 2n n2 2 n=1

and

∞ X n=1

ζ(2n) = ln(2π) − 1. n(2n + 1)

Many series involving the Riemann zeta function values have been developed by various ways in recent years [85]. Here we collect a few identities for your verification. P∞ n ζ(s+n) 1. (1 − 21−s )ζ(s) = n=1 (s) n! 2s+n , where (s)n = s(s + 1) · · · (s + n − 1).

Abel theorem continued

99

P∞ n ζ(s+n) 2. (1 − 21−s )ζ(s) = 1 − n=1 (−1)n−1 (s) n! 2s+n . P∞ n n−1 3. ψ(x) = −γ + n=2 (−1) ζ(n)x , where ψ is the digamma function and γ denotes Euler’s constant. P∞ ζ(2n) 1 4. n=1 (n+1)(n+2) = 2 . P∞ ζ(2n+1) 5. n=1 (2n+1)22n = ln 2 − γ. P∞ ζ(2n)−1 P∞ ζ(2n)−1 = ln 2, = 32 − ln π. 6. n=1 n=1 n n+1 P∞ ζ(2n+1)−1 13 = 12 − 13 γ − 12 ln 2 − 2 ln A, where A denotes the 7. n=1 2n+3 Glaisher-Kinkelin constant. P∞ αn+β 1 1 8. n=2 n(n+1) (ζ(n) − 1) = β − 2 (α + β)γ + 2 (α − β)(3 − ln(2π)), where α and β are constants. 9. Problem 11333 (Proposed by P. F. Refolio, 114(10), 2007). Show that  2 2(n2 −1)  n ! ∞ Y n −1 n+1 = π. n2 n−1 n=2 10. Problem 11793 (Proposed by I. Mez¨o, 121(7), 2014). Prove that ∞ ∞ X X ln(n + 1) ζ(n) 0 = −ζ (2) + (−1)n+1 , 2 n n −2 n=3 n=1

where ζ denotes the Riemann zeta function and ζ 0 denotes its derivative.

2.9

Abel theorem continued

Problem 11755 (Proposed by P. P. D´alyay, 121(2), 2014). Compute ∞ ∞ X (−1)n X 2n − 1 n=1

k=n+1

(−1)k−1 . 2k − 1

Discussion. There are many ways to attack this problem. Here we single out two. Denote the double series by S. 1. By symmetry, we have 1 S=− 2

2 ! ∞ ∞ ∞  X (−1)n−1 X (−1)k−1 X (−1)n−1 − . 2n − 1 2k − 1 2n − 1 n=1 n=1 k=1

100

Infinite Series 2. Using 1/(2k − 1) =

R1

x2(k−1) dx yields

0

∞ X k=n+1

(−1)k−1 = 2k − 1

1

Z 0

(−1)n x2n dx. 1 + x2

Consequently we convert the problem into the calculation of an integral: Z 1 1 x ln(1 + x) − x ln(1 − x) dx. S= 2 0 1 + x2 We now present two solutions based on the above observations. Solution P∞I. (−1)n−1 Since only converges conditionally, to justify the rearrangen=1 2n−1 ment of terms of the double series by Fubini’s theorem, we introduce an (x) = (−1)n−1 x2n−1 . For x ∈ (0, 1), define 2n−1 f (x) :=

∞ X n=1

The absolute convergence of of summation. Thus

ak (x).

k=n+1

P∞

an (x) now allows interchanging the order

n=1

f (x) =

∞ X

an (x)

∞ X

ak (x)

k−1 X

ak (x).

n=1

k=1

By symmetry, we find that f (x)

=

=

∞ X

1 2

an (x)

n=1

1 2

∞ X

ak (x) −

arctan (x) −

! a2k (x)

k=1

k=1 2

∞ X

∞ X

! a2k (x)

.

!

1 2

k=1

By Abel’s limit theorem, we obtain 1 f (1) = lim− f (x) = 2 x→1

 π 2 4

−

∞ X k=1

1 (2k − 1)2

=



π2 π2 − 16 8

 =−

π2 , 32

which implies that S = −f (1) =

π2 . 32 

Abel theorem continued

101

Solution II. We compute ∞ X k=n+1

(−1)k−1 2k − 1

∞ X

=

(−1)k−1 ∞ X

1

= 0

Z

x2(k−1) dx

0

k=n+1

Z

1

Z

! k−1

(−1)

2(k−1)

x

k=n+1 1

= 0

(−1)n x2n dx. 1 + x2

Then S

=

= = where we have used ∞ X n=1

Z 1 ∞ X (−1)n (−1)n x2n dx 2n − 1 0 1 + x2 n=1 ! Z 1 ∞ X 1 x2n dx 2 2n − 1 0 1+x n=1 Z 1 1 x(ln(1 + x) − ln(1 − x)) dx, 2 0 1 + x2

1 1 x2n = x(ln(1 + x) − ln(1 − x)). 2n − 1 2

Next, the substitution x =

1−t 1+t

S=−

1 2

yields Z 1 0

(1 − t) ln t dt. (1 + t)(1 + t2 )

By partial fractions 1−t 1 t = − , 2 (1 + t)(1 + t ) 1 + t 1 + t2 using the geometric series and integration by parts, we obtain Z 1 Z 1 ∞ X ln t n dt = (−1) tn ln t dt 0 1+t 0 n=0 = Z 0

1

t ln t dt 1 + t2

= =

∞ X n=0 ∞ X n=0 ∞ X n=0

(−1)n+1 (−1)n

Z

1 π2 = − ; (n + 1)2 12 1

t2n+1 ln t dt

0

(−1)n+1

1 π2 = − . 4(n + 1)2 48

dx

102

Infinite Series

In summary, we find that 1 S=− 2

  π2 π2 π2 − + = . 12 48 32 

Remark. We can avoid the absolute convergence in Solution I by considering the finite sum: N X SN := an bn , n=1

where an =

(−1)n 2n − 1

and

bn =

∞ X k=n+1

(−1)k−1 . 2k − 1

By Abel’s summation formula, we have ! ! N ∞ N N X X X X 1 2 an . an an − an bN + SN = 2 n=1 n=1 n=1 n=1 Letting N → ∞ yields that S = π 2 /32 immediately. P∞ Recall the dilogarithm function defined by Li2 (x) = k=1 xk /k 2 . In Solution I, we explicitly obtain ∞ X

a2k (x)

k=1



1 = 2

 1 4 Li2 (x ) − Li2 (x ) . 4 2

Following the idea Euler first used to show that ζ(2) = π 2 /6, we present another charming solution to this proposed problem. By Euler’s infinite product of sine, we have cos x − sin x =

∞  Y n=1

(−1)n 4x 1+ (2n − 1)π

 .

Replacing x by πx/4 yields cos

 πx  4

− sin

 πx  4

=

∞  Y n=1

1+

(−1)n x 2n − 1

 .

(2.30)

Now we compute the coefficient of x2 on both sides of (2.30). Clearly, h  πx   πx i [x2 ] cos − sin 4 4   πx 2  πx  1  πx 3 1 π2 2 = [x ] 1 − + ··· + − + ··· = − . 2 4 4 3! 4 32

A convergence test

103

On the other hand, we have [x2 ]

 X ∞ ∞  ∞ Y (−1)n x (−1)n X = 1+ 2n − 1 2n − 1 n=1 n=1

k=n+1

(−1)k . 2k − 1

Thus, we conclude the desired answer by equating the coefficient of x2 in (2.30). P∞A unified P∞ treatment of the summation of certain iterated series of the form n=1 m=1 an+m is studied in [44]. Under certain conditions, the double iterated series can be represented as the difference of two single series. We end this section with four more Monthly problems for additional practice. 1. Problem 11519 (Proposed by O. Furdui, 117(7), 2010). Find ∞ ∞ X X

(−1)n+m

n=1 m=1

Hn+m , n+m

where Hn denotes the nth harmonic number. 2. Problem 11682 (Proposed by O. Furdui, 119(10), 2012). Compute ∞ X

n

(−1)

n=0

∞ X (−1)k−1 n+k

!2 .

k=1

3. Problem 12134 (Proposed by P. Bracken, 126(8), 2019). Evaluate the series ! ! ∞ ∞ X X 1 1 n −1− . k2 2n n=1 k=n

4. Problem (Proposed by M. Tetiva, 127(6), 2020). Let γn = P12194 n − ln n + k=1 1/k, and let γ be Euler’s constant limn→∞ γn . Evaluate  ∞  X 1 . Hn − ln n − γ − 2n n=1

2.10

A convergence test

Problem 11829 (Proposed by P. Bracken, 122(3), 2015). Let an be a monotone P∞ decreasing sequence of real numbers that converges P∞ to 0. Prove that a /n < ∞ if and only if a = O(1/ ln n) and n n=1 n n=1 (an − an+1 ) ln n < ∞.

104

Infinite Series

Discussion. Using Abel’s summation formula, we have n n−1 X X an = Hk (ak − ak+1 ) + Hn an , n

k=1

k=1

Pk where Hk = i=1 1/i is the kth harmonic number. Hn ∼ ln n for suffiPSince ∞ ciently large n,Pby the comparison test, we see that n=1 (an −an+1 ) ln n < ∞ ∞ if and only if n=1 (an − an+1 )Hn < ∞. Our solution below is based on this observation. Solution. Let Sn =

n n X X ak (ak − ak+1 ) ln k. , Tn = k k=1

k=1

By the assumption of an , we see both Sn and Tn are monotone increasing. Let their limits be S and T , respectively. “ ⇒ ” Assume S < ∞. Since an is decreasing, we have, for all n ∈ N an Hn = an

n n X X 1 ak ≤ = Sn ≤ S. n k

k=1

k=1

This implies that an = O(1/Hn ) = O(1/ ln n) because Hn ∼ ln n for sufficiently large n. Moreover, Tn

= = ≤

n X k=1 n X k=2 n X k=2

=

ak ln k −

n X

ak+1 ln k

k=1

ak (ln k − ln(k − 1)) − an+1 ln n n

X ak−1 ak ≤ k−1 k−1 k=2

Sn−1 ≤ S.

This shows that T < ∞. “ ⇐ ” Assume that an = O(1/ ln n) and T < ∞. First, we show that an ln n is uniformly bounded. Observer that, for m > n ≥ 1, (an − am+1 ) ln n =

m X

(ak − ak+1 ) ln n ≤

k=n

m X

(ak − ak+1 ) ln k ≤ T.

k=n

Since limn→∞ an = 0, letting m → ∞ yields an ln n ≤ T

for all n ≥ 1.

A convergence test

105

Moreover, we have Sn − a1

=

n n X X ak ≤ ak (ln k − ln(k − 1)) k

k=2

=

n X

k=2

ak ln k −

k=2

=

n−1 X

n−1 X

ak+1 ln k

k=2

(ak − ak+1 ) ln k + an ln n

k=2

≤ T + T = 2T. This proves that S < ∞ as well.



Remark. A similar problem appears in the Monthly: Problem 11865 (Proposed by G. H. Chung, 122(9), 2015).PLet an be a ∞ monotone decreasing of nonnegative real numbers. Prove that n=1 an /n is finite if and only if limn→∞ an = 0 and ∞ X

(an − an+1 ) ln n < ∞.

n=1

In Problems 11829 and 11865, the monotonicity condition of an is vital for the conclusion. Here is an counterexample: an = ln n if n = k 2 , and an = 1/n otherwise. Finally, we provide an extension of the necessary part of the proposed problem: Let an be a monotone decreasing sequence of real numbers P∞ that converges to 0. If the sequence bn is such that n=1 an bn converges, then ! n X lim bk an = 0. n→∞

k=1

Proof. For any  > 0, there exists a m ∈ N such that for every n > m n X  ak bk < . |Rn | := 4 k=m+1

We now estimate (bm+1 + bm+2 + · · · + bn )an by using Abel’s summation formula. |(bm+1 + bm+2 + · · · + bn )an | 1 1 = an am+1 bm+1 + · · · + an bn am+1 an 1 1 1 = an Rm+1 + (Rm+2 − Rm+1 ) + · · · + (Rn − Rn−1 ) am+1 am+2 an     1 1 1 1 1 = an − Rm+1 + · · · + − Rn−1 + Rn am+1 am+2 an−1 an an

106 



 1 1 an − + ··· + 4 am+2 am+1    2 1  = an − < . 4 an am+1 2



≤

1 1 − an an−1



Infinite Series  1 + an

Since an → 0, there exists an N () > m such that for n > N () |(b1 + b2 + · · · + bm )an | ≤ Thus

n ! ! m X X bk an + bk an ≤ k=1

k=1

 . 2

n X

! bk

k=m+1

an ≤ .

In the proposed problem, taking bk = 1/k yields an Hn → 0 or an = o(1/ ln n). Without the monotonicity condition on an , we challenge the reader to prove the following assertion: Let an be a sequence of nonnegative real numbers that converges to 0. Then there positive numbers bn P∞ exists a sequence of P∞ such that nbn is nonincreasing, n=1 bn diverges, and n=1 an bn converges. The following Monthly problem, can be viewed as the complementary of the above problem, offers additional practice. Problem 12084 (Proposed by G. Stoica, 126(1), 2019). Pn Let a1 , a2 , . . . be a sequence of nonnegative numbers. Prove that (1/n) k=1 ak is unbounded if and if there exists a decreasing sequence b1 , b2 , . . . such that limn→∞ bn = Ponly P∞ ∞ 0, n=1 bn is finite, and n=1 an bn is infinity. Is the word “decreasing” essential?

2.11

A power series with an exponential tail

Problem 12012 (Proposed by O. Furdui and A. Sˆınt˘am˘arian, 124(10), 2017). Let k be a nonnegative integer. Find the set of real numbers x for which the power series  ∞   X n 1 1 1 e − 1 − − − ··· − xn k 1! 2! n! n=k

converges, and determine the sum. Discussion. Observe that, once we find the closed form of the sum  ∞  X 1 1 1 e − 1 − − − ··· − xn , 1! 2! n! n=k

(2.31)

A power series with an exponential tail

107

the proposed sum can be derived from the closed form of (2.31) by applying appropriate derivative operators. How can one obtain a closed expression for (2.31)? We either apply Abel’s summation formula as we did in Section 2.3 or we use the generating function approach as we did in Section 2.4. Solution. Let Sk (x) denote the proposed series. We first show that the power series converges for all real numbers x. Indeed, from Taylor’s theorem, there exists ξ ∈ (0, 1) such that   !   n n X n eξ 1 e|x|k |x|n−k e− xn ≤ |x|n ≤ · . k k (n + 1)! i! k!(n + 1) (n − k)! i=1 By the Weierstrass M-test, this implies that Sk (x) converges for all real numbers. In particular, it converges uniformly on any bounded subset of R. Thus, we are free to interchange summation and differentiation. Next, we claim that   e−ex ∞  X 1 1 1 1−x , if x 6= 1, xn = S0 (x) := e − 1 − − − ··· − e, if x = 1. 1! 2! n! n=0 (2.32) 1 1 1 Let an = e − 1 − 1! − 2! − · · · − n! and bk = xk . If x 6= 1, we have Bk = Pk k+1 )/(1 − x). Applying the limit version of Abel’s summation i=0 bi = (1 − x formula: ∞ ∞ X X ak bk = lim Bn an+1 + Bk (ak − ak+1 ), k=0

n→∞

k=0

we find   1 − xn+1 1 1 1 1 S0 (x) = lim − e − 1 − − − ··· − n→∞ 1−x 1! 2! n! (n + 1)! ∞ X 1 1 − xk+1 + 1 − x (k + 1)! k=0 ! ∞ ∞ X X 1 1 xk+1 e − ex = − = . 1−x (k + 1)! (k + 1)! 1−x k=0

k=0

The continuity of S0 (x) implies that  ∞  X 1 1 1 S0 (1) = e − 1 − − − ··· − = lim S0 (x) = e. x→1 1! 2! n! n=0 This proves (2.32). Finally, note that   n n−k dk n n−k x = n(n − 1) · · · (n − k + 1) x = k! x , dxk k

108

Infinite Series

which is equivalent to   n n xk dk n x = x . k k! dxk Thus, for x 6= 1, term by term differentiation gives   xk dk e − ex xk dk S (x) = Sk (x) = . 0 k! dxk k! dxk 1 − x

(2.33)

Note that

  1 j! dj for j ≥ 0. = dxj 1 − x (1 − x)j+1 Using the Leibniz rule yields    k−j k   j  X dk e − ex k d 1 d = (e − ex ) j k−j j dxk 1 − x dx 1 − x dx j=0 =

k   X k i! k!e x − e . (1 − x)k+1 i (1 − x)i+1 i=0

Thus, for x 6= 1, (2.33) becomes k

X 1 exk k x − x e k−j+1 (1 − x)k+1 j!(1 − x) j=0   k k j X (1 − x)  x e − ex = . (1 − x)k+1 j! j=0

Sk (x) =

To determine Sk (1), we compute the Taylor series of ex − e at x = 1, which is given by ∞ X e ex − e = (x − 1)i . i! i=1 Thus,

∞

X e e − ex = (x − 1)i−1 , 1−x i! i=1 which implies that Sk (1) = [(x − 1)k ]

∞ X e (x − 1)i−1 i! i=1

! =

e . (k + 1)!

There is an alternative derivation based on the method of generating functions. We begin with  ∞   ∞   X n n X n 1 1 1 Sk (x) = e x − 1 + + + ··· + xn k k 1! 2! n! n=k n=k  ∞   k X 1 x n 1 1 1 + = e − + + · · · + xn . (1 − x)k+1 k 1! 2! n! n=k

A power series with an exponential tail

109

To find a closed form of the sum in the above equation, we turn to its generating function and find that  ! ∞ ∞   X X n 1 1 1 G(t) := tk 1 + + + ··· + xn k 1! 2! n! k=0 n=k  X ∞   ∞  X 1 1 1 n k n = 1 + + + ··· + x t 1! 2! n! k n=0 k=0  ∞  X 1 1 1 = 1 + + + ··· + (x(t + 1))n 1! 2! n! n=0 =

ex(t+1) . 1 − x(t + 1)

Here in the last equality P∞ we have used the well-known generating function property: let f (t) = n=0 an tn . Then ∞ X n=0

(a0 + a1 + · · · + an )tn =

f (t) . 1−t

By the Cauchy product, we finally have ! k X ex k ext 1 k k x [t ] G(t) = [t ] = x e . x k−j+1 1−x 1 − 1−x t j!(1 − x) j=0  Remark. There is an elementary way to establish (2.32). In fact, we have   ∞ ∞ X X 1   xn S0 (x) = j! n=0 j=n+1 ! j−1 ∞ ∞ X X 1 X n 1 1 − xj = x = j! n=0 j! 1 − x j=1 j=1  e−ex 1−x , if x 6= 1, = e, if x = 1. P∞ We now show an extension for S0 (x): Let f (x) = n=0 an xn converge on (−R, R). Then   Z t ∞ X t t2 tn t n!an e − 1 − − − · · · − = et−x f (x) dx, (|t| < R). 1! 2! n! 0 n=0 (2.34) Let   ∞ X t t2 tn t y(t) := n!an e − 1 − − − · · · − . 1! 2! n! n=0

110

Infinite Series

Then ∞ X

  t2 tn−1 t t y (t) = a0 e + n!an e − 1 − − − · · · − 1! 2! (n − 1)! n=1   ∞ X t t2 tn t t = (e − 1)a0 + n!an e − 1 − − − · · · − 1! 2! n! n=1 0

t

+a0 +

∞ X

n!an

n=1

=

∞ X n=0

=

tn n!

 X  ∞ t2 tn t + n!an et − 1 − − − · · · − an tn 1! 2! n! n=0

y(t) + f (t).

Solving this linear differential equation subject to y(0) = 0 yields Z y(t) =

t

et−x f (x) dx.

0

From which taking f (x) = epx gives ∞ X n=0

  t2 tn et − ept t . = pn et − 1 − − − · · · − 1! 2! n! 1−p

Setting p = x and t = 1 we obtain (2.32) again. Furthermore, (2.34) can easily be extended as  Z t  ∞ X t2 tn t (n) t = xet−x f 0 (x) dx. nf (0) e − 1 − − − · · · − 1! 2! n! 0 n=0

2.12

An infinite matrix product

Problem 11739 (Proposed by F. Adams, A. Bloch and J. Lagarias, 120(9), 2013).   1 x Let B(x) = . Consider the infinite matrix product x 1 M (t) = B(2−t )B(3−t )B(5−t ) · · · =

Y

B(p−t ),

p

where the product runs over all primes, taken in increasing order. Evaluate M (2).

An infinite matrix product

111

Discussion. One way to proceed is to use matrix diagonalization. The symmetry of B(x) asserts that B(x) is diagonalizable. Let B(x) = P D(x)P −1 . Here P is an invertible constant matrix. Then ! Y P −1 M (t)P = P −1 D(p−t ) P. p

Q Thus, M (t) has a closed form as long as the infinite product p D(p−t ) has a closed form. Another way to proceed is to solve the recurrence for the n-partial finite product. Solution I. We prove the following more general result: For all t > 1,  ζ 2 (t)+ζ(2t) ζ 2 (t)−ζ(2t)  M (t) =

1  2

ζ(t) ζ(2t)

ζ(t) ζ(2t)

2

2

ζ (t)−ζ(2t) ζ(t) ζ(2t)

ζ (t)+ζ(2t) ζ(t) ζ(2t)

 ,

where ζ(s) is the Riemann zeta function. Note that the characteristic equation of B is det(λI − B) = (λ − 1)2 − x2 = 0. We find the eigenvalues of B are 1 ± x, the orthonormal bases for the eigenspaces are √    √  1/ √2 1/√2 λ = 1 − x, v1 = ; λ = 1 + x, v2 = . −1/ 2 1/ 2  Let P =

√ 1/ √2 −1/ 2

√  1/√2 . Then P −1 BP is diagonal: 1/ 2 P −1 BP =



1−x 0

0 1+x

 .

Moreover, P −1 M (t)P




P −1 B(2−t )P P −1 B(3−t )P P −1 B(5−t )P · · ·   1 − p−t 0 Y   = p 0 1 + p−t   Q −t 0 p (1 − p ) . =  Q −t 0 p (1 + p ) =

112

Infinite Series

Applying the Euler infinite product formula for the Riemann zeta function Y ζ(s) = (1 − p−s )−1 , p

we find Y

(1 − p−t ) ·

p

Y

(1 + p−t ) =

Y

p

(1 − p−2t ) =

p

and so



1 ζ(t)

0

0

ζ(t) ζ(2t)

 P −1 M (t)P = 

1 , ζ(2t)

  .

Finally, we conclude  M (t) = P P −1 M (t)P P −1 =


1   2

ζ 2 (t)−ζ(2t) ζ(t) ζ(2t)

ζ 2 (t)+ζ(2t) ζ(t) ζ(2t) 2

2

ζ (t)−ζ(2t) ζ(t) ζ(2t)

ζ (t)+ζ(2t) ζ(t) ζ(2t)

  .

In particular, when t = 2, since ζ(2) = π 2 /6, ζ(4) = π 4 /90, we find that   7 3 3  . M (2) = 2π 2 3 7  Solution II. For n ≥ 0, define Mn (t) =

n Y

B(p−t k )

 =

k=1

an bn



bn an

.

Then a0 = 1, b0 = 0 and an

=

an−1 + p−t n bn−1 ,

bn

=

p−t n an−1 + bn−1 .

By iteration, we have an + bn = (an−1 + bn−1 )(1 + p−t n )=

n Y

(1 + p−t k );

k=1

an − bn = (an−1 − bn−1 )(1 −

p−t n )

=

n Y

(1 − p−t k ).

k=1

An infinite matrix product

113

Since n Y

(1 − p−t k )→

k=1

n n 2 Y Y 1 − (p−t ζ(t) 1 k ) → , , (1 + p−t ) = k −t ζ(t) 1−p ζ(2t) k=1

k=1

as n → ∞, we find that lim an

=

lim bn

=

n→∞

n→∞

  1 1 ζ(t) + , 2 ζ(2t) ζ(t)   1 ζ(t) 1 − , 2 ζ(2t) ζ(t)

which implies that  1  2

M (t) =

ζ 2 (t)+ζ(2t) ζ(t) ζ(2t)

ζ 2 (t)−ζ(2t) ζ(t) ζ(2t)

2

2

ζ (t)−ζ(2t) ζ(t) ζ(2t)

ζ (t)+ζ(2t) ζ(t) ζ(2t)

  .

In particular, for t = 2, we see that lim an =

n→∞

21 , 2π 2

lim bn =

n→∞

9 2π 2

which gives  M (2) =

3  2π 2

7

3

3

7

 

again.



Remark. We collect three problems for additional practice.   x 1 1. Let B(x) = . Find a closed form for 1 x Pn =

n Y

B(k).

k=2

2. Problem 11685 (Proposed by D. Knuth, 120(1), 2013). Prove that  ∞  ∞ Y 1 1 X 1 1 + 2k . = + Qk−1 2j 2 2 − 1 j=0 (2 − 1) k=0 k=0 In other words, prove that     1 1 1 (1 + 1) 1 + 1+ 1+ ··· 3 15 255 1 1 1 1 = +1+1+ + + + ··· . 2 3 3 · 15 3 · 15 · 255

114

Infinite Series 3. Problem 11883 (Proposed by H. Ohtsuka, 123(1), 2016). For |q| > 1, prove that ∞ X k=0

(q 20 + q)(q 21

∞ 1 Y 1 1 = . k 1−2 2 q − 1 i=0 q i + 1 + q) · · · (q + q)

3 Integrations

Over the years, we have seen a great many elegant and striking integral problems and clever solutions published in the Monthly. Most of them contain mathematical ingenuities. In this chapter, we select 15 of these irresistible integrals. In presenting the combination of approaches required to evaluate these integrals, we have tried to follow the most interesting route to the results and endeavored to highlight connections to other problems and to more advanced topics. It is interesting to see the range of mathematical approaches this chapter exploits. These evaluations provide nice application of infinite series, gamma and beta functions, parametric differentiation and integration, Laplace transforms and contour integration.

3.1

A Lobachevsky integral

Problem 11423 (Proposed by G. Minton, 116(3), 2009). Show R ∞ that if n and m are positive integers with n ≥ m and n − m even, then 0 x−m sinn x dx is a rational multiple of π. Discussion. Since there are integer parameters m and n appearing in the problem, we are going to proceed by induction. To build “P (k + 1)” from “P (k),” integration by parts suggests we induct on m. Solution. We use induction on m. Let Z I(n, m) = 0

∞

sinn x dx. xm

First, for any positive odd number n = 2k + 1, we recall that   k 1 X 2k + 1 2k+1 k+j sin x = 2k (−1) sin(2k − 2j + 1)x, j 2 j=0

and

Z 0

∞

π sin(ax) dx = x 2

for a > 0. 115

116

Integrations

Hence I(2k + 1, 1) =

1 22k+1

k X

(−1)

k+j



j=0

2k + 1 j

 π

is a rational multiple of π. For m = 2, note that integration by parts yields Z ∞ sinn−1 x cos x dx. I(n, 2) = n x 0 Using the product to sum formula for sine and cosine, for n = 2k, we can expand sin2k−1 x cos x as 1 22k−1

k−1 X

(−1)k+j−1



j=0

2k − 1 j

 (sin(2k − 2j)x + sin(2k − 2j − 2)x)

so  I(2k, 2) =

k−2 X

k  (−1)k+j−1 22k−1 j=0



2k − 1 j

 +

1 2



2k − 1 k−1



 π

is also a rational multiple of π. For m ≥ 2, integration by parts twice leads to   Z ∞ 1 1 I(n, m + 1) = − sinn x d m 0 xm   Z ∞ n 1 = − sinn−1 x cos x d m(m − 1) 0 xm−1 n(n − 1) n2 I(n, m − 1) + I(n − 2, m − 1). = − m(m − 1) m(m − 1) When n − (m + 1) is even and nonnegative, the right-hand side is a rational multiple of π by the induction hypothesis. Therefore, the left-hand side is also such a multiple, which completes the proof.  Remark. Since     k X 2k k−j 2k + 1 (−1) = , j k j=0 this gives a more compact formula I(2k + 1, 1) =

π 22k+1

  2k . k

Similarly, I(2k, 2) =

π 22k−1



 2k − 2 . k−1

A Lobachevsky integral

117

Note that assumption on n − m even is vital for the conclusion because Z ∞ sin3 x 3 dx = ln 3 2 x 4 0 is not a rational multiple of π. In general, similar to the above arguments, when n ≥ m ≥ 2, we find that Z 0

Z 0

∞

∞

sin2n x dx = x2m−1

sin2n+1 x dx = x2m

n X

 2n (2k)2m−2 (−1) ln k; 22n−1 n − k (2m − 2)! k=1   n 2m−1 1 X m+k 2n + 1 (2k + 1) ln(2k + 1). (−1) 2n 2 (2m − 1)! n−k 1

m+k



k=0

Based on the Monthly Editorial Notes, Lobachevsky studied I(n, m) as early as 1842. Recently, a similar problem appeared as Mathematics Magazine Problem 2020 (Proposed by J. Sorel, 90(2), 2017). Find all natural numbers n such that the integral Z In :=

1

xn arctan x dx

0

is a rational number. By induction it was shown [84] that In is a rational number precisely when n = 4k + 3 with k = 0, 1, 2, . . .. A much more challenging problem appeared in the Monthly in 1967: Advanced Problem 5529 (Proposed by D. S. Mitrinovic, 74(8), 1967). Evaluate Z ∞ Y n sin kj (x − aj ) In := dx x − aj −∞ j=1 with kj , aj , j = 1, 2, . . . , n real numbers. The next year a solution was published in the form   n Y sin k (a − a ) j 1 j π In =  a − a 1 j j=2

(3.1)

under the assumption that k1 ≥ k2 ≥ . . . kn ≥ 0. But this solution, as Klamkin (1970) pointed out, can not be true since it is not symmetric in the parameters while In is. Djokovi´c and Glasser showed that Formula (3.1) holds true under the additional restrictions: k1 ≥ k2 + k3 + . . . + kn and all of the kj are positive. It appeared no simple general fix for (3.1). As a unsolved question, this problem is listed in Monthly several times and has disappeared in the later 1980’s. Recently, Borwein et. al. in a remarkable paper [18] completely set the case n = 3 and exhibit some more general structure of In .

118

Integrations

For the special case   Z ∞ x x Jn := sincx · sinc · · · sinc dx 3 2n + 1 −∞ where sincx := sin x/x. Using Mathematica, we obtain J0 = J1 = · · · = J6 = π, but

Z

∞

sincx · sinc

J7 = −∞

x 3

· · · sinc

x dx 15

467807924713440738696537864469 π < π. 467807924720320453655260875000 Based on these curious results, [18] gives the following generalization: Pn−1 Pn First Bite Theorem. If k=1 ak ≤ a0 < k=1 ak , then =

Z

∞

n Y

sinc(ak x) dx =

−∞ k=0

1 a0

  (a1 + a2 + · · · + an − a0 )n 1− π. 2n−1 n!a1 a2 · · · an

Finally, we end this section with six related problems for additional practice. 1. Let n and m be positive integers with 0 ≤ m < n. Show that Z ∞ sinn x cos(mx) dx xn 0 is a rational multiple of π. 2. Let n and m be positive integers such that n ≥ m > p, where p = (n − m)(mod 2). Let bxc be the floor function. Show that Z ∞ (−1)b(n−m)/2c π 1−p sinn x dx = m x 2n−p (m − 1)! 0   bn/2c−p X n (−1)k (n − 2k)m−1 lnp (n − 2k). k k=0

3. Let Sn be an series analog of Jn and define by Sn :=

∞ X

sincn (k)

for all n ≥ 1.

k=1

(a)Show that 1 Sn = − + rn π, 2 where rn is a rational number, 1 ≤ n ≤ 6.

Two log gamma integrals

119

(b)Show that the pattern of answers in (a) breaks at n = 7. With this insight, can you find a general formula of Sn which is analog to the First Bite Theorem? 4. Let

∞

Z Cn = 0

n x sin(5x) Y dx. cos x k k=1

Show that Cn = π/2 for all 1 ≤ n ≤ 82, but C83 < π/2. Can you reveal the principle behind these results? Similar to the First Bite Theorem above, can you find a formula for the following more general integral: Z ∞ n sin(mx) Y cos(ak x) dx? x 0 k=1

5. For positive integer m, evaluate Z ∞ sin(mx) m J0 (x) dx, x 0 where J0 (x) is the Bessel function of the first kind of order zero. 6. Calculate Z

∞

π1 := 0

Z

∞ Y

cos

x n

n=1 ∞

π2 :=

cos(2x) 0

∞ Y n=1

dx

cos

and

x n

dx.

It is known that π 4

π . 8 Although π2 is within 10−42 of π/8, a remarkable approximation, we do not have concise closed form expressions for π1 and π2 yet. As [10] pointed out, even asking for the numerical Q∞ value is
rather challenging because the oscillatory behavior of n=1 cos nx . π1
ln 2, Z ∞ γ γ e−αt ln Γ(t) dt, M (α) = + L(α) = − γ − ln α + α α α 0 where γ is Euler’s constant. Their derivations offer a remarkable applications of the gamma function, digamma function and complex integrals. The graph of M (α) generated by Mathematica is shown in Figure 3.1. It has a cusp at α = ln 2. The proposers of this problem investigated another branch of M (α) where 0 < α < ln 2 in [7] and found Z 1 α + ln(1 − e−α ) − γ − ln α α γ M (α) = + + e−αt ln Γ(t) dt. α 1 − e−α 1 − e−α 0

122

Integrations

FIGURE 3.1 The graph of M (α) This proposed problem indeed is Lemma 2.4 in the paper, which is used to confirm the continuity of M (α) and the jump of 4 for M 0 (α) at α = ln 2. Now, we end this section with five problems for additional practice. The last one gives an explicit expression of L(α) (Due to Dixit [35]). R1 1. Show that 0 t ln Γ(t) dx = 41 ln(2π)−ln A, where A is the GlaisherKinkelin constant. 2. Show that Z 1 Z 2 ln Γ(t) dt + 0

1

ln Γ(t) ln Γ(1 − t) dt =

0

3. Show that Z

∞

0



1 2 1 2 ln (2π) + π . 2 24

 1 1 − ln t dt = ln 2π. ψ(t + 1) − 2(t + 1) 2

4. Evaluate Z 0

1

e−αt ln Γ(t) dt

Z and

1

te−αt ln Γ(t) dt.

0

5. Let α > 0. Show that   ∞ X 1 1 2π ln n L(α) = − + 1 ln + 2α α 2 e −1 α α α + 4n2 π 2 n=1      iα iα γ + ln α 1 ψ +ψ − − . + 4 2π 2π α

Short gamma products with simple values

3.3

123

Short gamma products with simple values

Problem 11426 (Proposed by M. L. Glasser, 116(4), 2009). Find Γ(1/14) Γ(9/14) Γ(11/14) , Γ(3/14) Γ(5/14) Γ(13/14) where Γ denotes the usual gamma function, given by Γ(z) =

R∞ 0

tz−1 e−t dt.

Discussion. Recall the well-known Gauss multiplication formula, a long gamma product but with simple values: n−1 Y  k  (2π)(n−1)/2 √ . Γ = n n k=1

There is another less-known formula due to S´andor and T´oth [81]: for n ≥ 2, n 6= pr , a prime power, then   Y k Γ = (2π)φ(n)/2 , n gcd(k,n)=1

where φ(n) is the Euler totient function, the number of integers between 1 and n that are relatively prime to n. This singles out some factors from the Gauss product but remains simple values. For example, for n = 14, a product of six factors gives Γ(1/14) Γ(3/14) Γ(5/14)Γ(9/14) Γ(11/14) Γ(13/14) = (2π)3 . This problem suggests an even shorter gamma product still with simple values: Γ(1/14) Γ(9/14) Γ(11/14) = 4π 3/2 , which is not a consequence of either of the product formulas above. We will proceed with this problem by using the Legendre duplication formula, and will see that a large collection of short products exist which exhibit simple values for n = 2m with m odd. Solution. We show the desired ratio is 2. Indeed, rewrite the Legendre duplication formula as Γ(2z) Γ(1/2) = 22z−1 Γ(z) Γ(z + 1/2). Setting z = 1/14, 9/14 and 11/14 respectively, we find that Γ(1/7) Γ(1/2) = 2−6/7 Γ(1/14) Γ(4/7); Γ(9/7) Γ(1/2) = 22/7 Γ(9/14) Γ(8/7); Γ(11/7) Γ(1/2) = 24/7 Γ(11/14) Γ(9/7).

124

Integrations

Applying Γ(1 + z) = zΓ(z) yields Γ(8/7) = Γ(1/7)/7 and Γ(11/7) = 4Γ(4/7)/7. Therefore the product of the three equations above leads to Γ(1/14)Γ(9/14)Γ(11/14) = 4 (Γ(1/2))3 . Similarly, we have Γ(3/14)Γ(5/14)Γ(13/14) = 2 (Γ(1/2))3 . The ratio of 2 now follows from combining these two products.    Remark. Recall the definition of the Legendre symbol ap : Let p be an odd prime. Then     1 if a is a quadratic residue of p and a 6≡ 0 (mod p), a −1 if a is a quadratic nonresidue of p, =  p 0 a ≡ 0 (mod p). Using this notation, we offer the following generalization: Let p be a prime with p = 7 (mod 8). Then Y a∈O(2p)

  ( ap ) P(p−1)/2 k a Γ = 2 k=1 ( p ) , 2p

(3.4)

where O(n) is the set of odd positive integers less than n. For example, let p = 7, since               1 9 11 7 3 5 13 = = = 1, = 0, = = = −1, 7 7 7 7 7 7 7 (3.4) immediately deduces that 2 3 1 Γ(1/14) Γ(9/14) Γ(11/14) = 2( 7 )+( 7 )+( 7 ) = 2. Γ(3/14) Γ(5/14) Γ(13/14)

This extension (3.4) is also obtained by Tauraso (http://www.mat.uniroma2. it/~tauraso/AMM/AMM11426.pdf). We offer a composite solution as follows: Using p = 7 (mod 8), we have             2p − a −a −1 a a a (p−1)/2 = = · = (−1) =− . p p p p p p Let the left-hand side of (3.4) be P (p). Then P (p) =

Y a∈O(p)



( ap ) Γ(a/2p) . Γ((2p − a)/2p)

Short gamma products with simple values

125

Applying the Euler reflection formula and the Legendre duplication formula of the gamma function, if x 6= 1, 1/2, we have Γ(x) Γ(1 − x)

sin(πx) π Γ2 (2x) sin(π(1 − 2x)) = 21−4x 2 Γ (x + 1/2) sin(π(1 − 2x)/2) Γ2 (2x) Γ(x + 1/2)Γ(1/2 − x) = 21−4x 2 Γ (x + 1/2) Γ(2x)Γ(1 − 2x) Γ(2x)Γ(1/2 − x) = 21−4x . Γ(x + 1/2)Γ(1 − 2x) Γ2 (x)

=

Substituting x = a/2p yields Y

P (p) =

2

a (1− 2a p )( p )



a∈O(p)

( ap ) Γ(a/p)Γ((p − a)/2p) . Γ((p − a)/p)Γ((p + a)/2p)

Next, by the assumption on p, we have    −1       (p ± a)/2 2 p±a ±a a −(p2 −1)/8 = = (−1) =± . p p p p p Therefore, P (p)

Y

=

p−a a Γ( p ) (a/p)Γ( p ) ((p − a)/p) (p+a)/2 (p−a)/2 Γ( p ) ((p + a)/2p)Γ( p ) ((p − a)/2p) ! Qp−1 ( k ) p (k/p) k=1 Γ Qp−1 ( k ) Γ p (k/p)

a (1− 2a p )( p )

2

a∈O(p)

Y

= =

2(1− p )( p ) 2a

a∈O(p) P

2

a

k=1

2a a a∈O(p) (1− p )( p )

.

Note that (p−1)/2   X  a  (p−1)/2 X  p − 2k  X k = =− p p p k=1

a∈O(p)

k=1

and X a∈O(p)

a p

  a p

=

  (p−1)/2 p−1 X X 2k  2k  k k − p p p p

k=1

=

p−1 X k=(p+1)/2

k=1

  X   p−1 k k 2k 2k − p p p p k=1

!

126

Integrations (p−1)/2

=

X k=1

p−k p

(p−1)/2

= −

X k=1



p−k p

(p−1)/2

 −

X k=1

k p

  k p

  k . p

We finally obtain   (p−1)/2   (p−1)/2   (p−1)/2   X X X X  a k k k 2a =− +2 = . 1− p p p p p k=1

a∈O(p)

k=1

k=1

This proves (3.4) as claimed. The gamma function is probably one of the most ubiquitous functions in analysis. Recently Borwein and Corless [22] made an informative survey of papers on the gamma function published in the Monthly. Through those articles over one century, they provided a amazed cross-section of mathematics: Analysis, geometry, statistics, combinatorics, logic and number theory. We now end this section to compile two problems for additional practice: 1. (Due to Zucker, personal communication) Let n = 2m − 1 with m > 1. Prove that   m−1 Y  2k + n  1 Γ = 2m−1 π m/2 . Γ 2n 2n k=1

2. (Due to Nijenhuis [72]) Let n be an odd integer, and let A be the cyclic subgroup of Φ(2n) generated by n + 2, or any one of its cosets. Let v(n) denote the cardinality of A, and b(a) the number of elements of A that are larger than n. Prove that Y x = 2b(a) π v(n)/2 . Γ 2n x∈A

Comment. Notice that the number of sets A may not be unique. For example, when n = 31, we have six sets of A as follows: {1, 33, 35, 39, 47}, {3, 17, 37, 43, 55}, {5, 9, 41, 49, 51}, {7, 19, 25, 45, 59}, {11, 13, 21, 53, 57}, {15, 23, 27, 29, 61}. Among of them, each set, multiplying the member by 33 (mod 62), gives the next set, in circle order. For example, applying the formula to the first set with v = 5 and b = 4 yields           1 33 35 39 47 Γ Γ Γ Γ Γ = 24 π 5/2 . 62 62 62 62 62

Evaluate an integral by Feynman’s way

3.4

127

Evaluate an integral by Feynman’s way

Problem 11966 (Proposed by C. I. V˜alean, March, 2017). Prove that Z 1 π2 x ln(1 + x) (ln 2)2 dx = + . 1 + x2 96 8 0 Discussion. Let x = tan θ. The proposed integral becomes Z π/4 tan θ ln(1 + tan θ) dθ, 0

which does not make things easy. Integration by parts also seems hard to proceed. So we try to apply Richard Feynman’s “a different box of tools” [40] — that evaluate integrals by differentiation and integration with respect to a parameter. In this way, we are able to transform the transcendental integrands into rational functions. Solution I — By parametric differentiation. Let Z 1 x ln(1 + px) dx. I(p) = 1 + x2 0 The Leibniz rule yields Z 1 x2 I 0 (p) = dx. 2 0 (1 + x )(1 + px) In view of the partial fraction decomposition   1 −1 + px x2 1 = + , (1 + x2 )(1 + px) 1 + p2 1 + px 1 + x2 integration with respect to x gives I 0 (p) =

π 1 ln 2 p ln(1 + p) − + . p(1 + p2 ) 4 1 + p2 2 1 + p2

Since I(0) = 0, we find the desired integral Z 1 I(1) = I(0) + I 0 (p) dp 0  Z 1 ln(1 + p) π 1 ln 2 p = − + dp p(1 + p2 ) 4 1 + p2 2 1 + p2 0 Z 1 ln(1 + p) π2 (ln 2)2 = dp − + 2 16 4 0 p(1 + p )

128

Integrations 1

Z = 0

1

Z = 0

(1 + p2 − p2 ) ln(1 + p) π2 (ln 2)2 dp − + 2 p(1 + p ) 16 4 Z 1 ln(1 + p) p ln(1 + p) π2 (ln 2)2 dp − dp − + 2 p 1+p 16 4 0

1

ln(1 + p) π2 (ln 2)2 dp − I(1) − + . p 16 4 0 P∞ Recall that ln(1 + x) = n=1 (−1)n+1 xn /n. Then Z

=

Z

1

0

∞ Z 1 ∞ X X ln(1 + p) π2 (−1)n+1 pn−1 (−1)n+1 = dp = = . p n n2 12 n=1 0 n=1

This yields 1 I(1) = 2



π2 (ln 2)2 + 48 4

 =

(ln 2)2 π2 + 96 8

as claimed.



Solution II — By parametric integration. Let the proposed integral be I. Notice that Z 1 xdt ln(1 + x) = . 1 + xt 0 We have Z I

1

Z

1

Z

1

= 0

= = = = =

1

x2 dxdt 2 0 0 (1 + x )(1 + xt)  Z 1 Z 1 tx − 1 1 1 + dxdt 2 1 + x2 1 + tx 0 0 1+t   Z 1 1 t ln 2 π ln(1 + t) − + dt 2 2 4 t 0 1+t Z Z Z 1 ln 2 1 tdt π 1 dt ln(1 + t) − + dt 2 2 0 1 + t2 4 0 1 + t2 0 t(1 + t ) Z Z 1 1 (ln 2)2 π2 ln(1 + t) t ln(1 + t) − + dt − dt 4 16 t 1 + t2 0 0 Z 1 π2 ln(1 + t) (ln 2)2 − + dt − I. 4 16 t 0 Z

=

0

x2 dtdx (1 + x2 )(1 + xt)

Solving this for I we find that Z (ln 2)2 π2 1 1 ln(1 + t) (ln 2)2 π2 1 π2 π2 (ln 2)2 I= − + dt = − + = + . 8 32 2 0 t 8 32 2 12 96 8

Evaluate an integral by Feynman’s way

129 

Remark. Parametric differentiation and integration often provides a straightforward method to evaluate difficult integrals which conventionally require the more sophisticated method of contour integration. Sometimes it may lead to new proofs of classical results. For more examples, please refer to Chapter 19 in [29]. Here we present possibly one of the best solutions to Euler’s Basel problem [36] via parametric differentiation, which is due to Muzaffar [69]. His solution begins with Z π/2 I(x) = sin−1 (x sin θ) dθ, x ∈ (0, 1). 0

Differentiating under the integral sign gives Z π/2 Z π/2 sin θ sin θ 0 p p dθ. dθ = I (x) = 2 2 (1 − x ) + x2 cos2 θ 0 0 1 − x2 sin θ Substituting u = cos θ yields Z I 0 (x) =

1 du (1 − x2 ) + x2 u2  1 p 1  ln u + (1 − x2 )/x2 + u2 x 0   X ∞ 2n 1 1+x x ln = . 2x 1−x 2n +1 n=1 0

= =

1

p

Hence, Z I(x) = I(0) + 0

x

I 0 (t) dt =

∞ X n=1

x2n+1 1 = (Li2 (x) − Li2 (−x)), 2 (2n + 1) 2

where Li2 (x) is the dilogarithm function. Setting x = 1 we find Z π/2 ∞ 1 3 X 1 (Li2 (1) − Li2 (−1)) = = sin−1 (sin θ) dθ 2 4 n=1 n2 0 Z 1 1 sin−1 (t) 1 π2 √ = dt = (sin−2 t)2 = , 2 8 0 1 − t2 0 which implies that ζ(2) = π 2 /6 immediately. As a byproduct, integrating ∞ sin−1 (t) X (2t)2n−1 √
, = n 2n 1 − t2 n n=1

from 0 to 1/2, we recover another Euler’s beautiful formula ζ(2) = 3

∞ X

1


.

n2 2n n n=1

We now collect six additional problems for your practice.

130

Integrations R π/2 1. Another proof of ζ(2) = π 2 /6. Let I(x) = 0 tan−1 (x tan θ) dθ for x ∈ (0, 1). Show that     1 1+x I(x) = ln x ln + Li2 (x) − Li2 (−x) . 2 1−x 2. A Classical Problem. Show that √ Z ∞ 2 π ln xe−x dx = − (2 ln 2 + γ). 4 0 3. Putnam Problem 2005-A5. Evaluate Z 1 ln(1 + x) dx. 1 + x2 0 4. Problem 10884 (Proposed by Z. Ahmed, 108(6), 2001). Evaluate √ Z 1 arctan 2 + x2 √ dx. 2 2 0 (1 + x ) 2 + x Hint: Use arctan u = u

Z 0

1

dy . 1 + u2 y 2

5. Problem 11101 (Proposed by E. F. Skelton, 111(7), 2004). Show that   Z ∞ p b dx aπ √ a arctan √ = (ln(b + a2 + b2 ) − ln a) 2 2 2 2 2 a +x a +x 0 for a, b > 0. Hint: Use   Z b b a a du √ arctan √ = . 2 2 2 2 2 2 2 a +x a +x 0 a +x +u 6. Problem 11113 (Proposed by I. Sofair, 111(9), 2004). Evaluate Z ∞ Z ∞ −k√x2 +y2 e sin(ax) sin(bx) p Ik (a, b) = dxdy xy x2 + y 2 0 0 Ra in closed form for a, b, k > 0. Hint: Use sin(ax) = x 0 cos(xt) dt.

Three ways to evaluate a log-sine integral

3.5

131

Three ways to evaluate a log-sine integral

Problem 11639 (Proposed by O. Kouba, 119(4), 2012). Evaluate Z

π/2

(ln(2 sin x))2 dx.

0

Discussion. In general, it is difficult to determine whether or not an integral can be evaluated precisely. Thanks to Mathematica, we see this problem has the exact value π 3 /24. Once we “know” the answer analytically, it is a fairly simple matter to “prove” it. Indeed, this problem can be done in at least three different ways. The first method is based on the Fourier series of ln(2 sin(x/2)); the second is a direct calculation which involves the Euler beta function; the final solution relies on a contour integration. It is interesting to see how wide a range of mathematical topics these solutions exploit. Solution I — By Fourier series of ln(2 sin(x/2)). Recall that ∞  x  X cos nx . − ln 2 sin = 2 n n=1 Then

(3.5)

∞  X x cos nx cos mx ln2 2 sin = . 2 nm n,m=1

The orthogonality of cos nx on [0, π] implies that Z 0

π

 x dx ln2 2 sin 2

∞ X

=

= =

1 nm

n,m=1 ∞ X n=m=1 ∞ X

π 2

n=1

π

Z

1 nm

cos nx cos mx dx 0

Z

π

cos nx cos mx dx 0

1 π3 = . n2 12

Let t = x/2. Then Z 0

Hence Z 0

π

Z π/2  x ln2 2 sin dx = 2 ln2 (2 sin t) dt. 2 0

π/2

(ln(2 sin x))2 dx =

1 2

Z 0

π

ln2 (2 sin t) dt =

π3 . 24

132

Integrations 

Solution II — By the Euler beta function. Denote the integral by I. Observe that π/2

Z I

(ln 2 + ln sin x)2 dx

= 0

ln2 2 π + 2 ln 2 2

=

Z

π/2

Z

π/2

ln(sin x) dx + 0

ln2 (sin x) dx.

0

It is well-known that Z

π/2

ln(sin x) dx = − 0

Hence I=−

ln2 2 π+ 2

ln 2 π. 2

π/2

Z

ln2 (sin x) dx.

0

Now it suffices to show that Z π/2 1 3 ln2 2 π+ π . ln2 (sin x) dx = 2 24 0 Let t = sin2 x. Then Z π/2 ln2 (sin x) dx

=

0

=

1 2

Z

1 8

Z

1

0

√ ln2 ( t) √√ dt t 1−t

1

t1/2−1 (1 − t)1/2−1 ln2 t dt

0

On the other hand, recall Z

1

B(x, y) =

tx−1 (1 − t)y−1 dt =

0

Γ(x)Γ(y) . Γ(x + y)

Applying the Leibniz rule yields Z

π/2

ln2 (sin x) dx =

0

From Γ(1/2) =

√

1 d2 B(x + 1/2, 1/2) x=0 . 2 8 dx

π, the Legendre duplication formula gives

B(x + 1/2, 1/2) =

Γ(x + 1/2)Γ(1/2) (2x)Γ(2x)π = x . Γ(x + 1) 4 (xΓ(x))2

Three ways to evaluate a log-sine integral

133

Since xΓ(x)

= e−γx

∞  Y n=1

 =

1+

x −1 x/n e n

1 1 − γx + (γx)2 + o(x2 ) 2

Y ∞  n=1

1+

 x2 2 + o(x ) 2n2

! ∞ 1 2 1X 1 = 1 − γx + x2 + o(x2 ) γ + 2 2 n=1 n2   1 2 π2 = 1 − γx + γ + x2 + o(x2 ), 2 12 it follows   π3 2 B(x + 1/2, 1/2) = π − (2π ln 2)x + 2π ln 2 + x2 + o(x2 ). 6 Therefore, d2 1 B(x + 1/2, 1/2) = 4π ln2 2 + π 3 , x=0 dx2 3 and so Z

π/2

ln2 (sin x) dx =

0

ln2 2 1 3 π+ π . 2 24 

Solution III — By contour integration. Let z = reiθ . Then, on the unit circle 1 ln(1 − z) = ln(2 sin(θ/2)) + i(θ − π). 2

(3.6)

Integrating ln2 (1 − z)/iz along the contour shown in Figure 3.2, since there is no singularity on the contour, the residue theorem yields I ln2 (1 − z) dz = 0. (3.7) iz On the small arc, as  → 0, we have I I ln2 (1 − z) (1 − z) ln2 (1 − z) dz = dz iz iz(1 − z) ≤ max |(1 − z) ln2 (1 − z)/z| |1−z|=

2π → 0. 

Thus, by (3.6) and dz = ieiθ dθ = izdθ, (3.7) becomes Z 2π 1 (ln(2 sin(θ/2)) + i(θ − π))2 dθ = 0. 2 0

134

Integrations

FIGURE 3.2 The contour of integration The real part gives Z 2π Z 1 2π π3 . ln2 (2 sin(θ/2)) dθ = (θ − π)2 dθ = 4 0 6 0 The substitution θ = 2π − t yields Z 2π Z ln2 (2 sin(θ/2)) dθ = π

and so

π

ln2 (2 sin(t/2)) dt,

0 π

Z

ln2 (2 sin(θ/2)) dθ =

0

π3 . 12

Replacing θ/2 by x yields Z

π/2

(ln(2 sin x))2 dx =

0

π3 24

as claimed.



Remark. The Fourier series (3.5) was due to Euler. Here is a short elementary proof: ∞ X cos(nx) n n=1

=

∞ X enxi + e−nxi 2n n=1

1 2 1 − 2 1 − 2

= −

ln(1 − exi ) + ln(1 − e−xi )

ln(2 − 2 cos x)  x = ln 4 sin2 2  x = − ln 2 sin . 2

=




Three ways to evaluate a log-sine integral

135

In Solution I, as Jean-Pierre Grivaux did in the published solution, another path to evaluate this integral is via Parseval’s identity: Z 2π ∞  x 1 X 1 π2 1 ln2 2 sin dx = = . 2π 0 2 2 n=1 n2 12 Solution II may seem unnecessarily lengthy, however, it enables us to investigate the more general family of integrals Z π/2 I(p, q) = lnp sin x lnq cos x dx, (p, q ≥ 0). 0

Indeed, along the same lines as in Solution II, we have   1 ∂ p+q Γ(α + 1/2)Γ(β + 1/2) I(p, q) = 2p+q+1 ∂αp ∂β q Γ(α + β + 1) α=β=0   p+q −2(α+β) ∂ 2 Γ(2α + 1)Γ(2β + 1) π = . 2p+q+1 ∂αp ∂β q Γ(α + β + 1)Γ(α + 1)Γ(β + 1) α=β=0 In particular, we have Z π/2 ln sin x ln cos x dx = 0 π/2

Z

ln2 sin x ln cos x dx = −

0

and Z

ln2 2 1 π − π3 , 2 48

π/2

ln2 sin x ln2 cos x dx =

0

π 2



ln3 2 1 π + πζ(3), 2 8

 1 4 π + ln4 2 − ζ(3) ln 2 . 160

For q = 0, we can also find I(p, 0) as we did in Solution III. Integrating lnp (1 − z) on the contour in Figure 3.2 and taking the real part yields [p/2]

X k=0

(−1)k



p 2k

π/2

Z

x2k lnp−2k (2 sin x) dx = 0.

0

This implies the special values Z π/2 3 ln 2 3 ln3 2 π − π − πζ(3); ln3 sin x dx = − 8 2 4 0 Z π/2 2 19 5 ln 2 3 ln4 2 ln4 sin x dx = π + π + π + 3 ln 2πζ(3). 480 4 2 0 Recently, Borwein and Straub [23] studied another generalized log-sine integral in the form Z θ

xk lnn−1−k (2 sin(x/2)) dx,

Ls(k) n (θ) := −

0

136

Integrations

which appeared in the -expansion of various Feynman diagrams in quantum mechanics. Explicit evaluation at π and multiples are obtained. For example, π/3

Z

ln2 (2 sin(x/2)) dx =

0

Z

7 3 π ; 108

π/3

x ln2 (2 sin(x/2)) dx =

0

17 4 π . 6480

Finally, we end this section by compiling some additional problems for practice. Rπ 1. Show that 0 ln2 (tan(x/4)) dx = π 3 /4. 2. Let α, β > −1 and integers p, q ≥ 0. Prove that π/2

Z 0

lnp sin x lnq cos x(sin x)2α+1 (cos x)2β+1 dx   ∂ p+q Γ(α + 1)Γ(β + 1) 1 . = p+q+1 2 ∂αp ∂β q Γ(α + β + 2)

3. Let α, β > −1. Prove that π/2

Z

ln sin x(sin x)2α+1 (cos x)2β+1 dx

0

=

Γ(α + 1)Γ(β + 1) (ψ(α + 1) − ψ(α + β + 2)), 4Γ(α + β + 2)

where ψ is the digamma function. 4. For a, b, p, q > 0, show that Z

π/2 2

2

2



2

ln(a cos x + b sin x) dx = π ln 0

Z 0

π

ln(a2 cos2 x + b2 sin2 x) 2π dx = ln 2 2 2 2 pq p cos x + q sin x

5. Let F (x) := 2−2x



a+b 2



aq + bp p+q

;  .

(2x)Γ(2x) . xΓ(x)

Show that, for |x| < 1/2, ln F (x) = −(γ + 2 ln 2)x +

∞ X k=2

(−1)k ζ(k)

2k − 1 k x , k

where ζ is the Riemann zeta function. Rπ 6. Let an = 0 lnn (sin x) dx. Show that (Due to Beumer [13]) Pn−1 (a) (n − 1)an = ln 2 an−1 + k=1 (1 − 2−k )ζ(k + 1)an−k−1 .

A log-product integral √ P∞ π Γ((x+1)/2) xn a (b) = n n=0 n! 2 Γ(x/2+1) . Rπ n 7. Let an = 0 ln (sin x) dx.

137

(a) Define D(s) =

∞ X (2k − 1)!! 1 . · (2k)!! (2k + 1)s

k=0

Show that an = (−1)n n!D(n + 1). (b) Let σk =

1 1 1 1 − k + k − k + · · · = (1 − 21−k )ζ(k). 1k 2 3 4

Show that π n an = (−1) 2

σ1 σ2 σ3 ··· σn

−1 σ1 σ2 ··· σn−1

0 −2 σ1 ··· σn−2

0 0 −3 ··· σn−3

··· ··· ··· ··· ···

0 0 0 ··· σ1



8. Problem 11418 (Proposed by G. Lamb, 116(3), 2009). Find Z ∞ 2 t sech2 t dt −∞ a − tanh t for complex a with |a| > 1. 9. Problem 11629 (Proposed by O. Oloa, 119(3), 2012). Let  2 Z 1 1 1 f (σ) = xσ + dx. ln x 1 − x 0 (a)Show that f (0) = ln(2π) − 3/2. (b)Find a closed form expression for f (σ) for σ > 0. 10. Problem 12051 (Proposed by P. Ribeiro, 125(6), 2018). Prove  ∞  X π3 π 2n 1 = + ln2 2. n 3 4 (2n + 1) 48 4 n n=0

3.6

A log-product integral

Problem 11993 (Proposed by C. I. V˜alean, 124(7), 2017). Prove that Z 1 log(1 − x)(log(1 + x))2 π4 dx = − . x 240 0

138

Integrations

Discussion. Since we are able to evaluate a number of integrals involving a single logarithmic function, we now aim to reduce the number of logarithms with different arguments in the integrand. To this end, invoking the logarithmic properties, we call the algebraic identity (a + b)3 + (a − b)3 = 2a3 + 6ab2 . Solution. Let the proposed integral be I. Appealing to the algebraic identity 6ab2 = (a + b)3 + (a − b)3 − 2a3 , with a = log(1 − x) and b = log(1 + x), we rewrite the proposed integral as I

1 6

=

1

Z 0

    1−x dx log3 (1 − x2 ) + log3 . (3.8) − 2 log3 (1 − x) 1+x x

Using the substitution x2 = t, we have Z Z 1 1 1 dt 3 2 dx = log3 (1 − t) log (1 − x ) x 2 t 0 0 Z 1 3 1 log u = du (let u = 1 − t) 2 0 1−u ∞ Z 1 X 1 n = u log3 u du 2 n=0 0 = −3

∞ X n=0

1 = −3ζ(4), (n + 1)4

where ζ is the Riemann zeta function. As a byproduct, we also get Z 1 dt = −6ζ(4). log3 (1 − t) t 0 Similarly, using the substitution x = Z

1

log 0

3



1−x 1+x



dx x

1−t 1+t ,

log3 t dt 2 0 1−t ∞ Z 1 X 2 u2n log3 u du Z

= =

we have

1

2

n=0

= −12

0

∞ X n=0

45 1 = − ζ(4), (2n + 1)4 4

A Putnam problem beyond

139

where we have used the fact that ∞ ∞ X X 15 1 1 −4 = (1 − 2 ) = ζ(4). 4 4 (2n + 1) n 16 n=0 n=1 Plugging these results into (3.8), since ζ(4) = π 4 /90, we find that   45 3 π4 1 −3ζ(4) − ζ(4) + 12ζ(4) = − ζ(4) = − I= 6 4 8 240 as desired.



Remark. The solution clearly presents a useful approach to evaluate integrals involving the product of logarithms. We collect a few more problems for your practice. 1. Show that Z

1

0

log2 x log2 (1 + x) 29 dx = 2ζ(2)ζ(3) − ζ(5). x 8

2. Show that Z 0

1

logn x log2 (1 + x) dx = 2(−1)n n! x ! ∞ X k Hk −n−2 (−1) n+2 + (1 − 2 )ζ(n + 3) . k

k=1

Can you find a closed form for Z 1 logn x logm (1 + x) dx x 0 for m, n ∈ N? 3. Prove that Z 1 log x log(1 − x) log2 (1 + x) 7 25 dx = ζ(2)ζ(3) − ζ(5). x 8 16 0

3.7

A Putnam problem beyond

Problem 11234 (Proposed by J. Brennan and R. Ehrenborg, 113(6), 2006). Let a1 , . . . , an and b1 , . . . , bn−1 be real numbers, with a1 < b1 < a2 . . . an−1 < bn−1 < an . Let h be an integrable function from R to R. Show that  Z ∞  Z ∞ (x − a1 ) · · · (x − an ) dx = h(x) dx. h (x − b1 ) · · · (x − bn−1 ) −∞ −∞

140

Integrations

Discussion Let a1 = −1, a2 = 1 and b1 = 0. The proposed problem becomes Putnam Problem 1968-B4 [4]: Prove that  Z ∞  Z ∞ 1 h x− dx = h(x) dx. x −∞ −∞ This Rresult goes back as far as Cauchy and can be proved as follows. Let ∞ I = −∞ f (x − 1/x) dx. Applying the substitution u = x − 1/x yields two branches:  p 1 u ± u2 + 4 , u ∈ R. x± = 2 Since   dx− u dx+ 1 u +1+ √ + = 1− √ = 1, du du 2 u2 + 4 u2 + 4 it follows that Z I

0

= −∞ ∞

Z =

−∞ ∞

Z

∞

f (u) dx− + f (u) dx+ 0   dx− dx+ + f (u) du du du

Z =

f (u) du. −∞

We demonstrate the above argument goes through for the general case. It is remarkable that Z ∞ Z ∞ f (u) dx = f (x) dx −∞

remains true when u=x+C −

−∞ n−1 X k=1

αk x − bk

(3.9)

with constant C and all the αk > 0 for 1 ≤ k ≤ n − 1. Solution. We will instead prove the above generalization and then verify that n−1

X αk (x − a1 ) · · · (x − an ) =x+C − (x − b1 ) · · · (x − bn−1 ) x − bk

(3.10)

k=1

satisfies αk > 0 for 1 ≤ k ≤ n − 1. To this end, without loss of generality, we assume that b1 < b2 < · · · < bn−1 . On each of the following intervals (−∞, b1 ), (b1 , b2 ), · · · , (bn−1 , ∞)

A Putnam problem beyond

141

the function u given by (3.9) is continuous and has limits −∞ and ∞ at the left and right endpoints, respectively. Thus the range of u on each of these intervals is R. Furthermore, from (3.9) we have ! n−1 X n x − u−C + bk xn−1 + · · · = 0. k=1

This implies that its solution has n branches satisfying x1 + x2 + · · · + xn = u − C +

n−1 X

bk .

k=1

Hence

d (x1 + x2 + · · · + xn ) = 1. du Thus the proposed integral Z b1 Z b2 Z I: = f (u) dx1 + f (u) dx2 + · · · −∞ Z ∞

=

b1

f (u) −∞ ∞

∞

f (u) dxn

bn−1

d (x1 + x2 + · · · + xn ) du du

Z =

f (u) du. −∞

Now we verify that αk > 0 for 1 ≤ k ≤ n − 1. Multiplying (3.10) by x − bk on both sides, then letting x = bk yields −αk =

(bk − a1 ) · · · (bk − ak )(bk − ak+1 ) · · · (bk − an ) . (bk − b1 ) · · · (bk − bk−1 )(bk − bk+1 ) · · · (bk − bn−1 )

In the above ratio, by assumption, there are n − (k + 1) + 1 = n − k negative factors in the numerator and (n − 1) − (k + 1) + 1 = n − k − 1 negative factors in the denominator. It follows that (bk − a1 ) · · · (bk − ak )(ak+1 − bk ) · · · (an − bk ) > 0, (bk − b1 ) · · · (bk − bk−1 )(bk+1 − bk ) · · · (bn−1 − bk ) for 1 ≤ k ≤ n − 1. αk =

 Remark. This problem displays a principle of invariant integral value. For example, for b > 0, Z ∞ Z ∞ 1 2 2 2 exp(−(x − b/x)2 ) dx exp(−x − b /x ) dx = 2b 2e 0 −∞ √ Z ∞ π 1 2 exp(−x ) dx = 2b . = 2b 2e 2e −∞

142

Integrations

Moreover, let f (x) = 1/(x2 + c2 )α , α > 1/2. Then α Z ∞ Z ∞ 2 1 du f (x) dx = 2α−1 c 1 + u2 −∞ 0 Z ∞ −1/2 1 t 1 = 2α−1 dt = 2α−1 B(α − 1/2, 1/2), α c (1 + t) c 0 where B(x, y) is the Euler beta function. Applying this invariant principle also greatly simplifies the original proof of the following formula in [14]: α α Z ∞  Z ∞  x2 1 dx = dx x4 + 2ax2 + 1 (x − 1/x)2 + 2(a + 1) −∞ −∞ Z ∞ p = f (x) dx (with c = 2(a + 1)) −∞

=

1 B(α − 1/2, 1/2). (2(a + 1))α−1/2

This master formula, together with α α 2 Z ∞ Z ∞ x + 1 dx x2 x2 dx = 4 2 4 2 x + 2ax + 1 x + 2ax + 1 xb + 1 x2 0 0 provides a method for unifying a large class of integrals. See [14] for details. Notice that the positivity of αk in (3.10) is necessary. Here is a counterexample: Let u = x + 1/x, then Z ∞ Z ∞ √ √ exp[−(x + 1/x)2 ] dx = πe−4 6= exp(−x2 ) dx = π. −∞

−∞

The following Liouville formula has been misquoted in several tables of integrals (For example, see [49], 3.257, p. 346): √ Z ∞  −p−1 πΓ(p + 1/2) 2 (ax + b/x) + c dx = . p+1/2 2ac Γ(p + 1) 0 This expression is true only for b < 0. Indeed, for b > 0, 4ab + c > 0, we have  p+1 Z ∞ Z  −p−1 x2 1 ∞ 2 (ax + b/x) + c dx = dx 2 −∞ a2 x4 + (2ab + c)x2 + b2 0 B(p + 1/2, 1/2) = . 2a(4ab + c)p+1/2 It is interesting to ask for what other function u(x) will the integral of f (u) be the same as the integral of f (x). Glasser suggested that the invariant principle is valid if (3.10) is replaced by u=x−

∞ X k=1

ak cot[(x − bk )−1 ].

A surface integral with many faces

143

We leave it to the reader to work out the details. Now, we end this section with four problems for additional practice. 1. For all a, b ∈ R, show that Z ∞ (2x + a + b)2 dx = 2π. 2 2 2 −∞ (x + a) (x + b) + (2x + a + b) 2. Let u = αx − β/x with α, β > 0. For any continuous function F , show that Z ∞ Z ∞ F (u) F (x) dx = (α + β) dx. 2 2 1+x x + (α + β)2 0 0 Use this result to evaluate Z ∞ 0

J0 (|αx − β/x|) dx, 1 + x2

where J0 (x) is the Bessel function of the first kind of zero order. Pn 3. Let u = k=1 ak /(x + bk ). Show that ∞

Z

f (u) dx = −∞

n X

Z

∞

ak

f (1/x) dx. −∞

k=1

4. Let u = x/(ex − 1) or x/(1 − e−x ). For α > 0, show that Z ∞ Z ∞
α
α ue−u dx = ue−u du. 0

3.8

0

A surface integral with many faces

Problem 11277 (Proposed by P. De, 114(3), 2007). Find Z

π/2 Z π/2

φ=0

θ=0

log(2 − sin θ cos φ) sin θ dθdφ. 2 − 2 sin θ cos φ + sin2 θ cos2 φ

Discussion. You may recognize that the proposed double integral is a surface integral over the first octant of the unit sphere in spherical coordinates. Thus, in rectangular coordinates, this integral becomes Z

1

Z

I= 0

0

√ 1−x2

ln(2 − x) 1 p dydx. 2 1 + (1 − x) 1 − x2 − y 2

144

Integrations

On the other hand, we can calculate this integral based on a lemma stated below. This lemma enables us to convert a class of double integrals into single integrals, which leads to a straightforward evaluation of I. Here we give two solutions based on the above discussions. Surprisingly, both approaches end with the calculation of the same integral Z 1 ln(1 + x) dx. 1 + x2 0 Solution I. Since I is a surface integral over the first octant of the unit sphere in spherical coordinates, converting it to rectangular coordinates yields Z

1

√

Z

1−x2

I= 0

0

1 ln(2 − x) p dydx. 1 + (1 − x)2 1 − x2 − y 2

For the inner integral, we have √

Z 0

1−x2

1 p

1 − x2 − y 2

−1



dy = sin

y √ 1 − x2

 √1−x2 π = . 2 0

Hence, π I= 2

Z 0

1

ln(2 − x) π dx = 1 + (1 − x)2 2

Z 0

1

ln(1 + x) dx. 1 + x2

There are a number of ingenious ways to evaluate this integral. One way is to take advantage of “symmetry.” Let x = tan α. We obtain ! √ Z 1 Z π/4 Z π/4 ln(1+x) 2 cos(π/4 − α) dx = dα ln(1+tan α) dα = ln 1+x2 cos α 0 0 0 Z π/4 Z π/4 π = ln 2 + ln cos(π/4 − α) dα − ln cos α dα 8 0 0 π ln 2, = 8 where the last two integrals are equal by symmetry about α = π/8. This immediately deduces π π π2 I = · ln 2 = ln 2. 2 8 16  Solution II. We begin with a lemma. Lemma. If f is integrable on [0, 1], then Z π/2 Z π/2 Z π 1 f (x) dx. f (sin θ cos φ) sin θ dφdθ = 2 0 0 0

(3.11)

A surface integral with many faces

145

Proof. Let S be the unit sphere in the first octant. The double integral can then be viewed as a surface integral over S with the angle θ measured from the z-axis. Thus, x = sin θ cos φ, y = sin θ sin φ, z = cos θ and kTθ × Tφ k = | sin θ|. This yields Z

π/2 Z π/2

Z Z f (sin θ cos φ) sin θ dφdθ =

0

0

f (x) dS. S

On the other hand, viewing the x-axis as the polar axis and φ as the angle, we have Z Z Z π/2 Z 1 Z π 1 f (x) dS = f (x)dxdφ = f (x) dx, 2 0 S 0 0 which proves the lemma. Replacing θ by π/2 − θ in (3.11), we also find Z Z π/2 Z π/2 π 1 f (x) dx. f (cos θ cos φ) cos θ dθdφ = 2 0 0 0 To use the lemma to evaluate I, let f (x) = ln(2 − x)/(2 − 2x + x2 ) in (3.11). Then Z Z π 1 ln(1 + x) π 1 ln(2 − x) dx = dx. I= 2 0 2 − 2x + x2 2 0 1 + x2 In view of Z 1 π ln(1 + x) dx = ln 2, 2 1 + x 8 0 again, we find that π2 I= ln 2. 16  Remark. (3.11) can be proved without using the surface integral. To this end, let x = sin θ cos φ in the inner integral. Then Z π/2 Z sin θ sin θ p f (sin θ cos φ) sin θ dφ = f (x) dx. 0 0 sin2 θ − x2 Using u = sin θ in the outer integral gives Z π/2 Z π/2 f (sin θ cos φ) sin θ dφdθ 0

Z

0 1

Z

= 0

0

u

√

u2

u √ f (x)dxdu − x2 1 − u2

146

Integrations Z

1

Z

Z

1

= 0

=

1 2

0

1

 u √ √ du f (x)dx (switch the integral order) u2 − x2 1 − u2 x Z 1  1 √ dv f (x)dx (set v = u2 ). √ v − x2 1 − v x2

Notice that applying the substitution v = (1 + x2 )/2 + s(1 − x2 )/2 yields Z 1 Z 1 1 ds √ √ dv = = π. √ 2 1−v 2 v − x 1 − s2 x −1 This proves (3.11) again. Along the same lines, (3.11) can be generalized Z π Z 2π f (a sin φ cos θ + b sin φ sin θ + c cos φ) sin φ dθdφ 0

0

Z

1

= 2π

p f (x a2 + b2 + c2 ) dx

−1

and ZZZ

Z

1

f (ax + by + cz) dV = π

p (1 − x2 )f (x a2 + b2 + c2 ) dx,

−1

B 3

where B is the unit ball in R . We leave the proofs to the reader. It is also worth noting that this proposed problem provides a prefect example of discovering mathematical truths through the use of computers. Bailey and Borwein evaluated the numerical value of I to exceedingly high precision and found the exact value π 2 ln 2/16 via the Inverse Symbolic Calculator, an online numerical constant recognition tool available at http://wayback.cecm.sfu.ca/projects/ISC/ISCmain.html To establish the result rigorously, let A(θ, φ) = 1 − sin θ cos φ. They then established an analytical evaluation via double series expansions and Wallis’s formula: Z π/2 Z π/2 ln(1 + A) I2 = sin θ dθdφ 1 + A2 0 0 ! ! Z π/2 Z π/2 X ∞ ∞ k+1 X m 2m kA · = (−1) (−1) A sin θ dθdφ k+1 0 0 m=0 k=0 Z π/2 Z π/2 ∞ X ∞ X (−1)m+k = A2m+k+1 sin θ dθdφ k + 1 0 0 m=0 k=0

=

∞ ∞ π X X (−1)m+k+1 , 2 m=1 (k + 1)(2m + k) k=0

where they used Z 0

π/2 Z π/2 0

Ak (θ, φ) sin θ dθdφ =

π . 2k + 1

A surface integral with many faces

147

Now, it suffices to show that ∞ X ∞ X m=1 k=0

π (−1)m+k+1 = ln 2. (k + 1)(2m + k) 8

Note that ∞ X ∞ X m=1 k=0

Z 1 Z 1 ∞ X ∞ X (−1)m+k+1 k m+k+1 = x dx · y 2m+k−1 dy. (−1) (k + 1)(2m + k) m=1 0 0 k=0

Invoking the geometric series twice gives Z 1Z 1 ∞ X ∞ X y (−1)m+k+1 = dxdy. 2 )(1 + xy) (k + 1)(2m + k) (1 + y 0 0 m=1 k=0

Integrating the inner integral with respect to x yields Z 1 ∞ X ∞ X (−1)m+k+1 ln(1 + y) π = dy = ln 2 2 (k + 1)(2m + k) 1+y 8 0 m=1 k=0

as desired. Experimental mathematics is a newly developed approach to mathematical discovery. Modern computers make this easier and have extended our study and research range. Recently Stenger [89] stated that “Experimental math has been very successful in mathematical research, but there are a couple of reasons why it might be even more successful in helping to solve Monthly problems.” To illustrate this, he looked at three particular methods and applied them to solve six Monthly problems. Note that Experimental math is primarily heuristic. It does guide us to an answer, but we still have to prove it. Here we provide two problems from SIAM Review for additional practice. 1. (Due to A. H. Nuttall) For α > 0, show that α Z π  sin x παα exp(x cot x) dx = . x Γ(1 + α) 0 The above result has been confirmed numerically to 15 decimal places for numerous values of α ∈ [0, 150]. 2. (Due to H. P. Robinson) Does Z

∞

(ζ(x) − 1) dx = lim 0

n→∞

! Z n n X 1 dx − ? ln k 0 ln x

k=2

It was found that both sides match to at least 43 decimal places.

148

3.9

Integrations

Evaluate a definite integral by the gamma function

Problem 11564 (Proposed by A. Stadler, 118(4), 2011). Prove that √ ! Z ∞ e−x (1 − e−6x ) 3+ 5 . dx = ln x(1 + e−2x + e−4x + e−6x + e−8x ) 2 0 Discussion. The usual integration techniques studied in the first year calculus will not work on this integral directly. Using the substitution t = e−x yields Z 1 Z 1 1 − t6 (1 − t2 )(1 − t6 ) − dt = − dt. 2 4 6 8 (1 − t10 ) ln t 0 (1 + t + t + t + t ) ln t 0 This suggests we consider the parametric integral Z 1 (1 − tp )(1 − t6 ) dt. (1 − t10 ) ln t 0 The key insight is to use parametric differentiation to remove the ln t from the denominator. Solution. We will prove a more general result: if p ≥ 0, α > 0 and β = p + α + 2, then Z ∞ −x Z 1 e (1 − e−px )(1 − e−αx ) (1 − tp )(1 − tα ) dx = − dt x(1 − e−βx ) (1 − tβ ) ln t 0  0 sin((p + 1)π/β) . (3.12) = ln sin(π/β) Once (3.12) is confirmed, the desired result follows upon letting p = 2, α = 6, β = 10. Hence, √ !   Z ∞ e−x (1 − e−6x ) sin(3π/10) 3+ 5 dx = ln = ln . x(1 + e−2x + e−4x + e−6x + e−8x ) sin(π/10) 2 0 To prove (3.12), let Z I(p) = − 0

1

(1 − tp )(1 − tα ) dt. (1 − tβ ) ln t

Evaluate a definite integral by the gamma function

149

We compute 1

Z

0

I (p)

= 0 1

Z =

tp (1 − tα ) dt (1 − tβ ) ∞ X tp (1 − tα ) tnβ dt

0

= =

n=0

∞ Z X n=0 ∞ X

tp+nβ (1 − tα ) dt

0



n=0

Recall that

1

1 1 − p + 1 + nβ p + α + 1 + nβ ∞

1 X Γ0 (x) =γ+ + − Γ(x) x n=1



1 1 − n+x n

 .

 .

This yields that 0

I (p)

= =

 ∞  1 X 1 1 − β n=0 n + (p + 1)/β n + (p + α + 1)/β  0  1 Γ ((p + α + 1)/β) Γ0 ((p + 1)/β) − . β Γ((p + α + 1)/β) Γ((p + 1)/β)

Hence,  I(p) = ln

Γ((p + α + 1)/β) Γ((p + 1)/β)

 + C.

Since I(0) = 0, it follows that  C = − ln and so

 I(p) = ln

Γ((α + 1)/β) Γ(1/β)

 ,

Γ((p + α + 1)/β)Γ(1/β) Γ((p + 1)/β)Γ((α + 1)/β)

 .

By the Euler reflection formula Γ(x)Γ(1 − x) =

π , sin πx

and β = p + α + 2, we have π , sin((p + 1)π/β) π Γ((p + α + 1)/β)Γ(1/β) = . sin(π/β)

Γ((p + 1)/β)Γ((α + 1)/β) =

(3.13)

150

Integrations

Substituting them into (3.13) yields (3.12).



Remark. The approach used to prove (3.12) enables us to evaluate a class of integral of this type, which were initially considered by Euler. The following problems offer additional practice. 1. Let p, q > −1 and p + q > −1. Then Z ∞ −x Z 1 e (1 − e−px )(1 − e−qx ) dx dx = (1 − xp )(1 − xq ) x ln x 0 0 p+q+1 . = ln (p + 1)(q + 1) 2. Let a, b, c > 0 and d > a + b + c. Then   Z ∞ (1 − e−ax )(1 − e−bx )e−cx Γ(c/d)Γ((a + b + c)/d) dx = ln . x(1 − e−dx ) Γ((a + c)/d)Γ((b + c)/d) 0 3. Let p, q > −1 and p + α, q + α > −1. Then   Z 1 p Γ(q + 1)Γ(p + α + 1) x − xq 1 − xα dx = ln . 1 − x ln x Γ(p + 1)Γ(q + α + 1) 0 4. Let p, q, α > 0. Then   Z 1 p−1 x − xq−1 dx Γ((p + α)/2α)Γ(q/2α) dx = ln . 1 + xα ln x Γ((q + α)/2α)Γ(p/2α) 0 5. Let p, q > 0. Then   Z 1 p−1 Γ((p + 2)/2n)Γ(q/2n) x − xq−1 1 − x2 dx = ln . 1 − x2n ln x Γ((q + 2)/2n)Γ(p/2n) 0 6. Let 0 < p, q < α. Then   Z ∞ p−1 sin(pπ/α) x − xq−1 dx dx = ln ; 1 − xα ln x sin(qπ/α) 0 Z ∞ p−1 h  pπ   qπ i x − xq−1 dx dx = ln tan cot . 1 + xα ln x 2α 2α 0 7. Problem 5035 (Proposed by Y. Matsuoka, 69(6), 1962). Let α be a fixed positive number. Prove that, as n → ∞,   Z π/2 α+1 1 /n(α+1)/2 tα cos2n t dt = Γ 2 2 0   1 α+5 − Γ /n(α+3)/2 + O(1/n(α+5)/2 ). 12 2 Hint: Consider the Lagrange reversion of a power series: tα = a1 sinα t + a2 sinα+2 t + a3 sinα+4 t + · · · .

Digamma via a double integral

151

8. Problem 12184 (Proposed by P. Perfetti, 127(5), 2020). Prove Z ∞ √ ln(x4 − 2x2 + 2) √ dx = π ln(2 + 2). 2 x x −1 1

3.10

Digamma via a double integral

Problem 11937 (Proposed by J. C. Sampedro, 123(9), 2016). Let s be a complex number not a zero of the gamma function Γ(s). Prove 1

Z

Z

0

1

0

Γ0 (s) (xy)s−1 − y dxdy = . (1 − xy) ln(xy) Γ(s)

Discussion. By the Euler reflection formula, we have Γ(z)Γ(1 − z) sin πz = π. This implies that Γ(z) 6= 0 for all z ∈ C. To ensure the proposed integral converges, we must assume that Re(s) > 0. To show the convergence of the double integral for Re(s) > 0, we rewrite it as Z

1

Z

− 0

0

1

1 − (xy)s−1 dxdy + 1 − xy ln(xy)

Z 0

1

Z

1

0

1 − y dxdy . 1 − xy ln(xy)

s−1

Notice that (1 − (xy) )/(1 − xy) has a finite limit as xy → 1, both functions R 1 R 1 dxdy xs−1 and y s−1 are integrable at 0, and 0 0 | ln(xy)| < ∞. Thus the first 1−y integral converges absolutely. Since 0 ≤ 1−xy ≤ 1 for x, y ∈ (0, 1), the second integral converges absolutely as well. There are a number of well-known series and integral representations for the digamma function ψ(s) = Γ0 (s)/Γ(s). This prompts us to work on transforming the double integral into either a series or an integral representation of the digamma function. We give one example of each in the following two solutions.

Solution I — by series. Recall that ∞

Γ0 (s) 1 X = −γ − − Γ(s) x n=1



1 1 − s+n n

 = −γ −

∞  X n=1

1 1 − s+n−1 n

 ,

152

Integrations

where γ is Euler’s constant. We now show the double integral precisely equals to the series above. Denote the double integral by S. Expanding 1/(1 − xy) into a geometric series yields Z 1Z 1 ∞ (xy)s−1 − y X (xy)n dxdy S = ln(xy) n=0 0 0 ∞ Z 1Z 1 X (xy)s−1 − y = (xy)n dxdy. ln(xy) 0 0 n=0 Here interchanging summation and integration is justified by the terms having the same sign. Since for 0 < xy < 1, Z ∞ (xy)k =− (xy)t dt, ln(xy) k we find that S

= − = −

∞ Z X

1

Z

1

∞

Z

0 n=0 0 Z ∞ ∞ X

(xy)t dt −

n+s−1

Z

n+s−1 n=0 ∞ Z ∞ X

1Z

0

Z

∞

 y(xy)t dt dxdy

n

1

(xy)t dxdydt −

Z

0

∞

n

Z

0

∞

dt dt − 2 (t + 1) (t + 1)(t + 2) n n+s−1 n=0    ∞ X n+2 1 − ln = − n+s n+1 n=0    ∞ X n+1 1 − ln . = − n+s−1 n n=1 = −

1

Z

Z

1

 y(xy)t dxdydt

0



Since the integrands in the first equality are nonnegative, we are able to reverse the order of integration. By the definition   1 1 γ = lim 1 + + · · · + − ln n , n→∞ 2 n we have γ=

  ∞  X n+1 1 − ln , n n n=1

and conclude that S = −γ −

∞  X n=1

1 1 − s+n−1 n

 =

Γ0 (s) . Γ(s) 

Digamma via a double integral

153

Solution II — by integral representation. The integral representation we have in mind is  Z 1  s−1 t Γ0 (s) 1 =− + dt. Γ(s) 1 − t ln t 0

(3.14)

That is a well-known formula of the digamma function ([49], 4.281.4, p. 583). We show a more general result: Let s be a complex number such that Re(s) > 0 and let n ∈ N. Then  Z 1Z 1 Z 1  s−1 (xy)s−1 − y n ln n! t 1 dxdy = − − + dt. (3.15) n 1 − t ln t 0 (1 − xy) ln(xy) 0 0 Let t = xy. Then x = t/y and dx = dt/y. We compute Z 1Z 1 (xy)s−1 − y n dxdy 0 0 (1 − xy) ln(xy) Z 1 Z y s−1 t − yn = dtdy 0 0 y(1 − t) ln t Z 1  s−1   Z 1 1 t n−1 = −y dy dt y 0 (1 − t) ln t t   Z 1 1 1 =− ts−1 ln t + (1 − tn ) dt n 0 (1 − t) ln t   Z 1  s−1 Z 1 t 1 1 − tn dt =− + dt − −1 . 1 − t ln t n(1 − t) ln t 0 0 Since

n−1 n−1 1 − tn 1 X k 1 X k −1= t −1= (t − 1), n(1 − t) n n k=0

k=0

we find that  Z 1 n−1 Z 1 n−1 X tk − 1 dt 1 1 X 1 1 − tn −1 = dt = ln(k+1) = ln(n!). n(1 − t) ln t n 0 ln t n n 0 k=0

k=0

This, together with (3.14), confirms (3.15) as claimed. Remark. Here we give an elementary proof of (3.14). In view of Z ∞ −t e − e−xt ln x = dt, t 0 we have 0

Γ (s)

Z

∞

= 0

Z = 0

∞

e−x xs−1 ln x dx Z ∞ −t  e − e−xt −x s−1 e x dt dx t 0



154 Z = =



Z

∞

Γ0 (s) = Γ(s)

Z

∞

∞

Z

∞

xs−1 e−x − e−t 0 0 0 Z ∞ dt 1 Γ(s) e−t − . s (1 + t) t 0

Hence, 0



Integrations  dt s−1 −x(1+t) x e dx t

e−t 1 − t t(1 + t)s

 dt.

The substitution t → et − 1 leads to  Z ∞  −t e e−st Γ0 (s) dt. = − Γ(s) t 1 − e−t 0 This, with the substitution t → e−t , yields (3.14). We collect a few problems for additional practice. 1. Let s > −1. Show that Z 1Z 1 (− ln(xy))s dxdy = Γ(s + 2)ζ(s + 2). 1 − xy 0 0 2. (Hadjicostas-Chapman formula [27]). Let Re(s) > −2. Show that   Z 1Z 1 1 (1 − x)[− ln(xy)]s dxdy = Γ(s + 2) ζ(s + 2) − . 1 − xy s+1 0 0 3. Let p, q > 0 and p 6= q. Show that   Z 1Z 1 1 Γ(q/2)Γ((p + 1)/2) xp−1 y q−1 dydx = ln . p−q Γ(p/2)Γ((q + 1)/2) 0 0 (1 + xy)(− ln(xy)) What happens if p = q? 4. Let s > 0. Show that Z 1Z 1 (xy)s−1 (1 − x) Γ0 (s) dxdy = ln s − . Γ(s) 0 0 (1 − xy)(− ln(xy)) In particular, for s = 1, this yields Sondow’s formula [83] Z 1Z 1 1−x γ= dxdy. 0 0 (1 − xy)(− ln(xy)) 5. Problem 11322 (Proposed by J. Sondow, 114(9), 2007). Let N be a positive integer. Prove that Z 1 Z 1 (x(1 − x)y(1 − y))N dydx x=0 y=0 (1 − xy)(− ln(xy)) 2 Z ∞  ∞ X N! = dt. t(t + 1) · · · (t + N ) t=n n=N +1

Another double integral

155

6. Problem 11331 (Proposed by R. Bagby, 114(10), 2007). Show that if k is a positive integer, then k+1 Z ∞ k X ln(1 + t) dt = (k + 1) aj ζ(j + 1), t 0 j=1 where ζ denotes the Riemann zeta function and aj is the coefficient Qk−1 of xj in x n=1 (1 − nx). Hint: The factor (k +1) before the series was missing in the original statement. Show the proposed integral is equivalent to Z 1 k−1 x (− ln x)k+1 dx. (1 − x)k+1 0 7. SIAM Review Problem 85-22 (Proposed by E. O. George and C. C. Rousseau). Evaluate s/2 Z 1Z u  uv Mn (s) = n(n − 1) (u − v)n−2 dvdu. (1 − u)(1 − v) 0 0 Here Mn (s) is the moment generating function for the mid-range of a random n-sample from the logistic distribution.

3.11

Another double integral

Problem 11650 (Proposed by M. Becker, 119(6), 2012). Evaluate Z ∞ Z ∞ 2 x2 − y 2 dydx. e−(x−y) sin2 (x2 + y 2 ) 2 (x + y 2 )2 x=0 y=x Discussion. Let I be the proposed integral. To reformulate the problem into an equivalent but manageable one, we switch to polar coordinates and obtain Z ∞ Z π/2 2 cos 2θ dθdr I = e−r (1−sin 2θ) sin2 (r2 ) r 0 π/4 Z ∞ −r2 2 2 Z π/2 2 e sin (r ) er sin 2θ cos 2θdθdr = r π/4 0 Z ∞ −r2 2 2 e sin (r ) 1 r2 sin 2θ π/2 = e π/4 dr r 2r2 0 Z 1 ∞ sin2 (r2 ) −r2 (e − 1) dr. = 2 0 r3

156

Integrations

Furthermore, the change of variable t = r2 yields Z 1 ∞ sin2 t −t (e − 1) dt. I= 4 0 t2

(3.16)

Thus it suffices to evaluate (3.16). This can be done in a variety of ways. We present three different calculations as below. Solution I. We calculate (3.16) by parametric differentiation. It is well-known that Z

∞

π sin2 t dt = . t2 2

0

To compute the integral tegral

R∞ 0

e−t sin2 t/t2 dt, we introduce the parametric in∞

Z

sin2 (αt) −t e dt. t2

F (α) = 0

By the Leibniz rule, we have ∞

Z

0

F (α)

sin(2αt) −t e dt; t

= 0

Z

00

F (α)

=

∞

2

e−t cos(2αt) dt,

0

where the justifications of differentiating under the integrals are ensured by the uniformly convergence on α. Integrating by parts leads to F 00 (α) =

2 . 1 + 4α2

Appealing to F (0) = F 0 (0) = 0, integrating F 00 (α) twice yields F (α) = α arctan(2α) −

1 ln(1 + 4α2 ). 4

Hence, Z

∞

F (1) = 0

sin2 t −t 1 e dt = arctan 2 − ln 5. t2 4

In summary, we conclude I=

1 1 1 arctan 2 − ln 5 − π. 4 16 8

Notice that arctan 2 + arctan(1/2) = π/2. We find a more compact form:   1 1 1 I=− ln 5 − arctan . 16 4 2

Another double integral

157 

Solution II. We calculate (3.16) by parametric integration. Rewrite Z 1 ∞ e−t − 1 sin2 t I= dt. 4 0 t t Since e−t − 1 =− t

Z

1

e−st ds

0

it follows that

1

and

sin2 t =

1 (1 − cos(2t)), 2

∞

e−st (1 − cos(2t)) dtds. t 0 0 R∞ For the inner integral, using 1/t = 0 e−tu du, for s > 0, we have  Z ∞ Z ∞ Z ∞ −st e (1 − cos(2t)) dt = e−(s+u)t (1 − cos(2t)) dt du t 0 0 0  Z ∞ s+u 1 − du = s + u 4 + (s + u)2 0 ! √ 4 + s2 . = ln s I=−

1 8

Z

Z

The interchange of order of integration is justified by nonnegativity of the integrands. Integrating by parts gives ! √   Z 1 1 4 + s2 1 1 1 I=− ln ds = − ln 5 − arctan . 8 0 s 16 4 2  Solution III. We calculate (3.16) by the Laplace transform. To this end, we show that  2  Z ∞ 2 2 s s2 + 4 sin t −st sin t (s) = e dt = arctan − ln . (3.17) L 2 2 t t s 4 s2 0 Once (3.17) is established, we have   2   2   1 sin t sin t 1 1 1 I= L (1) − L (0) = arctan 2 − ln 5 − π. 4 t2 t2 4 16 8 We now prove (3.17) by using parametric differentiation. Let  2  sin (pt) L(p) = L (s). t2

158

Integrations

Then   2       sin (pt) d sin(2pt) L (p) = L (s) = L (s) t2 dp t 2s . = 2L[cos(2pt)](s) = 2 4p + s2 d2 dp2

00

Since L(0) = L0 (0) = 0, we integrate this equation twice to get     2 2p s s + 4p2 L(p) = a arctan . − ln s 4 s2 (3.17) now follows from setting p = 1.



Remarks. It is interesting to see that this problem can be solved by Mathematica. Indeed, after passing to the polar coordinates, the command Integrate[Sin[r^2]^2*(Exp[-r^2] - 1)/(2 r^3), {r, 0, Infinity}] produces the output 1/16 (-2 \[Pi] + 4 ArcTan[2] - Log[5]) The approach by the Laplace transform seems go through for much board sense. To illustrate this method further, we present another solution to the following Monthly problem based on the Laplace transform. Problem 11322 (Proposed by J. Sondow, 114(9), 2007). Let N be a positive integer. Prove that 2 Z ∞ Z 1Z 1 ∞ X N! (x(1 − x)y(1 − y))N dydx = dt. (1 − xy)(− ln(xy)) t(t + 1) · · · (t + N ) n 0 0 n=N +1

Let

 fN (s) =

(1 − e−s )N 0

First, by induction, we show that Z ∞ FN (p) := L[fn ] = fN (s)e−ps ds = 0

for s ≥ 0 for s < 0. N! . p(p + 1) · · · (p + N )

(3.18)

Clearly, (3.18) hold for N = 0. In general, for p > 0, we have Z ∞ Z ∞ −s −ps FN (p)−FN +1 (p) = fN (s)e e ds = fN (s)e−(p+1)s ds = FN (p+1), 0

0

and so FN +1 (p) = FN (p) − FN (p + 1). By the induction hypothesis, we find that FN +1 (p)

= =

N! N! − p(p + 1) · · · (p + N ) (p + 1)(p + 2) · · · (p + N + 1) (N + 1)! . p(p + 1) · · · (p + N + 1)

Another double integral

159

Thus (3.18) holds for all integers N ≥ 0. Invoking the convolution properties of the Laplace transform, we have L[fN ∗ fN ](p) = FN2 (p) =



N! p(p + 1) · · · (p + N )

2 .

Integrating with respect to the parameter p gives Z 0

∞

e−ps fN ∗ fN (s) ds = s

∞

Z p



N! u(u + 1) · · · (u + N )

2 du.

On the other hand, invoking Fubini’s theorem, we find Z ∞ −ps Z ∞ Z ∞ −ps e e fN ∗ fN (s) ds = fN (t)fN (s − t) dtds s s 0 0 0 Z ∞ Z ∞ −ps e = fN (t)fN (s − t) dsdt s 0 0 Z ∞ Z ∞ −p(u+t) e fN (t)fN (u) dudt = s 0 0 Z 1Z 1 (xy)p−1 (1 − x)N (1 − y)N = dydx − ln(xy) 0 0 (let x = e−t , y = e−u ). Hence, for p > 0, we conclude that 1

Z

1

Z

0

0

(xy)p−1 (1 − x)N (1 − y)N dydx = − ln(xy)

Z p

∞



N! u(u + 1) · · · (u + N )

2 du.

Summing both sides for p from N + 1 to infinity yields   Z 1Z 1 ∞ X (1 − x)N (1 − y)N  dydx (xy)p−1  − ln(xy) 0 0 p=N +1

=

∞ X p=N +1

Z p

∞



N! u(u + 1) · · · (u + N )

2 du,

where the interchange the order of integration and summation is justified by the positivity of integrands. Appealing to the geometric series, this proves Z 0

1

Z 0

1

2 Z ∞ ∞ X (x(1 − x)y(1 − y))N N! dydx = du (1 − xy)(− ln(xy)) u(u + 1) · · · (u + N ) p

as desired.

p=N +1

160

Integrations We end this section with eight problems for additional practice. 1. Let Z f (s) =

1

xs



0

1 1 + ln x 1 − x

2 dx.

For s > 0, show that  f (s) = sψ(s + 1) + ln

2π(1 + s)s+1 e−2(s+3/4) Γ2 (1 + s)

 ,

where ψ is the digamma function. 2 Hint: Let g(t) = (1/(1 − e−t ) − 1/t) . Then Z L(s) := L[g](s) = 0

∞

g(t)e−st dt =

Z 0

1

xs−1



1 1 + ln x 1 − x

2 dx = f (s−1).

You can recover L(s) from showing that  Z ∞  2 −st 1 t e 2te−st 1 L00 (s) = + − dt = + (s − 1)ψ 00 (s). −t )2 −t s (1 − e 1 − e s 0 2. For |a| < 1/2, evaluate (Due to M. L. Glasser)   Z ∞ t ecos t sin cos(at − sin t) dt. 2 t 0 3. SIAM Review Problem 94-14 (Proposed by N. Ortner). For α > 0, evaluate Z ∞ x arctan(α/x) √ dx. 1 + x2 + 1 + x2 0 4. Problem 11275 (Proposed by M. Becker, 114(2), 2007). Find Z ∞ Z ∞ (x − y)2 ln((x + y)/(x − y)) dxdy. xy sinh(x + y) y=0 x=y Hint: By changing variables show that the double integral is equal to Z 1 2 π2 u ln u − du. 2 0 1 − u2 5. Problem 11953 (Proposed by C. V. V˘alean, 124(1), 2017). Evaluate Z ∞Z ∞ sin x sin y sin(x + y) dxdy. xy(x + y) 0 0

An integral with log and arctangent

161

Hint. Let sinc(x) = sin x/x. The Fourier transform gives sinc(x) = F(h)/2, where h is the indicator function of the interval [−1, 1]. Applying the Plancherel theorem Z ∞ Z ∞ F(f )(x)F(g)(x) dx = 2π f (t)¯ g (t) dt, −∞

−∞

shows that Z

∞

sinc(x)sinc(x + y) dx = πsinc(y). −∞

6. Problem 12070 (Proposed by C. V. V˘alean, 125(9), 2018). Prove π/4

π/4

cos x cos y(y sin y cos x − x sin x cos y) dxdy cos(2x) − cos(2y) 0 0 7ζ(3) + 4π ln 2 = . 64

Z

Z

7. SIAM Review Problem 65-5 (Proposed by N. Mullineux and J. R. Reed). Show that Z ∞Z ∞ π e−y cos(xy) p dydx = 2 . 2 2 2 2 α µy α + y + α + y −∞ 0 8. SIAM Review Problem 77-3 (A define integral of Bohr). Find the exact value of Z ∞ K= F (x)(F 0 (x) − ln x) dx, 0

where

Z

∞

F (x) = −∞

3.12

cos(xy) dy. (1 + y 2 )3/2

An integral with log and arctangent

Problem 12054 (Proposed by C. I. V˘alean, 125(6), 2018). Prove   Z 1 arctan x 1 + x2 π3 ln dx = . 2 x (1 − x) 16 0

162

Integrations

Discussion. The substitution x = tan θ or x = (1 − u)/(1 + u) doesn’t seem to lead anywhere, so we try to evaluate this integral directly by transforming the integrand into a power series. Solution. First, we show that ∞ X (−1)n−1 H2n 2n+1 x , 2n + 1 n=1

arctan x ln(1 + x2 ) = 2

(3.19)

where Hn is the nth harmonic number. To see this, recall that arctan x =

∞ X (−1)n 2n+1 x 2n + 1 n=0

and

ln(1 + x2 ) =

∞ X (−1)n−1 2n x . n n=1

The Cauchy product of these two series gives ∞ X

! n X 1 1 · x2n+1 arctan x ln(1 + x2 ) = (−1)n−1 2 2k 2n + 1 − 2k n=1 k=1  ∞ n  X X 1 1 1 n−1 = 2 (−1) + x2n+1 2n + 1 2k 2n + 1 − 2k n=1 k=1

=

×(using partial fractions) ∞ X (−1)n−1 H2n 2n+1 x , 2 2n + 1 n=1

which proves (3.19). Hence, 1

Z 0

arctan x ln(1 + x2 ) dx = x

Z 0

1

∞ X (−1)n−1 H2n 2n 2 x 2n + 1 n=1

! dx

∞ X (−1)n−1 H2n . =2 (2n + 1)2 n=1

(3.20)

The interchange of the order of summation and integration is justified as follows: For x ∈ [0, 1], using the estimate of the alternating series remainder yields ∞ H X (−1)n−1 H 2(N +1) 2n 2n x ≤ → 0, as N → ∞. 2n + 1 2N + 3 n=N

Next, we have Z 0

1

∞ Z 1 X (−1)n 2n arctan x ln(1 − x)2 dx = 2 x ln(1 − x) dx. x 2n + 1 n=0 0

An integral with log and arctangent Since Z 1 x

2n

ln(1 − x) dx

∞ X x2n+k k

1

Z −

=

0

163

0

!

k=1

Z ∞ X 1 1 2n+k x dx dx = − k 0 k=1

∞ X 1 1 − =− k(2n + k + 1) 2n + 1 k=1 k=1   1 1 × − k 2n + k + 1 1 − H2n+1 , 2n + 1 ∞ X

=

=

where the positivity of integrands justifies the interchange of the order of summation and integration, we have 1

Z 0

∞ X (−1)n H2n+1 arctan x ln(1 − x)2 dx = −2 . x (2n + 1)2 n=0

(3.21)

Finally, combining (3.20) and (3.21), we find that Z 0

1

arctan x ln x



1 + x2 (1 − x)2

=

∞ ∞ X X (−1)n H2n+1 (−1)n−1 H2n + 2 2 (2n + 1)2 (2n + 1)2 n=0 n=1

=

2

=

2

 dx

∞ X (−1)n (H2n+1 − H2n ) (let H0 = 0) (2n + 1)2 n=0 ∞ X n=0

π3 (−1)n = , 3 (2n + 1) 16

where we have used the well-known result ∞ X n=0

(−1)n π3 = . (2n + 1)3 32 

Remark. It is interesting to reveal one unsuccessful journey to evaluate this integral by the substitution x = tan θ. In this case, we have Z

1

arctan x ln x

0



1 + x2 (1 − x)2



Z

π/4

θ ln(1 − sin(2θ)) dθ. sin θ cos θ

dx = − 0

The change of variable 2θ → θ leads to Z 0

1

arctan x ln x



1 + x2 (1 − x)2

 dx = −

1 2

Z 0

π/2

θ ln(1 − sin θ) dθ. sin θ

164

Integrations

To compute Z

π/2

θ ln(1 − sin θ) dθ, sin θ 0 we initially introduced the parametric integral Z π/2 θ J(p) := ln(1 − p sin θ) dθ. sin θ 0

(3.22)

The Leibniz rule gives J 0 (p) =

π/2

Z 0

θ dθ. 1 − p sin θ

This integral may look simple but it is really hard to calculate. Even Mathematica does not provide a useful result. Then we turn to the power series of ln(1 − x) and obtain Z π/2 Z π/2 ∞ X 1 θ ln(1 − sin θ) dθ = θ sinn−1 θ dθ. sin θ n 0 0 n=1 For n ≥ 1, let Z

π/2

Sn =

θ sinn−1 θ dθ.

0

Clearly, S1 = π 2 /8, S2 = 1. For n ≥ 3, integrating by parts leads to the following reduction formula Sn =

n−2 1 + Sn−2 . 2 (n − 1) n−1

For k ≥ 1, this implies that S2k−1

= =

2k − 3 (2k − 3)(2k − 5) · · · 1 π 2 1 + + ··· + 2 2 [2(k − 1)] 2(k − 1)[2(k − 2)] 2(k − 1)2(k − 2) · · · 2 8 k−2 X n=0

(2k − 3)(2k − 5) · · · [2(k − n) − 1] 1 · 2(k − 1)2(k − 2) · · · [2(k − n − 1)] 2(k − n − 1)

(2k − 3)!! π 2 ; [2(k − 1)]!! 8 1 2(k − 1) 2(k − 1) 2(k − 2) · · · 2 + + ··· + 2 2 (2k − 1) (2k − 1)(2k − 3) (2k − 1)(2k − 3) · · · 3

+ S2k

= =

k−1 X n=0

2(k − 1)2(k − 2) · · · 2(k − n) 1 · . (2k − 1)(2k − 3) · · · [2(k − n) + 1] 2(k − n) + 1

However, we are unable to find the closed forms of the following series ∞ X 1 S2k 2k

k=1

and

∞ X k=1

1 S2k−1 . 2k − 1

An integral with log and arctangent

165

In summary, although we have Z π Z θ π π 1 3 ln(1 − sin θ) dθ = ln(1 − sin θ) dθ = − π 3 , sin θ 2 sin θ 8 0 0 currently we are unable to evaluate (3.22) in an elementary way. We now end this section with ten problems for additional practice. 1. Problem 10884 (Proposed by Z. Ahmed, 108(6), 2001). Evaluate √ Z 1 arctan 2 + x2 √ dx. 2 2 0 (1 + x ) 2 + x 2. (Coxeter’s Integral) Evaluate Z

π/2

 arccos

0

cos θ 1 + 2 cos θ

 dx.

Hint: One way to proceed is first to find p Z 1 arctan p + x2 p dx 2 2 0 (1 + x ) p + x for p > 1, then convert Coxeter’s integral into one special case of the above integral by using r 1−t arccos t = 2 arctan . 1+t 3. Problem 11152 (Proposed by M. Ivan, 112(5), 2005). Evaluate Z 0

1

ln(cos(πx/2)) dx. x(1 + x)

4. Problem 11961 (Proposed by M. Berindeanu, 124(2), 2017). Evaluate Z π/2 sin x √ dx. 1 + sin x 0 5. For 0 < α, β < 1, show that   Z 1Z 1 dxdy 1 (1 − α)(1 − β) = F , 2 2 2 (1 + α)(1 + β) α β (x + y) + (1 + xy) where

Z F (x) = 0

x

arctan t dt. t

166

Integrations 6. Problem 4826 (ProposedR by M. S. Klamkin and L. A. Shepp, x t 66(1), 1959). Let F (x) = 0 arctan dt. Express F (1) in terms of t √ F (2 − 3), then obtain a more rapidly converging expansion. Indeed, this problem can be traced back to Ramanujan. He established the astonishing identity sin(2x) sin(6x) sin(10x) + + + · · · = F (tan x) − x ln(tan x) 12 32 52 by showing their derivatives are equal. 7. Problem R x arctan t 4865 (Proposed by L. Lewin, 66(8), 1959). Let F (x) = dt. Prove that t 0 6F (1) − 4F (1/2) − 2F (1/3) − F (3/4) = π ln 2. 8. Define Γ(s, x) =

R∞ x

ts−1 e−t dt. Find Z

1

Z

1

ln Γ(s, x) dsdx. 0

9. Show that Z 0

Z 0

1

1

0


2n x2n ln x n π √ dx = 2n+1 2 1 − x2

(2n)!! x2n+1 ln x √ dx = 2 (2n + 1)!! 1−x

! 2n X (−1)k−1 − ln 2 , k

k=1

ln 2 +

2n+1 X k=1

! (−1)k−1 − ln 2 . k

10. SIAM Review Problem 84-8 (Proposed by J. A. Prove directly that    Z 1 2 1 1+x 2 arctan arctan(1/x) + ln − π π π 1−x 0   1 π √ = ln . 2 2 2

3.13

Morrison). 1 2



dx x

Another integral with log and arctangent

Problem 12158 (Proposed by H. Grandmontagne, 127(1), 2020). Prove Z 1 (ln(x))2 arctan(x) 21πζ(3) π 2 G π 3 ln(2) dx = − − 1+x 64 24 32 0

Another integral with log and arctangent 167 P∞ P∞ where ζ(3) is Ap´ery’s constant k=1 1/k 3 and G = k=0 (−1)k /(2k + 1)2 is Catalan’s constant. Discussion. Recall that in the previous section, when the standard substitution x = tan θ does not work, we turn to transforming the integrand into a power series. Since ∞ X (−1)n 2n+1 arctan x = x , 2n + 1 n=0 the Cauchy product of series gives     1 1 arctan x 2 3 = x+x + 1− x + 1− x4 1−x 3 3     1 1 1 1 5 + 1− + x + 1− + x6 + · · · 3 5 3 5 ! ! ∞ n−1 ∞ n X X (−1)k X X (−1)k 2n+1 x + x2n . = 2k + 1 2k + 1 n=1 n=0 k=0

k=0

Replacing x by −x yields ∞ arctan x X = 1+x n=0

Since Z 0

n X (−1)k 2k + 1

! x

2n+1

k=0

1

(ln x)2 xk dx =

2 (k + 1)3

−

∞ X

n−1 X

n=1

k=0

(−1)k 2k + 1

! x2n .

for all k ≥ 0,

the proposed integral is successfully reduced to the following series: ! Z 1 ∞ n X 1 X 1 (−1)k (ln(x))2 arctan(x) dx = 1+x 4 n=0 (n + 1)3 2k + 1 0 k=0 ! ∞ n−1 X X (−1)k 1 −2 . (2n + 1)3 2k + 1 n=1

(3.23)

k=0

However, this reformulation provides little insight because both double series can be viewed as generalizations of Euler sums which have not been studied in details yet. In the following, using integration by parts, we convert the proposed integral into a double integral. The presence of symmetry enables us to reduce the amount of calculation in arriving at the desired answer. At the end of this section, we will revisit the double series in (3.23). Solution. Rewrite the proposed integral as Z x  Z 1 Z 1 (ln(t))2 (ln(x))2 arctan(x) dx = arctan(x)d dt . 1+x 1+t 0 0 0

168

Integrations

Let

x

Z f (x) = 0

(ln(t))2 dt. 1+t

Then 1

Z f (1) = 0

Z 1 ∞ ∞ X X (ln(t))2 3ζ(3) 2(−1)n = dy = (ln(t))2 tn dt = . (−1)n 3 1+t (n + 1) 2 0 n=0 n=0

Applying the fundamental theorem of calculus, we have Z 1 3πζ(3) 3ζ(3) π (f (x) arctan(x))0 dx = f (x) arctan(x)|10 = · = . 2 4 8 0 On the other hand, since Z 1 Z 0 (f (x) arctan(x)) dx = 0

1

0

f (x) dx + 1 + x2

Z

1

(ln(x))2 arctan(x) dx, 1+x

0

we find that 1

Z 0

3πζ(3) (ln(x))2 arctan(x) dx = − 1+x 8

Let Z I := 0

1

Z 0

1

f (x) dx. 1 + x2

(3.24)

f (x) dx. 1 + x2

Using the substitution t = xy, we have Z 1 x(ln(xy))2 f (x) = dy. 1 + xy 0 Hence, Z I= 0

1

Z 0

1

x[(ln(x))2 + 2 ln(x) ln(y) + (ln(y))2 ] dydx = I1 + I2 + I3 . (3.25) (1 + x2 )(1 + xy)

First, we have Z I1

1

= 0

Z

1

= 0

Z

1

= 0

Z = 0

1

1

x(ln(x))2 dydx 2 0 (1 + x )(1 + xy)   Z (ln(x))2 1 x dy dx (1 + x2 ) 0 1 + xy   y=1 (ln(x))2 ln(1 + xy) dx y=0 (1 + x2 )

Z

(ln(x))2 ln(1 + x) dx. (1 + x2 )

Another integral with log and arctangent

169

Next, we have Z I3

1

= 0

Z

1

= 0

=

1

πy + 2 ln 2 − 4 ln(1 + y) (ln(y))2 dy 2) 4(1 + y 0 Z Z π 1 y(ln(y))2 ln(2) 1 (ln(y))2 dy + dy − I1 . 4 0 1 + y2 2 1 + y2 0

Z =

1

x(ln(y))2 dydx 2 0 (1 + x )(1 + xy)   Z 1 x dx (ln(y))2 dy 2 0 (1 + xy)(1 + x )

Z

Since Z

1

0

y(ln(y))2 dy 1 + y2

= =

∞ X n=0 ∞ X n=0

= Z 0

1

(ln(y))2 dy 1 + y2

= =

(−1)n

Z

1

(ln(y))2 y 2n+1 dy

0 ∞ 2(−1)n 1 X (−1)n = (2n + 2)3 4 n=0 (n + 1)3

3ζ(3) ; and 16 Z 1 ∞ X n (−1) (ln(y))2 y 2n dy n=0 ∞ X

2

n=0

0

(−1)n π3 = 2β(3) = , 3 (2n + 1) 16

P∞ where β(x) is the Dirichlet beta function defined by β(x) = n=0 (−1)n /(2n+ 1)x , we find that 3πζ(3) π 3 ln(2) + . (3.26) I1 + I3 = 64 32 Finally, by partial fractions, we have   x 1 x+y y = − . (1 + xy)(1 + x2 ) 1 + y 2 1 + x2 1 + xy Thus, in view of the symmetry, we have Z 1Z 1 x ln(x) ln(y) dydx I2 = 2 (1 + x2 )(1 + xy) 0 0   Z 1Z 1 1 x+y y = 2 − ln(x) ln(y) dydx 2 1 + x2 1 + xy 0 0 1+y

170

Integrations 1

Z =

Z

1

2 0

Z

0 1Z 1

−2 Z =

0 0 1Z 1

2 0

and so Z

0 1

Z

1

I2 = 0

x+y ln(x) ln(y) dxdy (1 + x2 )(1 + y 2 )

0

y ln(x) ln(y) dydx (1 + y 2 )(1 + xy)

x+y ln(x) ln(y) dxdy − I2 , (1 + x2 )(1 + y 2 ) x+y ln(x) ln(y) dxdy. (1 + x2 )(1 + y 2 )

In view of symmetry again, we have Z 1Z 1 x I2 = 2 ln(x) ln(y) dxdy 2 )(1 + y 2 ) (1 + x 0 0 Z 1 Z 1 x ln(x) ln(y) = 2 dx · dy. 2 1 + x 1 + y2 0 0 Since 1

Z 0

x ln(x) dx 1 + x2

= =

Z 0

1

ln(y) dy 1 + y2

= =

∞ X n=0 ∞ X

k=0 ∞ X

0

1

(ln(x))x2n+1 dx

(−1)n+1 π2 =− ; 2 4(2n + 1) 48 n=0 Z ∞ 2 X (−1)k ln(y)y 2k dy

π2 I2 = 2 · − 48 

Z 0

k=0

we find that

(−1)n

and

(−1)k+1 = −G, (2k + 1)2  · (−G) =

π2 G . 24

(3.27)

Combining (3.25), (3.26) and (3.27) into (3.24), we conclude Z 0

1

(ln(x))2 arctan(x) 3πζ(3) 21πζ(3) π 2 G π 3 ln(2) dx = −I = − − 1+x 8 64 24 32

as claimed. Remark. The key ideas used in the solution are ˆ to introduce f (x) that leads to the double integral (3.25); ˆ to evaluate the double integral I2 via symmetry.



Another integral with log and arctangent

171

We now go back to the double series in (3.23). For the first series, we rewrite it as ! ! n n−1 ∞ ∞ X X X 1 (−1)k 1 X (−1)k S1 := = . (n + 1)3 2k + 1 n3 2k + 1 n=0 n=1 k=0

k=0

By Abel’s summation formula, we have ! n−1 ! n−1 n X X (−1)k X (−1)k 1 S1 (n) := − 3 k 2k + 1 2k + 1 k=1

k=0

k=1

k X 1 3 i i=1

! .

Taking n → ∞ gives ∞

πζ(3) X (−1)k − S1 = 4 2k + 1

k X 1 i3 i=1

∞ X (−1)k I := 2k + 1

!

k=1

To evaluate

k=1

recall that

∞ X

(k)

H3 xk =

k=1

k X 1 3 i i=1

! .

,

Li3 (x) , 1−x

(3.28)

where Lin (x) is the polylogarithm function. Replacing x by −x2 in (3.28) and then integrating yields Z 1 Li3 (−x2 ) dx I = 1 + x2 0 Z ∞ Z ∞ Li3 (−x2 ) Li3 (−x2 ) = dx − dx 1 + x2 1 + x2 0 1 Z ∞ Z 1 Li3 (−x2 ) Li3 (−1/x2 ) = dx − dx. 1 + x2 1 + x2 0 0 This implies that Z ∞ Z 1 Li3 (−x2 ) Li3 (−x2 ) − Li3 (−1/x2 ) 2I = dx + dx. 1 + x2 1 + x2 0 0 By the identity Li3 (−x) − Li3 (−1/x) = −ζ(2) ln x − ln3 x/6, we find Z 2I

∞

= 0

Z = 0

∞

−2ζ(2) ln x − 4 ln3 x/3 dx 1 + x2 0 Z 1 Z Li3 (−x2 ) ln x 4 1 ln3 x dx − 2ζ(2) dx − dx. 2 1 + x2 3 0 1 + x2 0 1+x

Li3 (−x2 ) dx + 1 + x2

Z

1

172

Integrations

R 1 ln2 y For the first integral, in view of Li3 (x) = 21 0 x1−xy dy, exchanging the order of the integration, we have   Z 1 Z Z ∞ 1 ∞ x2 Li3 (−x2 ) 2 ln y − dx = dx dy 1 + x2 2 0 (1 + x2 )(1 + x2 y) 0 0   Z 1 Z 1 2 π 1 ln t = ln2 y − = −2π dt √ 4 y + y 1 +t 0 0 ∞ Z 1 X = −2π (−1)k tk ln2 t dt = −3πζ(3). k=0

Moreover, we have Z 1 0

Z 0

1

0

ln x dx 1 + x2

= −G,

ln3 x dx 1 + x2

=

∞ Z X

1

(−1)k x2k ln3 x dx

0

k=0 ∞ X

= −

k=0

(−1)k 6 = −6β(4), (2k + 1)4

where β(x) is the Dirichlet beta function. In summary, we obtain I = Gζ(2) + 4β(4) −

3πζ(3) . 2

Therefore, we finally find S1 =

7πζ(3) − Gζ(2) − 4β(4). 4

We challenge the reader to establish the following identity for the second series in (3.23): ! ∞ n−1 X X (−1)k 1 7πζ(3) π 3 ln 2 1 = + − β(4). 3 (2n + 1) 2k + 1 128 64 2 n=1 k=0

It is well-known that integrals of products of logarithmic and polylogarithmic functions are associated with Euler sums. For example, since Z 1 (m − 1)! xj−1 (ln x)m−1 dx = (−1)m−1 for j ≥ 1, m ≥ 2, jm 0 we have Z 1 k k X (−1)m−1 1 m−1 1 − x = (ln x) dx. jm (m − 1)! 0 1−x j=1

Another integral with log and arctangent

173

Hence S(m, n)

:=

  ∞ k X X 1  1  k n j=1 j m

k=1

= =

∞ (ln x)m−1 X 1 − xk dx 1−x kn 0 k=1 Z 1 (−1)m−1 (ln x)m−1 Lin (x) ζ(m)ζ(n) − dx, (m − 1)! 0 1−x

(−1)m−1 (m − 1)!

Z

1

where we have used the well-known identity Z 1 (ln x)m−1 dx = (−1)m−1 Γ(m)ζ(m) = (−1)m−1 (m − 1)!ζ(m). 1−x 0 We now end this section to collect some related problems for your practice. In particular, the first four are nice applications of the approach used in the solution. Let G be the Catalan constant. 1. Prove Z 0

1

ln(x) arctan(x) G ln 2 π 3 dx = − . 1+x 2 64

2. Prove x arctan(x) ln(1 − x2 ) π3 π ln2 2 dx = G ln 2 − − . 1 + x2 48 8

1

Z 0

3. Prove Z 1 0

arctan2 (x) ln(1 + x) π 3 ln 2 21πζ(3) 3ζ(2)G dx = + − . 1 + x2 384 256 16

4. Prove Z

1

0

π 3 ln 2 7πζ(3) (ln x)2 arctan(x) dx = − + β(4), x(1 + x2 ) 16 32

where beta function defined by β(x) = P∞ β(x)n is the Dirichlet x (−1) /(2n + 1) . n=0 5. Show that Z 0

1

ln xLi2 (x) 3 dx = − ζ(4). 1−x 4

6. Show that Z 1 n ln xLin+1 (x) (−1)n+1 n! dx = [ζ(2n + 2) − ζ 2 (n + 1)]. 1 − x 2 0

174

Integrations 7. Show that Z 1 ∞ ∞ X ln xLi2 (−x) π2 G (−1)n+1 X (−1)k dx = = . 1 + x2 n2 (2k + n + 1)2 48 0 n=1 k=0

8. Show that

1

Z 0

ln2 xLi3 (x) 9ζ(3)2 dx = ζ(6) − . 1+x 16

In general, show that Z 1 n (−1)n n! ln xLin+1 (x) dx = [ζ(2n + 2) − (1 − 2−n )2 ζ 2 (n + 1)]. 1+x 2 0 9. Show that Z

1

0

Li3 (−x) π 2 G 3πζ(3) dx = − − β(4). 1 + x2 12 128

10. Show that Z 1 n n! ln xLin+1 (−x) dx = (−1)n+1 n+1 (1 − 2−n )ζ(n + 1)β(n + 1). 2 1+x 2 0 11. Show that P∞ (a) n=1 P∞ (b) n=1 P∞ (c) n=1

(−1)n n2

P

n+1

(−1) n2 (−1)n+1 n2

2n−1 (−1)k k=1 k



= πG −

H2n−1 = πG − P  n−1 1 k=0 2k+1

12. Let A(m, n) =

29 16

27 16

ζ(3).

7 4

ζ(3).

ζ(3).

= πG −

∞ (m) X (−1)k+1 Hk . kn

(3.29)

k=1

Show that (ln x)n−1 Lim (−x) dx. x(1+x) 0 1 1 2 = 2 ζ(2) − 2 (ln 2) . = ζ(3) − 12 ln 2ζ(2); A(1, 2) = 58 ζ(3). 3 = 19 16 ζ(4) − 4 ln 2ζ(3). = 2ζ(5) − 78 ln 2ζ(4) − 83 ζ(2)ζ(3); A(1, 4)

(a) A(m, n) =

(−1)n Γ(n)

R1

A(1, 1) A(2, 1) A(3, 1) A(4, 1) 1 2 ζ(2)ζ(3). (f) Let m ≥ 2. Then

(b) (c) (d) (e)

(−1)m A(m, 1) = Γ(m)

Z 0

1

(ln x)m−1 ln 1−x

1+x 2

=


dx

59 32 ζ(5)

−

Another integral with log and arctangent =

175

m−2 m 1 X ζ(m + 1) − ln 2η(m) − η(k + 1)η(m − k), 2 2 k=1

where η(s) is called the Dirichlet eta function which is defined by ∞ X (−1)k+1 = (1 − 21−x )ζ(x). η(x) = kx k=1

(g) If n is even, then A(1, n) =

n/2−1 X 1+n 1 η(1 + n) − ζ(1 + n) − η(2k)ζ(1 + n − 2k). 2 2 k=1

13. Let A(m, n) be defined as (3.29). Let Lin (x) be the polylogarithm function. Show that 7 1 1 2 4 (a) A(1, 3) = 11 4 ζ(4) − 4 ln 2ζ(3) + 2 (ln 2) ζ(2) − 12 (ln 2) − 2Li4 (1/2). 7 1 2 4 (b) A(2, 2) = − 51 16 ζ(4) + 2 ln 2ζ(3) − (ln 2) ζ(2) + 6 (ln 2) + 4Li4 (1/2). (c) A(2, 3) = − 58 ζ(2)ζ(3) − 11 32 ζ(5). (d) Open question: Determine the closed form of A(m, 2n + 1) for m ≥ 1, n ≥ 2.

14. Recall the Witten zeta function (see Section 2.1) W(r, s, t) =

1

X n,m=1

nr ms (n

+ m)t

.

(a) For r, s > 1 and t > 0, show that W(r, s, t) =

1 Γ(t)

Z

1

Lir (x)Lis (x) 0

(− ln x)t−1 dx. x

(b) Show that W(s + 1, s, 1) = ζ 2 (s + 1)/2 for s ≥ 1. (c) Show that W(2, 2, 1) = 2ζ(2)ζ(3) − 3ζ(5). (d) Introduce the multivariate zeta function ζ(r, s) defined by ζ(r, s) :=

X n>m

1 . nr m s

Can you represent W(r, s, t) in terms of the multivariate zeta function values?

176

Integrations

3.14

An orthonormal function sequence

Problem 11850 (Proposed by Z. Ahmed, 122(6), 2015). Let r   1 dn 2 1 . (1 + x2 )n/2 n An (x) = π n! dx 1 + x2 R∞ Prove that −∞ Am (x)An (x) dx = δ(m, n) for nonnegative integers m and n. Here, δ(m, n) = 1 if m = n, and otherwise δ(m, n) = 0. Discussion. In Problem 10777 (107(1), 2000), the current proposer asked to evaluate  n   Z ∞ m  d 1 d 1 I(m, n) := dx. dxm 1 + x2 dxn 1 + x2 0 The published solution [2] is based on integration by parts and the Fourier transform: Z ∞ Z ∞ p 1 −|t| and f (t)g(t) dt = F[f (t)](x)F[g(t)](x) dx. F[ π/2e ](x) = 1 + x2 −∞ −∞ The same argument does not work on the current problem because of the nonconstant weights in An . To evaluate the proposed integral we need to find dn 2 another expression for dx n (1/(1 + x )), and so for An . Solution. The expression for the nth derivative of 1/(1 + x2 ) in our mind is, for n ≥ 0,   dn 1 (−1)n n! sin(n + 1)θ , (3.30) = dxn 1 + x2 (1 + x2 )(n+1)/2 √ 1 where θ = arcsin(1/ 1 + x2 ). Indeed, let f (x) = 1+x 2 . Applying the Leibniz 2 rule to (1 + x )f (x) = 1 yields (x2 + 1)f (n+2) (x) + 2(n + 2)xf (n+1) (x) + (n + 2)(n + 1)f (n) (x) = 0, where f (k) indicates the kth derivative of f (x). Let f (n) (x) = (−1)n n!an . Then (1 + x2 )an+2 − 2xan+1 + an = 0, subject to a0 =

2x 1 , a1 = . 1 + x2 (1 + x2 )2

√ Now let θ = arcsin(1/ 1 + x2 ). Solving this difference equation gives an =

sin(n + 1)θ , (1 + x2 )(n+1)/2

n = 0, 1, 2, . . . .

(3.31)

An orthonormal function sequence

177

Hence, r An = (−1) and so Z ∞ Am (x)An (x) dx = (−1)

n

m+n

−∞

2 sin(n + 1)θ , π (1 + x2 )1/2

2 π

Z

∞

−∞

sin(m + 1)θ sin(n + 1)θ dx. 1 + x2

Appealing to x = tan θ and dx = sec2 θdθ, we have Z ∞ Z π/2 2 Am (x)An (x) dx = (−1)m+n sin(m + 1)θ sin(n + 1)θ dθ. π −π/2 −∞ Finally, since {sin nx} is orthogonal on (−π/2, π/2), i.e.,  Z π/2 0, if m 6= n, sin(m + 1)θ sin(n + 1)θ dθ = π −π/2 2 , if m = n, this implies that Z

∞

Am (x)An (x) dx = δ(m, n). −∞

 Remark. Notice that the above argument leads to another solution to Problem 10777. In fact, let f (x) = 1/(1 + x2 ). By (3.31), for n ≥ 0, we have f (n) (0) = n! cos(nπ/2), f (n) (∞) = 0. To evaluate I(m, n), we consider two cases: Case I. When m + n is even. Without loss of generality, we assume that m > n. Integrating by parts yields Z ∞ I(m, n) = f (m) (x)f (n) (x) dx = −f (m) (0)f (n−1) (0) − I(m + 1, n − 1). 0

Repeatedly using this reduction formula we find Z ∞ Z I(m, n) = (−1)n f (m+n) (x)f (x) dx = (−1)n 0

Since

∞

f (m+n) (x)f (x) dx.

0

1 = 1 + x2

Z

∞

e−t cos(xt) dt,

0

and cos(k) (x) = cos(x + kπ/2), we have Z ∞ f (m+n) (x) = tm+n e−t cos(xt + (m + n)π/2) dt 0 Z ∞ = (−1)(m+n)/2 tm+n e−t cos(xt) dt, 0

178

Integrations

and so I(m, n)

=

(−1)(m−n)/2

Z

∞

0

=

(−1)(m−n)/2

Z

∞

0



∞

 tm+n e−t cos(xt) dt dx 0 Z  ∞ cos(xt) m+n −t t e dx dt. 1 + x2 0 1 1 + x2

Z

The interchange the order of integrations is justified by the dominate convergence theorem. Notice the inner integral is the well-known Laplace integral which is given by Z ∞ cos(xt) π dx = e−t . 2 1+x 2 0 We finally find that (m−n)/2 π

I(m, n) = (−1)

2

Z

∞

tm+n e−2t dt = (−1)(m−n)/2

0

(m + n)!π . 2m+n+2

Case II. When m + n is odd. We assume that m > n again. Integrating by parts (m − n) times (with u = f (n) , v = f (m−1) initially) yields I(m, n) = (−1)m

m−n X

f (m−k) (0)f (n+k−1) (0) − I(m − (m − n), n + (m − n)).

k=1

In view of I(m, n) = I(n, m), we conclude I(m, n) =

m−n (−1)m X (sin((m + n)π/2) 4 k=1

− (−1)k sin((m − n)π/2))(m − k)!(n + k − 1)!. In particular, we have I(2n + 2, 2n + 1) = 0, I(2n + 1, 2n) = −[(2n)!]2 /2. It is also interesting to see that we can establish (3.30) by partial fractions. In fact, since   dn n! 1 = (−1)n , n dx x+a (x + a)n+1 we have dn dxn



1 1 + x2

 = =

  1 dn 1 1 − 2i dxn x − i 1 + i   n! 1 1 (−1)n − . 2i (x − i)n+1 (x + i)n+1

√ Now let θ = arcsin(1/ 1 + x2 ). Then x = cot θ and x − i = cot θ − i =

2ie−iθ 1 −iθ = e . eiθ − e−iθ sin θ

An definite integral Quickie

179

Similarly, we have x + i = cot θ + i =

2eiθ 1 iθ = e . eiθ − e−iθ sin θ

Since sin2 θ = 1/(1 + x2 ), we conclude that    n+1  1 θ i(n+1)θ dn n sin −i(n+1)θ = (−1) n! e − e dxn 1 + x2 2i (−1)n n! sin(n + 1)θ = . (1 + x2 )(n+1)/2

3.15

An definite integral Quickie

Problem 4212 (Proposed by H. F. Sandham, 53(4), 1946). Evaluate Z

2

∞

e−x dx. (x2 + 1/2)2

0

Discussion. In the MAA journal Mathematics Magazine, a problem submitted as a Quickie should have an unexpected, succinct solution. Thus a Quickie provides a double challenge to the reader — to solve the problem and to devise a solution that is clear, concise and surprising. This problem was proposed in 1946. Two published solutions ([38], 1947) were elegant, but not unexpected. Later this problem was solved by using parametric differentiation. Is there any special approach apparently unrelated to the problem provide the quickness sought? A recurrence relation! Solution. We consider one more general variation: For nonnegative integer n, let Z

∞

In := 0

2

e−x dx. (1 + 2x2 )n

Then the proposed integral is 4I2 . Applying integration by parts, we have Z In − In+1

∞

= 0

2

2x2 e−x dx (1 + 2x2 )n+1

Z ∞ 2 2 e−x 4x2 e−x = dx − (n + 1) dx (1 + 2x2 )n+1 (1 + 2x2 )n+2 0 0 Z ∞ 2 [(1 + 2x2 ) − 1]e−x = In+1 − 2(n + 1) dx (1 + 2x2 )n+2 0 = In+1 − 2(n + 1)In+1 + 2(n + 1)In+2 . Z

∞

180

Integrations

This implies that In + 2nIn+1 − 2(n + 1)In+2 = 0. This enables us to evaluate I2 without knowing I1 . Indeed, Z 1 1 ∞ −x2 1√ I2 = I0 = e π. dx = 2 2 0 4 √ Thus the proposed integral equals π.



Remark. Here we list two problems for your practice. R∞ 2 1. Let sinh x · sinh y = 1. Show that 0 y dx = π4 . 2. Problem 11933 (Proposed by J. M. Pacheco Pn and A. Plaza, 123(8), 2016). For positive integer n, let Hn = k=1 1/k. Prove Z 0

1

1 dx· x+1

Z 0

1

x+1 dx · · · 2 x +x+1

Z 0

1

1 xn−2 + · · · + x + 1 dx ≥ . n−1 x + ··· + x + 1 Hn

4 Inequalities

We select 10 inequality problems that have appeared in the Monthly over the past half century. We will encounter many familiar characters, among them: Cauchy’s inequality, L’Hˆ opital’s monotone rule, Majorization and Convexity, Hilbert’s identity, Hardy’s inequality, and differential equations. What makes our trip worthwhile is the realization that these old chestnuts still have something new to tell us. They all appear here in novel and surprising ways and you should add them to your inequality tool box.

4.1

An inequality from Klamkin

Problem E2483 (Proposed by M. S. Klamkin, 81(6), 1974). Let x be nonnegative and let m, n be integers with m ≥ n ≥ 1. Prove that (m + n)(1 + xm ) ≥ 2n

1 − xm+n . 1 − xn

Discussion. For x = 1, the right-hand side of the inequality is understood as a limit where the inequality actually becomes an equality. Also, for x = 0, the inequality becomes m + n ≥ 2n, which is true since m ≥ n. Moreover, notice that when m = n, the inequality becomes equality again, and replacing x by 1/x in the inequality yields the same inequality after multiplying by xm . Thus, it suffices to prove the inequality for x ∈ (0, 1) and m > n. There are many verifications for this inequality. We single out three. The first proof belongs to the proposer. He worked backward — assume the inequality holds then arrive at an equivalent inequality that can be easily proven. The second proof is due to Allen Stenger, then an undergraduate student at Penn State. He transformed the inequality into an area comparison. The last one is based on the following: L’Hˆ opital’s Monotone Rule (LMR): Let f, g : [a, b] → R be continuous functions that are differentiable on (a, b) with g 0 (x) 6= 0 on (a, b). If f 0 /g 0 is

181

182

Inequalities

increasing (decreasing) on (a, b), then the functions f (x) − f (b) g(x) − g(b)

and

f (x) − f (a) g(x) − g(a)

are correspondingly increasing (decreasing) on (a, b). Solution I — by M. S. Klamkin. Based on the above discussion, we can assume that√x ∈ (0, 1) and m > n. Let m + n = r, m − n = s. Then r > s > 0. Let t = x so that 0 < t < 1. The desired inequality becomes r(1 + tr+s ) ≥ (r − s)

1 − t2r . 1 − tr−s

Rearranging this inequality gives 1 − t2r 1 − t2s ≥ . rtr sts Substituting t = e−y yields sinh(sy) sinh(ry) ≥ , ry sy

(4.1)

where now y ∈ (0, ∞). Since ∞ ∞ X sinh x 1 1 1 X = x2n+1 = x2n , x x n=0 (2n + 1)! (2n + 1)! n=0

this implies that sinh x/x is strictly increasing. Hence (4.1) is established, and so is the proposed inequality.  Solution II — by A. Stenger. Assume that x ∈ (0, 1) and m > n. Let y = xn and r = m/n > 1. Then 0 < y < 1 and the desired inequality is equivalent to (r + 1)(1 + y r ) ≥

2(1 − y r+1 ) . 1−y

Rearranging this inequality gives 1 − y r+1 1 (1 − y)(1 + y r ) ≥ . 2 r+1

(4.2)

Consider the function f (x) = xr on [y, 1]. Since f is concave up, the region under y = xr on [y, 1] is contained in the trapezoid with vertices (y, 0), (y, y r ), (1, 1) and (1, 0). The standard area formulas conclude that Z 1 1 1 − y r+1 Area of the trapezoid = (1 − y)(1 + y r ) ≥ xr dx = . 2 r+1 y

An inequality from Klamkin

183

This proves (4.2) and so the proposed inequality.  Solution III — by L’Hˆ opital’s monotone rule (LMR). As above, we assume that x ∈ (0, 1) and m > n. The desired inequality is equivalent to 2n (1 − xn )(1 + xm ) ≥ . (4.3) m+n 1−x m+n Let f (x) = (1 − xn )(1 + xm ), g(x) = 1 − xm+n . Then 1 f 0 (x) −nxn−1 +mxm−1 −(m+n)xm+n−1 = = [nx−m −mx−n +(m+n)]. g 0 (x) −(m + n)xm+n−1 m+n To show that f 0 (x)/g 0 (x) is decreasing for 0 < x < 1, it suffices to prove that nx−m − mx−n + (m + n) is decreasing for 0 < x < 1. This follows from [nx−m − mx−n + (m + n)]0 = −mnx−m−1 + nmx−n−1 = −mnx−m−1 (1 − xm−n ) < 0. Thus, LMR deduces that F (x) :=

f (x) − f (1) f (x) (1 − xn )(1 + xm ) = = g(x) − g(1) g(x) 1 − xm+n

is decreasing for 0 < x < 1. Moreover, applying L’Hˆopital’s rule yields 2n (1 − xn )(1 + xm ) = . m+n x→1 1−x m+n

F (1) = lim− F (x) = lim x→1

Now (4.3) follows from the fact that F (x) ≥ F (1) immediately.



Remark. Consider

(1 + xn )(1 − xm ) . 1 − xm+n Along the same lines in the Solution III, applying the LMR, we see that G(x) is increasing for 0 < x < 1. Thus, G(x) :=

(1 + xn )(1 − xm ) 2m . ≤ lim− G(x) = 1 − xm+n m+n x→1 This, together with (4.3), yields the double inequalities 2n

1 − xm+n 1 − xm+n m n ≤ (m + n)(1 + x ) ≤ (m + n)(1 + x ) ≤ 2m . 1 − xn 1 − xm

Let m+n = r, m−n = s. As we did in the Solution I, Klamkin’s inequality becomes m (m−n)/2 xm − 1 ≥ x . xn − 1 n Replacing x by x/y yields  m 1/(m−n) x − ym n √ · ≥ xy. n n x −y m

184

Inequalities

When n = 1, the left-hand side is often called the Stolarsky mean. It is wellknown that  m 1/(m−1) x−y x − ym 1 √ xy ≤ ≤ · . ln x − ln y x−y m It is interesting to see that LMR provides a powerful method for proving the monotonicity of a large class of quotients. The LMR first appeared in Gromov and Taylor’s work [51] for volume estimation in differential geometry. Since then, the LMR and its variants have been used in approximation theory, quasi-conformal theory and probability. But, for most analysis readers, the LMR is not as well-known as it should be. To popularize and promote this monotonicity rule, we present a proof of the LMR and two more examples within the realm of elementary analysis. The reader can see the wide applicability of the LMR and find more applications in exercises. Proof. We may assume that g 0 (x) > 0 and f 0 (x)/g 0 (x) is increasing for all x ∈ (a, b). By the mean value theorem, for a given x ∈ (a, b) there exists c ∈ (a, x) such that f 0 (c) f 0 (x) f (x) − f (a) = 0 ≤ 0 , g(x) − g(a) g (c) g (x) and so f 0 (x)(g(x) − g(a)) − g 0 (x)(f (x) − f (a)) ≥ 0. Therefore,  0 f (x) − f (a) f 0 (x)(g(x) − g(a)) − g 0 (x)(f (x) − f (a)) = ≥ 0. g(x) − g(a) (g(x) − g(a))2 This shows that (f (x) − f (a))/(g(x) − g(a)) is increasing on (a, b) as desired. Similarly, we can show that (f (x) − f (b))/(g(x) − g(b)) is increasing. So far we have assumed that a and b are finite. This rule can be extended easily to the case where a or b is infinity. Our first example is to confirm the monotonicity of (xp − 1)/(xq − 1) for p > q > 0, x ≥ 1 by using the LMR. To see this, let f (x) = xp − 1 and g(x) = xq − 1. Consider  f (x)/g(x), if x 6= 1, G(x) = p/q, if x = 1. We have f 0 (x)/g 0 (x) = (p/q)xp−q , which is increasing on (1, ∞) as long as p > q > 0. Hence, by the LMR, G(x) is increasing on (1, ∞). In particular, G(x) > G(1) = p/q. This yields the following bonus inequality: xp − 1 xq − 1 > , p q

for x > 1, p > q > 0.

(4.4)

An inequality from Klamkin

185

Our second example comes from Wilker’s inequality, which appeared as Monthly Problem E3306. Wilker asked for a proof that 

sin x x

2 +

tan x > 2, x

0 0, f 00 (x)/g 00 (x) is increasing on (0, π/2). Thus, using the LMR twice and noticing that f (0) = f 0 (0) = g(0) = g 0 (0) = 0, we deduce that f 0 (x)/g 0 (x) and so F (x) = f (x)/g(x) is increasing on (0, π/2). Thus (4.5) follows immediately from the fact that F (x) ≥ F (0). We now end this section with some problems for additional practice: x

1. Notice that f (x) = a x−1 is increasing for a > 1, x > 0 from (4.4). Show that f (x) is log-convex (i.e., ln f (x) is convex), and then prove that  1/(m−n)  m 1/(m−n) xm−n − y m−n x − ym n √ xy ≤ ≤ . · (m − n)(ln x − ln y) xn − y n m 2. Let x ≥ 0, m ≥ k ≥ n ≥ 1 and 2k ≥ m + n. Then (m + k + n)(1 + xm )(1 + xk ) ≥ 3n

1 − xm+k+n . 1 − xn

3. Let x ≥ 0, min(1/x, x) ≤ y ≤ max(1/x, x) and m ≥ k ≥ n ≥ 1. Show that

186

Inequalities m+n k

(a) (m + k + n)(1 + xm )(1 + y k ) ≥ 3n 1−x1−xn y , if 2k ≥ m + n; k+n m

(b) (m + k + n)(1 + xk )(1 + y m ) ≥ 3n 1−x1−xny , if k + n ≥ m. 4. For 0 < x < π/2, prove that
3 (a) sinx x > cos x. 2x x (b) sin x + tan x > 3. Comment. Using t2 ≥ 2t − 1 with t = x/ sin x in the second inequality yields  x 2 x + > 2. sin x tan x 5. Problem 11770 (Proposed by S. P. Andriopoulos, 121(3), 2014). Prove, for real numbers a, b, x, y with a > b > 1 and x > y > 1, that  (x+y)/2   a+b a+b a x − by > ln . x−y 2 2 6. Problem 11009 (Proposed by D. Borwein, J. Borwein and J. Rooin, 100(3), 2003). Let a > b ≥ c > d > 0. Prove that f (x) =

ax − bx cx − dx

is increasing and convex for all x. Comment. One may show that f (x) is increasing based on the LMR. Without loss of generality, assume that d = 1, otherwise divide the ratio by dx . Now, let y = cx , α = ln b/ ln c, β = ln a/ ln c. Then β > α ≥ 1. One can rewrite f (x) as F (y) :=

yβ − yα y−1

y ∈ (0, 1) ∪ (1, ∞).

7. In Wilker’s inequality, prove that
sin x 2 x + tan x x x3 tan x

−2

≥

16 , (0 < x < π/2). π4

The initial published solution by Jean Anglesio is relatively long. Another shorter solution (still about one page long) by David Callan was published half year later. 8. Kober’s inequality claims that cos x ≥ 1 − 2x/π for 0 < x < π/2. Prove the following refinement: 1−

π−2 2 2 2 x+ x(π − 2x) ≤ cos x ≤ 1 − x + 2 x(π − 2x). 2 π π π π

Knuth’s exponential inequality

187

9. For 0 < x < 1, prove that  πx  π 4 x x < tan < . π 1 − x2 2 2 1 − x2 10. (Quarter Circle Lemma) For all m ≥ 1, show that ∞ X n=1

 m 1/2 1 < π. m+n n

Hint. In the quarter circle, consider the area of√a triangle which √ m, n − 1), and is√similar to the triangle with vertices (0, 0), ( √ ( m, n). 11. Problem 11308 (Proposed by O. Furdui, 114(7), 2007). Let n be a positive integer. For 1 ≤ i ≤ n, let xi be a real number in (0, π/2) and ai be a real number in [1, ∞). Prove that 2ai Y ai n  n  Y xi xi + > 2. sin xi tan xi i=1 i=1 12. Problem 11869 (Proposed by G. Stoica, 122(9), 2015). Prove that |y ln y − x ln x| ≤ |y − x|1−1/e for 0 < x < y ≤ 1. Comment. The exponent 1 − 1/e is the best possible.

4.2

Knuth’s exponential inequality

Problem 11369 (Proposed by D. Knuth, 115(6), 2008). Prove that for all real t, and all α ≥ 2, eαt + e−αt − 2 ≤ (et + e−t )α − 2α .

(4.6)

Discussion. It is easy to see that (4.6) holds for t = 0 or α = 2. So we can assume that t 6= 0 and α > 2. By introducing f (t, α) = [(et + e−t )α − 2α ] − (eαt + e−αt − 2), the published solution by Dale [34] established the positivity of f (t, α) based on  Z t sinh x f (t, α) = α (ex + e−x )α − (eαx − e−αx ) dx. cosh x 0

188

Inequalities

We will present two more elementary proofs as follows. First, let x = et and F (x) = (x + x−1 )α − xα − x−α . Then (4.6) becomes F (x) ≥ F (1). Since F (x) = F (x−1 ), it suffices to show that F is decreasing for 0 < x < 1. The second proof is based on the LMR. We will see that both proofs are almost algorithmic in nature, using only Bernoulli’s inequality. Solution I. We show that F 0 (x) ≤ 0 for 0 < x < 1, which implies that F (x) is decreasing for 0 < x < 1. Since F 0 (x) = α(x + x−1 )α−1 (1 − x−2 ) − αxα−1 + αx−α−1 , F 0 (x) ≤ 0 is evidently equivalent to (1 + x2 )α−1 ≥

1 − x2α . 1 − x2

Applying Bernoulli’s inequality, (1 + t)p ≥ 1 + pt

for t ≥ −1, p ≥ 1,

we have (1 + x2 )α−1 ≥ 1 + (α − 1)x2 . Thus, 1 − x2α 1 + (α − 1)x ≥ =1+ 1 − x2 2



1 − x2(α−1) 1 − x2



x2

as long as α−1≥

1 − x2(α−1) 1 − x2

or x2(α−1) ≥ 1 + (α − 1)(x2 − 1). This follows from Bernoulli’s inequality again: x2(α−1) = (1 + (x2 − 1))α−1 ≥ 1 + (α − 1)(x2 − 1).  Solution II. We show that (4.6) holds for all t > 0. Let x = et . Then x > 1 and (4.6) is equivalent to (x2 + 1)α − x2α − 1 ≥ 2α − 2. xα To apply the LMR, let f (x) = (x2 + 1)α − x2α − 1 and g(x) = xα . Then f 0 (x) 2x(x2 + 1)α−1 − 2x2α−1 = . g 0 (x) xα−1

Knuth’s exponential inequality Define

189

 α−1 f 0 (x) 1 = 2x x + − 2xα . g 0 (x) x

G(x) =

To show that G(x) is increasing for x > 1, it suffices to prove that G0 (x) ≥ 0 for x > 1. Indeed, α−1    2(α − 1) 1 α− 2 − 2αxα−1 . G0 (x) = 2 x + x x +1 Using Bernoulli’s inequality, we have 

1 x+ x

α−1 =x

α−1



1 1+ 2 x

α−1

≥ xα−1 + (α − 1)xα−3 .

Therefore, for x > 1 and α ≥ 2, G0 (x) ≥

2(α − 1)(α − 2)xα−3 (x2 − 1) ≥ 0. x2 + 1

Now, the LMR deduces that f (x) − f (1) (x2 + 1)α − x2α − 2α + 1 = g(x) − g(1) xα − 1 is increasing for x > 1. On the other hand, L’Hˆopital’s rule implies that lim

x→1

f (x) − f (1) (x2 + 1)α − x2α − 2α + 1 = lim = 2α − 2. x→1 g(x) − g(1) xα − 1

Thus, (x2 + 1)α − x2α − 2α + 1 f (x) − f (1) = ≥ 2α − 2. g(x) − g(1) xα − 1 This proves (4.6) as desired.



Remark. When we search for a general context for (4.6), we find a path connected to two important principles in the theory of inequalities — Majorization and Schur convexity. Since Majorization and Schur convexity are not as well-known as they should be, we will take a few paragraphs to explain them. We say an n-tuple x = (x1 , x2 , · · · , xn ) is majorized by another n-tuple y = (y1 , y2 , · · · , yn ) (written as x ≺ y) if M1 x1 ≥ x2 ≥ · · · ≥ xn and y1 ≥ y2 ≥ · · · ≥ yn , M2 x1 + x2 + · · · + xk ≤ y1 + y2 + · · · + yk for all 1 ≤ k ≤ n − 1, M3 x1 + x2 + · · · + xn = y1 + y2 + · · · + yn .

190

Inequalities

For example, we have (1, 1, 1, 1) ≺ (2, 1, 1, 0) ≺ (3, 1, 0, 0) ≺ (4, 0, 0, 0). ¯ = (x1 + Moreover, for any x = (x1 , x2 , · · · , xn ), with the arithmetic mean x x2 + · · · + xn )/n, ¯, · · · , x ¯ ) ≺ (x1 , x2 , · · · , xn ) ≺ (x1 + x2 + · · · + xn , 0, · · · , 0). (¯ x, x

(4.7)

Next, if D ⊂ Rn and f : D → R, we say that f is Schur convex (concave) on D if f (x) ≤ (≥) f (y) for all x, y ∈ D and x ≺ y. Schur provided the following criterion to check whether a function is Schur convex or concave: Schur’s Criterion: Let f : (a, b)n → R be continuously differentiable and symmetric. Then f (x) is Schur convex (concave) on (a, b)n if and only if   ∂f ∂f (xj − xk ) − ≥ (≤) 0 ∂xj ∂xk for all 1 ≤ j < k ≤ n and all x ∈ (a, b)n . In particular, for xj > 0 for 1 ≤ j ≤ n, we see that f (x) = x1 x2 · · · xn is Schur concave because   ∂f ∂f f (x) (xj − xk ) − = −(xj − xk )2 ≤ 0. ∂xj ∂xk xj xk In view of (4.7), the Schur concavity of f yields  n x1 + x2 + · · · + xn n ¯, · · · , x ¯) = x ¯ = f (¯ x, x ≥ f (x) = x1 x2 · · · xn . n This is the well-known AM-GM inequality, which is a consequence of Jensen’s inequality. In general, Schur provided the following powerful theorem that shows that almost every invocation of Jensen’s inequality can be replaced by the Schur convexity. Schur’s Majorization Inequality. Let φ : (a, b) → R be a convex function. Then the function f : (a, b)n → R defined by f (x1 , x2 , · · · , xn ) =

n X

φ(xj )

j=1

is Schur convex. Thus, for x, y ∈ (a, b)n with x ≺ y, we have n X j=1

φ(xj ) ≤

n X j=1

φ(yj ).

(4.8)

Knuth’s exponential inequality

191

We now present a few concrete examples to illustrate how majorization and Schur convexity are used in practice. First, let f be convex on (0, ∞) and xj ∈ (0, ∞). Applying (4.8) to (4.7) yields  nf

x1 + x2 + · · · + xn n

 ≤

n X

f (xj ) ≤ f (x1 + x2 + · · · + xn ) + (n − 1)f (0).

j=1

The left-hand side inequality indicates that Jensen’s inequality is a special case of (4.8). Next, we consider Problem 11139 (Proposed by G. Bennett, 112(3), 2005). Show that if p > 1 or p < 0, then 1p + 3p 1p + 3p + 5p 1p + 3p + · · · + (2n − 1)p 1p < < < · · · < < ··· . 3p 5p + 7p 7p + 9p + 11p (2n + 1)p + · · · + (4n − 1)p (4.9) Here we see that the first inequality in (4.9), rephrased as 5p + 7p ≤ 3p + 9p , follows at once from the facts that xp is convex on (0, ∞) for p > 1 or p < 0 and (7, 5) ≺ (9, 3). The second inequality, after some simplifications, becomes 9p + 11p + 27p + 33p ≤ 5p + 15p + 25p + 35p , which follows from the majorization (33, 27, 11, 9) ≺ (35, 25, 15, 5). Along the same lines, we can establish the entire (4.9). Finally, we return to Knuth’s inequality (4.6). Rewrite (4.6) as (et )α + (e−t )α + 2α ≤ (et + e−t )α + 1α + 1α . Naturally, we have φ(x) = xα , which is convex for x > 0 and α > 1, and {x1 , x2 , x3 } = {et , e−t , 2},

{y1 , y2 , y3 } = {et + e−t , 1, 1}.

(4.10)

Unfortunately, here x ≺ y is not valid. Failure occurs at M2 for k = 2. Hardy, Littlewood, and P´ olya [54] proved that (4.8) is valid for all convex function φ : (0, ∞) → R if and only if hypotheses M1–M3 hold. Thus, it is reasonable to consider new forms of majorization by replacing the hypotheses M1–M3, wherein the inequality (4.8) is assumed to hold for classes of functions, other than for all convex ones. Recently, Bennett [11] characterized the p-power functions by replacing M1–M3 with B1 x1 + x2 + x3 = y1 + y2 + y3 , B2 x21 + x22 + x23 = y12 + y22 + y32 ,

192

Inequalities

B3 max{x1 , x2 , x3 } ≤ max{y1 , y2 , y3 }, and proved that (4.8) is valid for φ(x) = xp (p ≥ 2 or 0 ≤ p ≤ 1) if and only if the above hypotheses B1–B3 hold. This establishes (4.6) immediately because (4.10) satisfy the hypotheses B1–B3. It is interesting to see that his proof for sufficiency is inventoried in his following earlier proposed problem. Monthly Problem 11397 (115(10), 2008). Let a, b, c, x, y, z be positive numbers such that a + b + c = x + y + z and abc = xyz. Show that if max{x, y, z} ≥ max{a, b, c}, then min{x, y, z} ≥ min{a, b, c}. Without the regular majorization, this fact ensures no loss of generality in assuming that x > a ≥ b > y ≥ z > c in his proof. We end this section with a few problems for additional practice. 1. Let α ≥ 1. (a) If x ∈ (0, ∞), show that (1 + x)α ≥ xα + 2α − 1. (b) If a, b ∈ (0, ∞), show that aα + bα ≤ (a + b)α − (2α − 2) min(aα , bα ). 2. SIAM Problem 68-1, A Network Inequality (Proposed by J. C. Turner and V. Conway). If p + q = 1, 0 < p < 1, and m, n > 1 are positive integers, give a direct analytic proof of the inequality (1 − pm )n + (1 − q n )m > 1. 3. (Due to G. Bennett) Let a, b, c, x, y, z be positive numbers. Then the inequality ap + bp + cp ≤ xp + y p + z p is valid whenever p ≥ 1, and it reverses direction whenever p < 1, if and only if the following three conditions are satisfied: (a) a + b + c = x + y + z, (b) abc = xyz, (c) max{a, b, c} ≤ max{x, y, z}. 4. (Ratio Principle) Let an and bn both be positive sequences. Show that if the sequence an /bn is increasing then same goes for the ratio of the sum: a1 + a2 + · · · + an . b1 + b2 + · · · + bn 5. Problem 11139 Revisited. The inequality (4.9), tipped upside down and with 1 added, can be rewritten as 1p + 3 p 1p + 3p + 5p + 7p 1p + 3p + 5p + 7p + 9p + 11p ≥ ≥ ≥ ··· . 1p 1p + 3p 1p + 3p + 5p (4.11)

Tight bounds for the normal distribution

193

Using the majorizations shows 1p + 3 p 5p + 7p 9p + 11p ≥ ≥ ≥ ··· . p p 1 3 5p Use this to prove (4.11). 6. Problem 12013 (Proposed by D. Stoner, 124(10), 2017). Suppose that a, b, c, d, e, and f are nonnegative real numbers that satisfy a + b + c = d + e + f . Let t be a real number greater than 1. Prove that at least one of the inequalities at + bt + ct > dt + et + f t , (ab)t + (bc)t + (ca)t > (de)t + (ef )t + (f d)t , and (abc)t > (def )t is false. 7. Problem 12065 (Proposed by H. Lee, 125(8), 2018). Let n be a positive integer, and let x1 , . . . , xn be a list of n positive real numbers. For k ∈ {1, . . . , n}, let yk = xk (n + 1)/(n + 1 − k) and let 1/k

zk =



1/k

k Y

(k!)  yj  k+1 j=1

.

Prove that the arithmetic mean of x1 , . . . , xn is greater than or equal to the arithmetic mean of z1 , . . . , zn , and determine when equality holds.

4.3

Tight bounds for the normal distribution

Problem 10611 (Proposed by Z. Sasv´ari, 104(5), 1997). Find the largest value of a and the smallest value of b for which the inequality √ √ 1 + 1 − e−ax2 1 + 1 − e−bx2 < Φ(x) < 2 2 holds for all x > 0, where 1 Φ(x) = √ 2π

Z

x

−∞

e−y

2

/2

dy.

194

Inequalities

Discussion. Since Z

0

e

−y 2 /2

Z dy =

∞

e

−∞

−y 2 /2

r dy =

0

π , 2

the stated inequalities are equivalent to √ √ 1 − e−ax2 1 − e−bx2 < f (x) < , 2 2 where Z x 2 1 e−y /2 dy. f (x) = √ 2π 0

(4.12)

Solution. First, we show that a = 1/2 and b = 2/π are the best possible constants for which (4.12) holds. Indeed, if the right-hand inequality of (4.12) holds for all x > 0, then ! √ √ 1 − e−bx2 b 1 0< − f (x) = −√ x + O(x3 ) 2 2 2π as x → 0, which implies that b ≥ 2/π. Similarly, notice that, for large x, Z ∞  Z ∞ 2 2 1 f (x) = √ e−y /2 dy − e−y /2 dy 2π x Z0 ∞  Z ∞ 1 1  −y2 /2  −y 2 /2 d e = √ e dy + y 2π 0 x ! 2 1 e−x /2 1 −x2 /2 −√ e +O . = 2 x3 2πx If the left-hand inequality of (4.12) holds for all x > 0, then √ 2 2 2 1 − e−ax2 1 1 0 < f (x) − e−x /2 + O = e−ax + O(e−2ax ) − √ 2 4 2πx

2

e−x /2 x3

!

2

as x → ∞. Dividing each side by ex /2 yields a ≤ 1/2. Next, we show that (4.12) hold for all x > 0 when a = 1/2 and b = 2/π. To this end, we transform the single integral into a double integral: Z xZ x 2 2 1 f 2 (x) = e−(u +v )/2 dudv. 2π 0 0 Let D = [0, x]2 , D1 = {(u, v) : 0 ≤ u, 0 ≤ v, u2 + v 2 ≤ x2 }, and D2 = {(u, v) : 0 ≤ u, 0 ≤ v, u2 + v 2 ≤ (4/π)x2 }. Then we have the inequalities ZZ ZZ 2 2 2 2 1 1 e−(u +v )/2 dudv < e−(u +v )/2 dudv 2π 2π D1 D ZZ 2 2 1 ≤ e−(u +v )/2 dudv, 2π D2

Tight bounds for the normal distribution

195

where the left-hand inequality holds because D1 ⊂ D; the right-hand because D and D2 have the same area and 2

+v 2 )/2

≤ e−(2/π)x ,

2

+v 2 )/2

≥ e−(2/π)x ,

e−(u e−(u

2

for (u, v) ∈ D − D2 ;

2

for (u, v) ∈ D2 − D.

Evaluating the above outer integrals via polar coordinates, we obtain 1 − e−x 2

2

2

/2

< f 2 (x)
0, p 6= 1). Γ(1 + 1/p) 0

196

Inequalities

Here the coefficient Γ(1 + 1/p) comes from the substitution t = y p and Z ∞ Z p 1 ∞ 1/p−1 −t 1 e−y dy = t e dt = Γ(1/p) = Γ(1 + 1/p). p p 0 0 We now collect some inequalities related to Mills’s ratio and F (x), and leave the proofs to the reader. 1. (Komatu’s inequality) For x > 0 and R(x) defined by (4.13), show that 2 2 √ < R(x) < √ . x2 + 4 + x x2 + 2 + x √ Also show that the upper bound can be replaced by 4/( x2 + 8 + 3x). 2. For x > 0, can one find the best possible constants a and b such that 2 2 √ < R(x) < √ ? 2 2 x +a+x x +b+x q√ 1 3. Show that if x > ( 5 − 1)/2, then R(x) < √1+x ; and R(x) > 2 √ 2 1+x 2. This indicates that x(2+x2 ) if x > R(x) ' √

1 , 1 + x2

as x → ∞.

4. (Approximation of Mills’ ratio by rational functions, due to O. Kouba). Define two sequences Pn (x), Qn (x) as (P0 (x), P1 (x)) = (1, x), Pn+1 (x) = xPn (x) + nPn−1 (x); (Q0 (x), Q1 (x)) = (0, 1), Qn+1 (x) = xQn (x) + nQn−1 (x). Show that 2

2

(a) For all n ≥ 0, Pn (x) = e−x /2 (ex /2 )(n) , where f (n) indicates the nth derivative of f . (b) For all n ≥ 0,   Z ∞ 1 (t − x)2 Pn (x) = √ tn exp − dt. 2 2π −∞ (c) For all n ≥ 0, x > 0, Q2n (x) Q2n+1 (x) < R(x) < . P2n (x) P2n+1 (x) Comment.√ Let Hn (x) be √ the nth Hermite polynomial. Then Pn (x) = −i √ H (ix/ 2) with i = −1. 2 n

An inequality due to Knopp

197

5. (An extension of (4.12), due to H. Alzer [5]). If p > 1, for all x > 0, prove that 

p

1 − e−ax

1/p

  p 1/p < F (x) < 1 − e−bx ,

where a = 1, b = [Γ(1 + 1/p)]−p . If 0 < p < 1, the above inequalities hold with the values of a and b switched. Comment. Alzer’s proof involves a very clever analysis of the function Z x p p I(p, x) = e−y dy − Γ(1 + 1/p)[1 − e−x ]1/p . 0

While reading through his proof, one may find that the monotone arguments acquire new layers of meaning. 6. Show that  ln

1 1 − e−ax



Z

∞

≤ x

e−y dy ≤ ln y



1 1 − e−bx



are valid for all x > 0 if and only if a ≥ eγ and 0 < b ≤ 1, where γ is Euler’s constant. 7. Problem 11219 (Proposed by R. A. Strubel, 113(4), 2006). Prove that when n is a positive integer and s is a real number greater than 1 s ∞  X n 1 + n(ζ(s) − 1) < ≤ nζ(s). n+k k=0

Hint: Use 

4.4

n n+k

s =

1 Γ(s)

Z

∞

ts−1 e−t exp(−kt/n) dt.

0

An inequality due to Knopp

Problem 11145 (Proposed by J. Zinn, 112(4), 2005). Find the least c such that if n ≥ 1 and a1 , . . . , an > 0 then n X k=1

Discussion.

k Pk

j=1

1/aj

≤c

n X k=1

ak .

(4.14)

198

Inequalities

We notice that the summand in the left-hand side of (4.14) is the harmonic mean of a1 , . . . , ak . Here (4.14) indicates that the sum of the harmonic means is bounded by the sum of the original sequence. Naturally, we apply the AMHM inequality to the summands on the left-hand side of (4.14) and find that n X k=1

k Pk

j=1

1/aj

≤

k n n n X X X 1 X 1 aj = . aj k j=1 k j=1

k=1

k=j

Here we fall short of our goal because the upper bound diverges. To overcome this failure, we rewrite the Cauchy-Schwarz inequality as    k k X X 1  (b1 + b2 + · · · + bk )2 ≤  b2j aj  , (4.15) a j=1 j j=1 and deduce that n X k=1

k Pk

j=1

1/aj

 n X k  ≤  P

k j=1

  k X  b2j aj  2 

j=1 bj   n n X k X  = aj b2j  P 2  , k j=1 k=j j=1 bj k=1

where bj , for j = 1, 2, . . . , k, can be viewed as parameters. Thus, to prove (4.14), we just need to choose bj , j = 1, 2, . . . , n such that   n X k   b2j  P 2  k k=j j=1 bj is bounded. Next, we test the bound c by letting ak = 1/k. When we substitute this sequence into (4.14), we see that it implies n X k=1

n X 1 2 ≤c k+1 k

for all n ≥ 1.

k=1

Since the harmonic series diverges, for the bound in (4.14) to hold in general, we must have c ≥ 2. Solution. We show that c = 2 is the best possible bound in (4.14). Repeating our foregoing calculation in the discussion, applying (4.15) with bk = k, we have  
1 1 1 + + ··· + 12 a1 + 22 a2 + · · · + k 2 ak . (1 + 2 + · · · + k)2 ≤ a1 a2 ak

An inequality due to Knopp

199

Since 1 + 2 + · · · k = k(k + 1)/2, the above inequality is equivalent to

1 a1

1 a2

+

k + ··· +

1 ak

≤

k X 4k j 2 aj . k 2 (k + 1)2 j=1

Summing over k gives n X 1 k=1 a1

+

1 a2

k + ··· +

≤

1 ak

n X k=1

k n n X X X 4k 2k 2 2 . j a = 2 j a j j 2 (k + 1)2 k 2 (k + 1)2 j=1 k j=1 k=j

Using partial fractions, n X k=j

n

X 2k + 1 = 2 2 k (k + 1) k=j



1 1 − k2 (k + 1)2

 =

1 1 − , j2 (n + 1)2

so we find that n X k=1

k

≤2

Pk

1 i=1 ai

n X j=1



 n n X X 1 1 1 ≤2 j 2 aj  2 − aj . − j (n + 1)2 k 2 (k + 1)2 j=1 k=j

This completes the proof of (4.14).



Remark. We see that (4.15) has played a key role in the above proof. Indeed, the flexibility of the parameters in (4.15) turns out to have a remarkable number of variations. Here, we single out one particularly charming case. With aj = j 2 for 1 ≤ j ≤ n, we get      n n n 2 X X X 1 π   (b1 + b2 + · · · + bn )2 ≤  j 2 b2j  ≤ j 2 b2j  . 2 j 6 j=1 j=1 j=1 Furthermore, with aj = t + j 2 /t for 1 ≤ j ≤ n, let S :=

n X

b2j , and T :=

j=1

n X

j 2 b2j .

j=1

(4.15) leads to 

2

(b1 + b2 + · · · + bn ) ≤ Cn where Cn =

n X j=1

2

2

T tS + t

 ,

n

X 1 t = . 2 t + j /t j=1 t2 + j 2

Since t/(t +x ) is decreasing in x ∈ (0, ∞) for all t > 0, invoking the right-end Riemann sum yields Z n Z ∞ t t π Cn ≤ dx ≤ dx = . 2 + x2 2 + x2 t t 2 0 0

200

Inequalities

Thus,   π T (b1 + b2 + · · · + bn )2 ≤ tS + , for all t > 0. 2 t p In particular, setting t = T /S yields √ (b1 + b2 + · · · + bn )2 ≤ π ST , or (b1 + b2 + · · · + bn )4 ≤ π 2 (b21 + b22 + · · · + b2n )(b21 + 22 b22 + · · · + nb2n ). (4.16) This is known as Carlson’s inequality. Based on the Monthly Editorial comment, the inequality (4.14) with c = 2 is actually a special case of the following inequality: For p > 0, !p  p X ∞ ∞ X p+1 k apk . ≤ Pk p 1/a j j=1 k=1

k=1

This inequality was first established by Knopp [59]. With c = 4, (4.14) appeared as the Putnam Problem 1964-A5. It is interesting to see the historic evolution of several famous mean inequalities during the early twentieth century. The tale begins with David Hilbert in 1906. In his research on integral equations, Hilbert was led to determine the convergence of some double series. For P this purpose, he established P∞ ∞ Hilbert’s inequality. If m=1 a2m < ∞ and n=1 b2n < ∞, then !1/2 ∞ !1/2 ∞ ∞ X ∞ X X X am bn ≤C a2m b2n (4.17) m + n m=1 n=1 n=1 m=1 with the bound C = 2π. Hilbert’s original proof was based on trigonometric integrals including the following remarkable formula:  N  N X X 1 1 + am bn n+m n−m n=1 m=1 !2 Z π N X 1 k = t (−1) (ak sin kt − bk cos kt) dt. (4.18) 2π −π k=1

Here, when n = m, 1/(n − m) is treated to be zero. Five years later, Schur provided a new proof that exploited the theory of analytic functions, and showed that (4.17), as well as its integral analogue, actually holds with the best possible bound C = π. But none of these proofs satisfied with Hardy’s desire: Simple and elementary. He aimed to derive (4.17) from the CauchySchwarz inequality only. Indeed, Hardy (1915) proved that the convergence of any of the three series 2 ∞  ∞ X ∞ ∞ X X X An am an an An (2) (3) (1) n n m +n n=1 n=1 m=1 n=1

An inequality due to Knopp

201

Pn implies that of the others, where An = k=1 ak . In 1920, he finally established Hardy’s inequality. If a1 , a2 , . . . is a sequence of nonnegative real numbers, then for p > 1, !p  Pk p X ∞ ∞ X p j=1 aj ≤ apk . (4.19) k p−1 k=1

k=1

Two years later, Carleman revealed the geometric mean case: Carleman’s inequality. If a1 , a2 , . . . is a sequence of positive real numbers, then ∞ ∞ X X √ k a1 a2 · · · ak ≤ e ak . (4.20) k=1

k=1

The proofs of (4.17), (4.19), and (4.20) perfectly illustrate how to introduce parameters to squeeze the bounds, and have served as a benchmark for new ideas and methods. For example, P´olya’s elegant proof of (4.20) used little more than the AM-GM inequality, but demonstrated where to use it most effectively. These inequalities have been generalized and applied in analysis and differential equations. For the reader interested in learning more about these inequalities, please refer to Michael Steele’s fascinating book [88]. With the Cauchy-Schwarz inequality as the initial guide, this lively, problem-oriented book will coach one toward mastery of more classical inequalities, including those of H¨ older, and (4.14)–(4.20). We now end this section with some problems for additional practice. 1. Let ak ≥ 0 for 1 ≤ k ≤ n. Show that √ n a1 a2 · · · an ≤

2 n(n − 1)

X

√

ai aj .

1≤i 0 for all k ≥ 1. If k=1 αk apk < ∞, prove that !  Pk p X ∞ ∞ X p j=1 αj aj αk apk . ≤ αk Pk p − 1 α j=1 j k=1

k=1

1/p

Comment. Replacing ak by ak and letting p → ∞ in the above inequality, since ! Pk 1/p p Pk j=1 αj aj αk 1/ j=1 αj α1 α2 lim a · · · a ) , = (a P 1 2 k k p→∞ j=1 αj P∞ this yields the Weighted Carleman’s inequality: If k=1 αk ak < ∞, then ∞ ∞ Pk X X αk 1/ j=1 αj 1 α2 αk (aα a · · · a ) ≤ e αk ak . 1 2 k k=1

k=1

9. Let a1 , a2 , . . . be a positive real number sequence. Show that √ k

a1 a2 · · · ak ≤

k e X (2j − 1)aj . 2k 2 j=1

Use this term-wise bound to prove (4.20).

An inequality due to Knopp

203

10. (A refinement of Carleman’s inequality, due to Chen [31]) Let a1 , a2 , . . . be a positive real number sequence. Show that   ∞ ∞ n X X X √ b j k 1 −  ak , a1 a2 · · · ak ≤ e j (1 + k) j=1 k=1

k=1

where n is any positive integer and bj is given by n

b1 =

1 1 1 X bi , bn+1 = − . 2 (n + 1)(n + 2) n + 1 i=1 n − i + 2

11. Problem 6663 (Proposed by W. Janous, 98(6), 1991). Show that 2 N  X 1 + x + x2 + · · · + xj−1 j

j=1

< (4 ln 2)(1+x2 +x4 +· · ·+x2N −2 )

for 0 < x < 1 and all positive integers N ; also show that the constant 4 ln 2 is the least possible. Comment. Notice that if we drop the factor ln 2 in the bound, the stated inequality is a direct application of (4.19) with p = 2. Here the bound 4 in (4.19) is the best possible in general, since ln 2 = 0.693 . . ., this problem provides us a special case in which the bound 4 can be improved. 12. Problem 11202 (Proposed by G. Bennett, 113(2), 2006). Prove P∞ that if an is a sequence of positive numbers with n=1 an < ∞, then for all p ∈ (0, 1) lim n1−1/p (ap1 + ap2 + · · · + apn )1/p = 0.

n→∞

13. Problem 12004 (Proposed by M. Omarjee, 124(8), 2017). Let a1 , a2 , . . . be a strictly increasing sequence of realPnumbers satisfying ∞ an ≤ n2 ln n for all n ≥ 1. Prove that the series n=1 1/(an+1 − an ) diverges. 1/p

14. (Independent Study) Replacing aj by aj and p by 1/p in (4.19), we get ! Pk  1/p X p 1/p ∞ ∞ X 1 j=1 aj ≤ ak k 1−p k=1

k=1

for 0 < p < 1. The summand on the left-hand side is often called pth power mean or H¨ older mean of a1 , a2 , . . . , ak . Thus, this inequality, Hardy’s inequality, and Carleman’s inequality are unified as ∞ X k=1

M (a1 , a2 , . . . , ak ) ≤ C

∞ X k=1

ak ,

(4.22)

204

Inequalities where M is a mean and C is some finite positive constant. Can you offer a characterization of mean M (for example, in the class of symmetric, increasing, Jensen concave) and also a formula for the best possible constant C satisfying (4.22)?

4.5

A discrete inequality by integral

Problem 11680 (Proposed by B. Bogosel and C. Lupu, 119(10), 2012). Let x1 , . . . , xn be nonnegative real numbers. Show that n X xi i i=1

!4 ≤ 2π 2

n n X xi xj X xi xj . i + j i,j=1 (i + j)3 i,j=1

Discussion. Inspired by Hilbert’s identity (4.18), we represent the terms in the proposed inequality as integrals and expect that they can be estimated efficiently. To illustrate how this approach is used in practice, we consider Problem E1682 (Proposed by V. R. Rao, 71(3), 1964), which asked to prove the following reverse Hilbert-type inequality: n X ai i i=1

!2 ≤

n X

ai aj . i + j−1 i,j=1

R1 Rewrite ai /i = 0 ai xi−1 dx. The Cauchy-Schwarz inequality enables us to prove this inequality in one line: n X ai i i=1

!2

Z = 0

n 1X i=1

!2 ai x

i−1

dx

Z ≤ 0

1

n X i=1

!2 ai x

i−1

dx =

n X

ai aj . i + j−1 i,j=1

To apply this approach to the proposed inequality, we need to choose a function f that represents the three sums in the inequality as integrals. The search for a suitable function of f can Pn take various paths, but (4.18) offers the insight and suggests that f (x) = i=1 xi e−ix . Finally, we need to estimate the corresponding integral inequality. Solution.

A discrete inequality by integral

205

Following the foregoing discussion, let f (x) = ∞

Z

f (x) dx

=

0

Z

∞ 2

f (x) dx

=

0

Z

∞

x2 f 2 (x) dx

0

=

n X

Z

∞

e−ix dx =

xi

i=1 n X i,j=1 n X

Pn

0

i=1

n X xi i=1

∞

Z xi xj

e

xi e−ix . We compute

−(i+j)x

0

i

,

n X xi xj , dx = i +j i,j=1

∞

Z

x2 e−(i+j)x dx = 2

xi xj 0

i,j=1

n X

xi xj . (i + j)3 i,j=1

Apply the integral version of Carlson’s inequality: If f ≥ 0 on [0, ∞) such that f, xf ∈ L2 ([0, ∞)), then ∞

Z

4 Z 2 f (x) dx ≤ π

0

∞ 2

Z

f (x) dx

0

∞

x2 f 2 (x) dx,

0

which matches the proposed inequality exactly.



Remark. The above solution offered a powerful example: Once a quantity is represented as an integral, it may be reshaped into alternative forms, and so it can be estimated more efficiently. We elaborate on this idea via two more examples. Example 1. We want to find a simple bound for the nth derivative of sin x/x. Observe that   d4 sin x sin x 2 cos x 12 sin x 24 cos x 25 sin x = + − − + . 4 dx x x x2 x3 x4 x5 It is hard to see that this expression will be bounded by 1/5. However, using sin x = x

Z

1

cos(xt) dt, 0

for all n ∈ N, we have n   Z 1 Z 1  d sin x nπ  1 n ≤ cos xt + dt ≤ tn dt = . t dxn x 2 n + 1 0 0 Example 2. Problem 11769 (Proposed by P. P. D´alyay, 121(4), 2014). Let ai , bi (1 ≤ i ≤ n) be positive real numbers. Show that  2  1/2 n n n n X X X X √ ai  aj ak aj ak aj ak   −2 ≤2 2 . 2 b (b + b ) (b + b ) (b + bk ) 3 j j k j k j j=1 j,k=1

j,k=1

j,k=1

In the original statement of this problem, it was missing the

√

2. Here we quote

206

Inequalities

a brilliant solution due to Roberto Tauraso (http://www.mat.uniroma2.it/ ~tauraso/AMM/AMM11769.pdf), who used the same idea of proving Carlson’s equality (4.16). Define f : (0, ∞) → (0, ∞) by f (x) =

n X

aj e−bj x .

j=1

Then, for m ∈ {0, 1, 2}, Z

∞

C :=

f (x) dx = 0

Dm

1 := m!

Z

n X ai , b j=1 j

∞

xm f 2 (x) dx =

0

n 1 X aj ak . m! (bj + bk )m+1 j,k=1

By the Cauchy-Schwarz inequality, for all t > 0 Z ∞ 2 2 Z ∞ 1 (x + t)f (x) dx f (x) dx = x+t 0 Z0 ∞  Z ∞  dx 2 2 ≤ (x + t) f (x) dx (x + t)2 Z Z 0∞ Z 0∞ 1 ∞ 2 2 2 2 x f (x) dx. = t f (x) dx+2 xf (x) dx + t 0 0 0 It follows that for any t > 0, C 2 − 2D1 ≤ tD0 +

2 D2 . t

In particular, letting t = (2D2 /D0 )1/2 , which gives the minimum value over all t of tD0 + 2D2 /t, yields √ C 2 − 2D1 ≤ 2 2(D0 D2 )1/2 , which is equivalent to the proposed inequality. We now end this section with a few problems for additional practice. 1. Let n be a positive integer. Use induction to prove that, for x > 0, Z 1 √ √ (−1)n dn (1 − t)n−1 t−1/2 cos( xt ) dt. cos( x ) = 2n dxn 2 (n − 1)! 0 Then show that

n d √ ≤ n! . cos( x) dxn (2n)!

2. Problem 11746 (Proposed by P. P. Dalyay, 120(10), 2013). Let f be a continuous function from [0, ∞) to RR such that the following R∞ ∞ integrals converge: S = 0 f 2 (x) dx, T = 0 x2 f 2 (x) dx, and U =



An inequality by power series 207 √ R∞ 4 2 x f (x) dx. Let V = T + T 2 + 3SU . Given that f is not 0 identically 0, show that 4 Z ∞ π 2 S(T + V )2 |f (x)| dx ≤ . 9V 0 3. Problem 11819 (Proposed by C. Lupu, 122(2), 2015). Let f be a continuous, nonnegative function on [0, 1]. Show that   Z 1 Z 1 Z 1 xf 2 (x) dx . x2 f (x) dx f 3 (x) dx ≥ 4 0

0

0

Comment. This problem can be generalized in the following manner: Let f, g be continuous and nonnegative functions on [0, 1]. If α, β ≥ 0, then Z 1  Z 1  f α+β (x) dx g α+β (x) dx 0

0

Z

1

≥

 Z f (x)g (x) dx α

1

β

0

 f β (x)g α (x) dx .

0

4. Problem 11840 (Proposed by G. Stoica, 122(5), 2015). Let z1 , z2 , . . . , zn be complex numbers. Prove that n !2 !2 n n n X X 2 X X Rezk . zk ≥ |Rezk | − |zk | − k=1

k=1

k=1

k=1

5. Problem 11925 (Proposed by L. Giugiuc, 123(7), 2016). Let n be an integer with n ≥ 4. Find the largest k such that for any list a of n real numbers that sum to 0, 2 3   n n X X  a3j  . a2j  ≥ k  j=1

4.6

j=1

An inequality by power series

Problem 11989 (Proposed by S. P. Andriopoulos, 124(6), 2016). Let x be a number between 0 and 1. Prove   ∞ Y 1 1 (1 − xn ) ≥ exp − . 2 2(1 − x)2 n=1 Discussion. Taking the logarithm of each side of the proposed inequality yields ∞ X n=1

ln(1 − xn ) ≥

1 1 − . 2 2(1 − x)2

208

Inequalities

As usual, we let f (x) =

∞ X

ln(1 − xn ) −

n=1

1 1 . + 2 2(1 − x)2

Since f (0) = 0, the proposed inequality can be proved by showing that f 0 (x) = −

∞ X 1 nxn−1 + ≥0 n 1 − x (1 − x)3 n=1

for all x ∈ (0, 1).

Unfortunately, the preceding expression of f 0 (x) makes it hard to determine 0 whether or not fP (x) ≥ 0. Now we will carry the analysis through the power ∞ ≥ 0 for series: If f (x) = n=1 an xn on (−1, 1) with an ≥ 0 for all n, then f (x)P all x ∈ (0, 1). It is interesting to see that an arithmetic function σ(n) = d|n d, which denotes the sum of all positive divisors of n, comes to the rescue. Solution. P∞ For 0 < x < 1, since 1/(1 − x) = n=0 xn , differentiating gives ∞ ∞ X X 1 n−1 = nx = (n + 1)xn . (1 − x)2 n=1 n=0

Invoking

Pn

k=1

k = n(n + 1)/2 yields

∞ ∞ n X n + 1 n X xn X 1 1 k. − + = x = 2 2(1 − x)2 2 n n=1 n=1 k=1

On the other hand, since σ(n) = ∞ X

ln(1 − xn )

=

−

n=1

P

d|n

d, we have

∞ ∞ X ∞ ∞ X X X (xn )k nxnk =− k nk n=1 n=1 k=1

=

−

∞ X m=1

k=1

∞ X 1 X σ(m) m xm =− x , k m m=1 k|m

where the interchange ofP the order of summations is justified by the positivity n of all summands. Since k=1 k ≥ σ(n), it follows that ! ∞ n ∞ X X 1 1 X 1 n ln(1 − x ) − + = k − σ(n) xn ≥ 0 2 2 2(1 − x) n n=1 n=1 k=1

for all 0 < x < 1. This proves the proposed inequality as desired.



Remark. Along the same lines, we can establish a similar reversed inequality:   ∞ Y 1 1 n (1 − x ) ≤ exp − . 2 2(1 − x2 ) n=1

An inequality by power series

209

Indeed, for 0 < x < 1, this immediately follows from −

∞ X

ln(1 − xn )

=

n=1

∞ X ∞ ∞ ∞ X X (xn )k 1 X nk x = k k n=1 n=1 k=1

∞ X

k=1

x x2 ≥ k k(1 − x ) 2(1 − x2 ) k=1   1 1 . = − 1− 2 1 − x2

=

k

The procedure used in the preceding solution is very useful. The next two examples will recapitulate and elaborate on this idea. Example 1. (Nesbitt’s inequality) For a, b, c > 0, Show that a b c 3 + + ≥ . b+c c+a a+b 2 Because the left-hand side of the stated inequality is homogenous, without loss of generality, we assume that a + b + c = 1. So it suffices to prove that, for a, b, c ∈ (0, 1), a b c 3 + + ≥ . 1−a 1−b 1−c 2 Using geometric series and Jensen’s inequality, we conclude that a b c + + 1−a 1−b 1−c

=

∞ X

(an + bn + cn )

n=1 ∞ X an + bn + cn 3 n=1  n ∞ X a+b+c ≥ 3· 3 n=1   ∞ 3 X 1 3 = 3· = . 3 2 n=1

=

3·

Example 2. (Wilker’s inequality revisited, see (4.5) and Problem 7 in Section 4.1) Here we present a power series proof of the following improved Wilker’s inequality: If x ∈ (0, π/2), then 16 3 x tan x < π4



sin x x

2 +

tan x 8 3 −2< x tan x, x 45

where both 16/π 4 and 8/45 are the best possible constants in (4.23). Let
sin x 2 x + tan sin 2x 1 2 cot x x x −2 F (x) := = + 4− . 3 5 x tan x 2x x x3

(4.23)

210

Inequalities

Notice that F (0) = lim+ F (x) = x→0

8 45

and

F (π/2) = 16/π 4 .

Thus, to prove (4.23), it suffices to show that F (x) is strictly decreasing on (0, π/2). But F 0 (x) = −

4 6 cos x 2 5 sin 2x cos 2x + − 5+ 4 + , 2x6 x5 x x sin x x3 sin2 x

which does not render F 0 (x) < 0 immediately clear. Instead we consider the power series of F (x). Recall that sin 2x

=

∞ X (−1)n 22n+5 2n+5 4 x ; 2x − x3 + 3 (2n + 5)! n=0

cot x

=

∞ X 22n+4 Bn+2 2n+3 1 1 − x− x , x 3 (2n + 4)! n=0

where Bn is the nth Bernoulli number. We obtain F (x)

=

∞ X n=0

=

22n+4 [2(2n + 5)Bn+2 + (−1)n ] x2n (2n + 5)!

8 2 16 8 8 − x + x4 + x6 + · · · . 45 945 14175 467775

Invoking that 22n−1 · π 2n Bn , (2n)! by induction, for n > 2, we find that ζ(2n) =

2(2n + 5)Bn+2 =

4 · (2n + 5)! 4 · (2n + 5)! ζ(2n + 4) > > 1. 2n+4 (2π) (2π)2n+4

Let √ 8 8 16 8 G(x) := F ( x) = − x+ x2 + x3 + · · · . 45 945 14175 467775 00 0 2 0 Then √ G (x) √ ≥ 0, and so G (x) is increasing on (0, π /4). Since G (x) = 0 F ( x)/(2 x) and   16 1 10 G0 (π 2 /4) = F 0 (π/2) = 4 1 − 2 < 0, π π π

it follows that G0 (x) < 0 on (0, π 2 /4), and so F 0 (x) < 0 on (0, π/2). Thus, for x ∈ (0, π/2), 8 16 = F (0) > F (x) > F (π/2) = 4 , 45 π which proves (4.23) as desired. Now we compile a few problems for additional practice.

An inequality by power series

211

1. For a, b ∈ (0, 1) and p, q > 0 with 1/p + 1/q = 1, show that q p pq bp ab ap + ≥ + ≥ . and p q p 2 1−a 1−b 1 − ab p(1 − a ) q(1 − bq )2 (1 − ab)2 2. Problem 11692 (Proposed by C. Lupu and S. Spataru, 120(2), 2013). Let a1 , a2 , a3 , a4 be real number in (0, 1) with a4 = a1 . Show that  3  3 X X 3 1 1 1 ≥ . + + 1 − a1 a2 a3 1 − a3k 1 − a2k ak+1 1 − ak a2k+1 k=1

k=1

3. CMJ Problem 1119 (Proposed by Spiros P. Andriopoulos, 49(1), 2018). For any number x with 0 < x < 1, prove that ∞ X n=1

xn < ln 1 + x + x2 + · · · + xn



1 1−x

 .

4. Problem 11504 (Proposed by F. Holland, 117(5), 2010). Let N be a positive integer and x a positive real number. Prove that !m N N −m+1 X X xk 1 ≥ 1 + x + · · · + xN . m! k m=0 k=1

5. Refinement of Shafer-Fink’s inequality. For 0 ≤ x ≤ 1, show that 1 7 3x 1 5 π−3 √ x + x ≤ sin−1 x − . ≤ 180 189 2 2 + 1 − x2 6. Hyperbolic analogue of Wilker’s inequality: For x > 0, show that  2 sinh x tanh x 8 3 + −2> x tanh x. x x 45 7. A Wilker-type inequality: For x ∈ (0, π/2), show that    x 2 2 3 x 2 16 x sin x < + −2< − x3 sin x, (4.24) 45 sin x tan x π π3 where both 2/45 and 2/π − 16/π 3 are the best possible constants in (4.24). P∞ P∞ 8. Let the power series f (x) = n=0 an xn and g(x) = n=0 bn xn converge for |x| < R. If bn > 0 and an /bn is strictly increasing for all n = 0, 1, . . ., show that f (x)/g(x) is also strictly increasing on (0, R).

212

Inequalities 9. Let n ∈ N. Assume that f is a real function defined on (a, b) such that f (k) (a+ ), f (k) (b− ) exist for k = 0, 1, . . . , n. If f (n) (x) is increasing on (a, b), prove that ! n−1 n−1 X f (k) (aa ) X (b − a)k f (k) (a+ ) 1 k − (x−a) + f (b ) − (x−a)n k! (b − a)n k! k=0

k=0

> f (x) >

n X f (k) (a+ ) (x − a)k . k!

k=0

Comment. Applying this to F (x) in Example 2 above yields a refinement of (4.23):   2  8 2 sin x tan x 8 − x x3 tan x < + −2 45 945 x x   8 8 2 α < − x + x3 tan x, 45 945 14174 where α = (480π 6 − 40320π 4 + 3628800)/π 8 = 17.15041 . . . . 10. Let x ∈ (0, π/2) and p ≥ 1. Show that 2p  p   tan x x 2p  x p sin x + > 2. + > x x sin x tan x What is the smallest p such that this inequality holds?

4.7

An inequality by differential equation

Problem 11923 (Proposed by O. Kouba, 123(7), 2016). Let fp be the function on (0, π/2) given by fp (x) = (1 + sin x)p − (1 − sin x)p − 2 sin(px). Prove fp > 0 for 0 < p < 1/2 and fp < 0 for 1/2 < p < 1. Discussion. Since fp (0) = 0, we naturally proceed to show that fp0 (x) > 0 for 0 < p < 1/2 and fp0 (x) < 0 for 1/2 < p < 1 on (0, π/2). However, notice that fp0 (x) = p(1 + sin x)p−1 cos x + p(1 − sin x)p−1 cos x − 2p cos(px). We see that the analysis of positivity and negativity of fp0 (x) is even harder than fp (x) itself, although the following graph offers an affirmative answer.

An inequality by differential equation

213

FIGURE 4.1 The graph of fp (x) with p = 1/3, 2/3, and 1/2, respectively The graphs with different p-values are shown in Figure 4.1. We see that fp (x) changes sign when p = 1/2. The published solution by Mumbai brought to mind Hadamard’s famous dictum ([52], p. 123): “The shortest and best way between two truths of the real domain often passes through the imaginary one.” Using the binomial series and de Moivre’s formula, Mumbai established fp (x) = 4 cos2p (x/2)

 ∞  X 2p tan4k+1 (x/2). 4k + 3

k=0

Since  ak :=

2p 4k + 3

 =

1 (2p) (2p − 1)(2p − 2) · · · (2p − 4k − 2), {z } | (4k + 3)! 4k − 2 terms

the desired inequality follows from the fact that ak > 0 if 0 < 2p < 1 and ak < 0 if 1 < 2p < 2.1 Here we present a real-variable proof via the method of variation of constants in differential equations. Consider the second-order inhomogeneous ordinary differential equation L[y] = y 00 + p(x)y 0 + q(x)y = g(x),

x ∈ (a, b).

(4.25)

1 Several solutions in this book have illustrated how complex-variables can be used to provide quick proofs of a wide variety of problems, especially for the series and integration. Lax and Zalcman’s beautiful slim book [64] offered us an extended meditation on Hadamard’s dictum. Their discussion begins with the shortest proof of the fundamental theorem of algebra and concludes with Newman’s proof of the prime number theorem.

214

Inequalities

Let y1 (x) and y2 (x) be linearly independent solutions of the corresponding homogeneous equation of (4.25), and suppose, without loss of generality, that their Wronskian W (y1 , y2 ) = 1. Define K(x, t) := y2 (x)y1 (t) − y1 (x)y2 (t). Then the solution of (4.25) satisfying y(a) = y 0 (a) = 0 is given by Z x y(x) = K(x, t)g(t) dt. a

In the following solution we will show that fp (x) displays such an integral representation. Solution. We show that indeed fp , fp0 > 0 for 0 < p < 1/2 and fp , fp0 < 0 for 1/2 < p < 1. First, we prove that fp (x) satisfies a second-order ordinary differential equation with constant coefficients. To this end, we compute fp0 (x)

=

p(1 + sin x)p−1 cos x + p(1 − sin x)p−1 cos x − 2p cos(px),

fp00 (x)

=

p(p − 1)(1 + sin x)p−2 cos2 x − p(1 + sin x)p−1 sin x + p(p − 1)(1 − sin x)p−2 cos2 x − p(1 − sin x)p−1 sin x + 2p2 sin(px)

=

−p2 (1 + sin x)p + p(2p − 1)(1 + sin x)p−1 + p2 (1 − sin x)p − p(2p − 1)(1 − sin x)p−1 + 2p2 sin(px),

where we have used the identity: cos2 x = 2(1 ± sin x) − (1 ± sin x)2 . Thus, we find that fp00 (x) + p2 fp (x) = g(x), (4.26) where   g(x) = p(2p − 1) (1 + sin x)p−1 − (1 − sin x)p−1 . For x ∈ (0, π/2) and 0 < p < 1, xp−1 is a decreasing function because its derivative is negative. This implies that (1 + sin x)p−1 − (1 − sin x)p−1 < 0, and so g(x) > 0 if 0 < p < 1/2 and g(x) < 0 if 1/2 < p < 1. It is well-known that y 00 + p2 y = 0 has linearly independent solutions cos(px) and sin(px). Let 1 1 y1 (x) = √ cos(px), y2 (x) = √ sin(px). p p Then W (y1 , y2 ) = 1 and K(x, t) = y2 (x)y1 (t) − y1 (x)y2 (t) = =

1 sin(p(x − t)), p

1 (sin(px) cos(pt) − cos(px) sin(pt)) p

An inequality by differential equation

215

which is positive for 0 < t < x < π/2 and 0 < p < 1. In view of fp (0) = fp0 (0) = 0, we find the unique solution of (4.26) as Z x fp (x) = K(x, t)g(t) dt (0 < x < π/2). (4.27) 0

Now the proposed inequality follows from the positivity and negativity properties of g. From (4.27) and K(x, x) = 0 we have Z x Z x ∂ fp0 (x) = K(x, t)g(t) dt = cos(p(x − t))g(t) dt. 0 ∂x 0 Since cos(p(x − t)) > 0 for 0 < t < x < π/2 and 0 < p < 1, from the property of g(x), we conclude that fp0 (x) > 0 if 0 < p < 1/2 and fp0 (x) < 0 if 1/2 < p < 1 as well.  Remark. Another interesting approach for the analysis of fp (x) is to apply the following Maximum Principle: Let f : [a, b] → R be a function satisfying f 00 (x) + p(x)f 0 (x) ≥ 0,

for all x ∈ (a, b),

where p(x) is bounded on [a, b]. Then f (x) achieves its maximum and minimum at the end points. In this case, we have fp (0) = 0, fp (π/2) = 2p − 2 sin(pπ/2). Since f1/2 (π/2) = f1 (π/2) = 0 and d d2 π2 fp (π/2) = 2p ln 2 − π cos(pπ/2), 2 fp (π/2) = 2p (ln 2)2 + sin(pπ/2) > 0 dp dp 2 we conclude that min {fp (x)} = 0 for 0 < p < 1/2;

x∈[0,π/2]

max {fp (x)} = 0 for 1/2 < p < 1.

x∈[0,π/2]

See Figure 4.2. Now the proposed problem follows from the maximum principle. The above solution illustrates the power of the maximum principle in the qualitative analysis of some differential equations or differential inequalities. For more details and examples on this subject, we refer the reader to Protter and Weinberger’s excellent book [76]. We now conclude this section with a few problems for additional practice. 1. SIAM Problem 86-17 (Proposed by C. L. Frenzen). The function  −1/2 1 − e−t f (t) = t is for |t| < 2π and has a Taylor series of the form f (t) = Panalytic ∞ n b t . Prove that n=0 n b0 = 1, (−1)n+1 b2n > 0, (−1)n b2n+1 > 0 for all n ∈ N. Hint: Let t = 4xi. Then f (t) = (x cot x)1/2 + ix(x cot x)−1/2 .

216

Inequalities

FIGURE 4.2 The graph of fp (π/2) for 0 < p < 1 2. Problem 11024 (Proposed by V. R¨adulescu, 110(5), 2003). Consider a continuous function g : (0, ∞) → (0, ∞) such that for some α > 0, lim

x→∞

g(x) = ∞. x1+α

Let f : R → (0, ∞) be a twice-differentiable function for which there exist a > 0 and x0 ∈ R such that for all x ≥ x0 , f 00 (x) + f 0 (x) > ag(f (x)). Prove that limx→∞ f (x) exists and is finite, and evaluate the limit. Comment: It is interesting to see what happens when the superlinear growth assumption on g is replaced by linear growth. 3. Problem 11137 (Proposed by V. R¨adulescu, 112(2), 2005). Let φ be a continuous positive function on the open interval (A, ∞), and assume that f is a C 2 -function on (A, ∞) satisfying the differential equation f 00 (t) = (1 + φ(t))(f 2 (t) − 1)f (t). (a)Given that there exists a ∈ (A, ∞) such that f (a) ≥ 1 and f 0 (a) ≥ 0, prove that there is a positive constant K such that f (x) ≥ Kex whenever x ≥ a. (b)Given instead that there exists a ∈ (A, ∞) such that f 0 (a) < 0 and f (x) ≥ 1 for x > a, prove that there is a positive constant K such that f (x) ≥ Kex for x ≥ a. (c)Given that f is bounded on (A, ∞) and that there exists α > 0 such that φ(x) = O(e−(1+α)x ), prove that limx→∞ ex f (x) exists and is finite.

Bounds for a reciprocal of log sum

217

Comment: The nonlinearity f (f 2 − 1) goes back to the GinzburgLandau theory arising in superconductivity. 4. Problem 11135 (Proposed by R. Redheffer, 112(2), 2005). Let A and b be continuous nondecreasing functions from [0, ∞) into [0, ∞). Let λ, µ ∈ [0, 1] with µλ < 1. Let Z t ρ = 1/(1 − µλ) and B(t) = b(s)ds. 0

Show that if v : [0, ∞) → R is continuous, v(0) ≤ 0, and Z t v(t) ≤ tA(t) + b(s)v(s) ds + λv(µt) 0

on [0, ∞), then for t ≥ 0, v(t) ≤ ρtA(t)eρB(t) . 5. Let p be a continuous real-valued function on R and let f satisfy the differential equation f 00 (x) + p(x)f 0 (x) − f (x) = 0. If f has more than one zero, show that f (x) ≡ 0.

4.8

Bounds for a reciprocal of log sum

Problem 11847 (Proposed by M. Bencze, 122(6), 2015). Prove that for n ≥ 1, n

n(n + 1)(n + 2) X 1 n n(n + 1)(n + 2) < < + . 2 3 4 3 ln (1 + 1/k) k=1 Discussion. Since we cannot find the exact sum that appears in the proposed inequality, it is natural to estimate the sum by applying the well-known inequality x < ln(1 + x) < x, 1+x

for x > 0.

This leads us to n X k=1

k2
+ , 6 4 3 so (4.28) is too vague to achieve the desired bounds. Notice that the proposed inequality depends on a positive integer n, hence induction may lead to a proof. More than that, working backward, we find an insight to tighten the loose bounds in (4.28) from the proof. Indeed, the inductive step would require 1 (n + 1)(n + 2)(n + 3) n(n + 1)(n + 2) + 2 > . 3 3 ln (1 + 1/(k + 1)) This is equivalent to  ln 1 +

1 k+1

 0, ln(1 + x) < √

x . 1+x

Similarly, to prove the right inequality we need x < ln(1 + x). 1 + x/2 Thus, the sharpened inequality x x < ln(1 + x) < √ 1 + x/2 1+x will lead us quickly to the proof of the proposed inequality. Solution. To prove (4.29), let f (x) = ln(1 + x) −

x . 1 + x/2

For x > 0, we have f 0 (x) =

1 4 x2 − = > 0. 2 1 + x (2 + x) (1 + x)(2 + x)2

(4.29)

Bounds for a reciprocal of log sum

219

Thus f (x) is strictly increasing on (0, ∞) and so f (x) > f (0) = 0. This proves the left-hand side inequality of (4.29). Similarly, let x − ln(1 + x). g(x) = √ 1+x For x > 0, we have √ x+2 1 x+2−2 1+x √ √ − . = 2(1 + x) 1 + x 1 + x 2(1 + x) 1 + x √ Thus g 0 (x) > 0 as long as x + 2 − 2 1 + x > 0. This can be asserted by g 0 (x) =

√
0 1 > 0, (x > 0). (x + 2) − 2 1 + x = 1 − √ 1+x Therefore, g(x) is also strictly increasing on (0, ∞) and so g(x) > g(0) = 0, which proves the right-hand side inequality of (4.29). Now we square (4.29), and then take the reciprocal to get 1+x 1 (x + 2)2 < 2 < . 2 x 4x2 ln (1 + x) Taking x = 1/k yields k(k + 1)
0, Sp (x) == (2x)p−1/2πΓ(p+1) 0 tet −1 Jp−1/2 (xt) dt, where Jα denotes the ordinary Bessel function of order α. (b) For p > 0, √ Z ∞ π Γ(p + 1/2) Sp (x) dx = ζ(2p). Γ(p + 1) 0

(c) A function f : (0, ∞) → R is said to be completely monotonic if (−1)n f (n) (x) ≥ 0 for all x ≥ 0 and n ∈ N. Show that Sp (x) is completely monotonic and log-convex for each x > 0.

222

4.9

Inequalities

Log-concavity of a partial binomial sum

Problem 11985 (Proposed by Donald Knuth, 124(6), 2017). For fixed s, t ∈ N with s ≤ t, let       n n n an = + + ··· . s s+1 t Prove that this sequence is log-concave, namely that a2n ≥ an−1 an+1 for n ≥ 1. Discussion. Notice that the binomial sequence since
n 2


k

n k−1


=

n k+1

n k




is log-concave for fixed n and 0 ≤ k ≤ n

(k + 1)(n − k + 1) > 1. k(n − k)

It is also known that the linear transformation n   X n xk , n = 0, 1, 2, . . . yn = k k=0

preserves log-concavity, i.e., the log-concavity of {xn } implies thatPof {yn }.
In n particular, with xn ≡ 1, which is log-concave, we find that zn = k=0 nk = 2n , the sum of nth row of Pascal’s triangle, is log-concave. Notice here an is only the partial sum of the entries in the nth row of Pascal’s triangle. To show that a2n ≥ an−1 an+1 , we use the q-log-concavity technique, which was first suggested by Stanley ([87], also available at http://math.mit.edu/ ~rstan/pubs/pubfiles/72.pdf). Define fn (q) =

t   X n k q . k

(4.30)

k=s

We show that fn2 (q) − fn−1 (q)fn+1 (q) has nonnegative coefficients. Thus a2n ≥ an−1 an+1 immediately follows from setting q = 1. Solution. Let fn (q) be defined by (4.30). Applying the identity yields

n k




=

n−1 k−1




+

n−1 k




   t  X n−1 n−1 fn (q) = + qk (shifting the index in the first sum) k−1 k k=s      t  X n−1 s n − 1 t+1 n−1 k = q − q + (q + 1) q s−1 t k k=s     n−1 s n − 1 t+1 = q − q + (q + 1)fn−1 (q). s−1 t

Log-concavity of a partial binomial sum

223

Next, let ∆n (q) = fn2 (q) − fn−1 (q)fn+1 (q). We show that all coefficients in ∆n (q) are nonnegative. Indeed,      n−1 s n − 1 t+1 ∆n (q) = fn (q) q − q + (q + 1)fn−1 (q) s−1 t      n n t+1 − fn−1 (q) qs − q + (q + 1)fn (q) s−1 t      n−1 n = fn (q) − fn−1 (q) q s s−1 s−1      n n−1 + fn−1 (q) − fn (q) q t+1 t t       t X n−1 n n n−1 = − q k+s s−1 k s−1 k k=s     t   X n n−1 n−1 n + − q k+t+1 . t k t k k=s

Since, for i ≤ j ≤ n,             n−1 n n−i n n n−j n n n n−1 = ≥ = , i j n i j n i j i j applying this inequality with (i, j) = (s − 1, k) and (i, j) = (k, t) yields that all coefficients in ∆n (q) are nonnegative, which proves the desired inequality.  Remark. Log-concave sequences occur naturally in combinatorics, algebra, analysis, geometry, computer science, probability, and statistics. There has been a considerable amount of research devoted to this topic in recent years. It provides rich materials which are suitable for undergraduate research. We refer the reader to Stanley [87] and Brenti [24]. Motivated by this proposed problem, let {a(n, k)}0≤k≤n be a triangular array of nonnegative numbers. We consider the linear transformation yn =

n X

a(n, k)xk ,

n = 0, 1, 2, . . . .

(4.31)

k=0

It is interesting to determine the conditions on {a(n, k)}0≤k≤n under which (4.31) preserves the log-concavity property. By taking the special log-concavity sequences, we find the following necessary conditions. If (4.31) preserves log-concavity, then for m ∈ N, q > 0, 1. the column sequence {a(n, m)}n≥m is log-concave; Pn 2. the row-sum sequence an = k=0 a(n, k) is log-concave; and Pn 3. the sequence fm (n, q) = k=m a(n, k)q k is log-concave for n ≥ m;

224

Inequalities 4. the diagonal sequence {a(n, n)}n≥0 is log-concave.

The proofs are left to the reader. Here we compile some problems for addition practice. 1. Let c(n, k) and S(n, k) be Stirling numbers of the first kind and the second kind, respectively. Show that both c(n, k) and S(n, k) are log-concave. 2. Let f : [a, b] → R be a nonnegative and continuous function. Define Z b In = f n (x) dx, n ≥ 1. a

Show that {In }n≥2 is log-concave. Comment. Recall that Legendre polynomials Pn (x) in integral form: for x ≥ 1, Z π p 1 (x + x2 − 1 cos θ)n dθ. Pn (x) = π 0 Thus, for fixed x ≥ 1, {Pn (x)}n≥0 is log-concave. 3. Let 1, a1 , a2 , . . . be a positive log-concave sequence. Define ! ∞ ∞ X X xn xk bn = exp ak . n! k! n=0 k=1

Show that {bn } is log-convex (i.e., b2n ≤ bn−1 bn+1 for all n) and {bn /n!} is log-concave. 4. Let Dn be the number of derangements on n objects. Show that {Dn }n≥3 is log-convex. 5. A Motzkin path of length n is a lattice path in the (x, y)-plane from (0, 0) to (n, 0) with steps (1, 1) (up), (1, −1) (down) and (1, 0) (level), never falling below the x-axis. Define Mn = |{the set of all Motzkin paths of length n}|. It is known that

X n Mn = Ck , 2k k≥0

2n n




where Cn = /(n + 1) is the nth Catalan number. Show that {Mn } is log-convex. Pn 6. (Newton’s Real Roots Theorem). If the polynomial k=0 ak xk has positive coefficients and all of its roots are real, then {an }0≤k≤n is log-concave. Comment. The q-log-concavity technique is based on this theorem, which provides an approach for proving the log-concavity of a sequence.

A bound of divisor sums related to the Riemann hypothesis

225

7. Let i, j be nonnegative integers. Show that the linear transformation  n  X n+i yn = xk , n = 0, 1, 2, . . . k+j k=0

preserves log-concavity. 8. Define   n   X n k n−1 m fm (n, q) = q = (q + 1)fm (n − 1, q) + q . k m−1 k=m

Show that 2 fm (n, q) − fm (n − 1, q)fm (n + 1, q)       n X n−1 n−1 n−1 n−1 = − q k+m . m−1 k−1 m−2 k k=m

9. (A test for (4.31) to preserve log-concavity). Let {a(n, k)}0≤k≤n be a triangular array of nonnegative numbers. Define   2a(n, k)a(n, t − k)     −a(n − 1, k)a(n + 1, t − k)  −a(n + 1, k)a(n − 1, t − k), for k < t/2; ak (n, t) = 2 a (n, k)     −a(n − 1, k)a(n + 1, t − k)    −a(n + 1, k)a(n − 1, t − k), for t even and k = t/2 and bt/2c

Am (n, t) =

X

ak (n, t).

k=m

Show that (4.31) preserves log-concavity if and only if Am (n, t) ≥ 0 for all 2m ≤ t ≤ 2n. Comment.
i−n
A explicit example other than Pascal’s triangle is { nk j−k }, where i, j ∈ N, i ≥ j.

4.10

A bound of divisor sums related to the Riemann hypothesis

Problem 10949 (Proposed by J. Lagarias, 109(6), 2002). Let Hn = Pn 1/j. Show that for each positive integer n, j=1 X d ≤ Hn + 2 exp(Hn ) ln(Hn ), (4.32) d|n

226

Inequalities

with equality only for n = 1. Discussion. Before Monthly published this problem, the proposer’s article [62] appeared at https://arxiv.org/abs/math/0008177. In this paper, he proved that the Riemann hypothesis is equivalent to X σ(n) = d ≤ Hn + exp(Hn ) ln(Hn ), (4.33) d|n

which is derived from Robin’s criterion [79]: The Riemann hypothesis is true if and only if X σ(n) = d ≤ eγ n ln(ln n) for all n ≥ 5041, d|n

where γ is Euler’s constant. The current proposed problem is a weaker version of (4.33). Pn Clearly, σ(n) ≤ k=1 k = n(n + 1)/2. But, the proposed upper bound is only on the scale of n ln(ln n) since Hn ∼ ln n. To get a feel for achieving the required upper bound, we derive a similar upper bound by an elementary argument. In addition to σ(n), we presume three more arithmetic functions that are defined using number-theoretic properties in some way. ω(n) = the number of distinct prime divisors of n; τ (n) = the number of positive divisors of n, including 1 and n; φ(n) = the number of positive integers not exceeding n that are relatively prime to n. These functions can explicitly be evaluated in terms of the prime factorization of n. Let n = pa1 1 · · · pakk , where pi are distinct primes and ai ∈ N for 1 ≤ i ≤ k. Then ω(n) = k and τ (n) =

X

1=

k  Y

1−

i=1

σ(n) =

X

(ai + 1);

i=1

d|n

φ(n) = n

k Y

d=

d|n

1 pi

 ;

k Y pai i +1 − 1 . pi − 1 i=1

Note that k

σ(n) Y pai i +1 − 1 = . n pai (pi − 1) i=1 i

A bound of divisor sums related to the Riemann hypothesis Since

227

pi − 1/pai i pai i +1 − 1 pi = < , ai pi (pi − 1) pi − 1 pi − 1

we find that σ(n) ≤ n

k Y i=1

pi pi − 1

! .

(4.34)

In particular, if ω(n) = 5, in view of the fact that x/(x − 1) is decreasing, we obtain σ(n) 2 3 5 7 11 77 ≤ · · · · = . n 1 2 4 6 10 16 On the other hand, we have σ(n)

=

k Y

(pai i + pai i −1 + · · · + 1) ≥

=

(pai i + pai i −1 )

i=1

i=1 k Y

k Y

pai i



i=1

1 1+ pi

 =n

k  Y i=1

1 1+ pi

 .

Using the well-known inequality k Y

(1 + αi ) ≥ 1 +

i=1

k X

αi

i=1

for all αi ≥ 0, we have k

σ(n) Y ≥ n i=1

  k X 1 1 1+ ≥1+ . pi p i=1 i

Since the sum of reciprocals of all primes is divergent, this implies that σ(n)/n has no finite upper bound. For proceeding with (4.32), the above line of reasoning suggests that we establish that Y pi ≤ C ln(ln n) pi − 1 p|n

for some positive constant C. Along the way, we presume some approximation formulas from [80], and use Mathematica extensively to do numerical verifications. Solution. We prove (4.32) by two steps. First, we show that, for n ≥ 7, σ(n) < 2.59n ln(ln n). The proof of (4.35) consists of the following three facts:

(4.35)

228

Inequalities

F1 Let pn be the nth prime. If n ≥ 39, then ln pn
31, then  Y 28 1 < ln(ln n). 1+ p 15 p|n

F3 If n ≥ 13, n 6= 42 and 210, then Y p p|n

p−1

< 2.59 ln(ln n).

To prove F1, we invoke a sharp explicit estimate ([80], Corollary 1, (3.5)): π(x) >

x ln x

for x ≥ 17,

(4.36)

where π(x) denotes the number of primes not exceeding x. Since p7 = 17, letting x = pn in (4.36) yields pn < n ln pn

for n ≥ 7.

√ Notice that pn < n pn for n ≥ 2. Taking the logarithm of both sides gives ln pn < 2 ln n. Thus, pn < 2n ln n for n ≥ 7. Since 2n ln n ≤ n3/2 for n ≥ 75, the above inequality leads to pn < n3/2 for n ≥ 75. Using the Mathematica command Negative[Table[n^3/2 - Prime[n], {n, 7, 74}]] we can easily verify that pn < n3/2 for n ≥ 7. Thus, pn
165. By the known inequality   Y p 1 < eγ ln x 1 + 2 , p−1 ln x p≤x 4.002
1018 , it suffices to prove F3 for k ≤ 16 and n < 5 · 1017 . Notice that Y p Y p < < 7.5 < 2.59 ln(ln n) p−1 p−1 p≤p16

p|n

for n ≥ 108 . Since

Q

Y p|n

p≤p9

> 2 · 108 , if n < 108 , then k ≤ 8. Moreover, we have

Y p p < < 5.9 < 2.59 ln(ln n) p−1 p−1 p≤p8

A bound of divisor sums related to the Riemann hypothesis 231 Q for n ≥ 3 · 104 . Since p≤p6 = 30030 > 3 · 103 , if n < 3 · 104 , then k ≤ 5. As we did in the discussion, we have Y p|n

Y p p 77 < ≤ < 2.59 ln(ln n) p−1 p−1 16 p≤p5

for n ≥ 1000. If n < 1000, k ≤ 4, and we have Y p|n

35 p ≤ < 2.59 ln(ln n) p−1 8

if n ≥ 300. If n < 300, then k ≤ 3. For n < 300, n 6= 210 = 2 · 3 · 5 · 7, we have Y p ≤ 3.75 < 2.59 ln(ln n) p−1 p|n

if n ≥ 71. Finally, we use Mathematica to verify F3 all the way from 70 to 31 (n = 42 is an exception). This proves F3. Now (4.35) follows from (4.34) and F3 immediately. Next we show that, for n ≥ 3, exp(Hn ) ln(Hn ) ≥ eγ n ln(ln n).

(4.38)

To this end, notice that    X Z n Z n n  n Z n X 1 1  X  bxc 1 dx Hn −1 = − = 1 dx = dx. = 2 2 k n x x x2 1 1 k k=1

1≤k≤x

k=1

Hence

Z Hn = 1 + 1

n

x − {x} dx = ln n + 1 − x2

Z

n

1

{x} dx, x2

where {x} denotes the fractional part of x. Recall that Z ∞ {x} γ =1− dx. x2 1 We obtain that

Z

∞

Hn = ln n + γ + n

{x} dx. x2

This implies Hn > ln n + γ, which upon exponentiating yields exp(Hn ) ≥ eγ n. Combining this with ln(Hn ) ≥ ln(ln n) for n ≥ 3 proves (4.38).

232

Inequalities

Since 2.59 < 2eγ , we see that (4.32) is valid for n ≥ 7. The remaining cases 1 ≤ n ≤ 6 now follow by numerical verification via Mathematica.  Remark. There are lots of unsolved problems in mathematics, but none more important and intriguing than the Riemann hypothesis. To herald the new century, Hilbert in 1900 challenged mathematicians with a list of 23 problems (the Riemann hypothesis is the eighth on the list) that he believed should set the course for the mathematical explorers of the twentieth century. To date, 17 of the 23 questions have been answered. Those who discovered the solutions make up “the honors class” and their solutions have transformed the mathematical landscape and indeed have become fundamental to the development of twentieth century mathematics. But the Riemann hypothesis remains unsolved. Exactly 100 years later, the Clay Mathematics Institute published a list of the seven unsolved problems that they predicted would be the most important questions of the twenty-first century. The official problem description on Riemann hypothesis was written by Enrico Bombieri, one of the greatest mathematicians of the age (see “Problems of the Millennium: The Riemann Hypothesis,” available at https://www.claymath.org/ sites/default/files/official_problem_description.pdf). Of the unsolved problems on Hilbert’s list, the Riemann hypothesis was the only one to be included on the new list. Moreover, the Riemann hypothesis and the other six problems have a reward of one million dollars each. Now, finding a proof will make you both famous and rich! Various mathematicians have made some headway toward a proof. Terence Tao arranged the existing facts together in “The Riemann hypothesis in various settings” (see https://terrytao.wordpress.com/2013/07/19/ the-riemann-hypothesis-in-various-settings/). This proposed problem provided an elementary gateway to the Riemann hypothesis. Its major appeal is that anyone with rudimentary exposure to number theory can play with it. Recently, Veen and Craats’s book [92] offered a nice introduction of the Riemann hypothesis for mathematically talented secondary school students. The resolution of the Riemann hypothesis is most likely to require a different level of mathematics. Now we collect some problems for additional practice. They may reveal new perspectives on the resolution of the proposer’s equivalent theorem. 1. Show that X 1 − ln(ln x) < 6 p prime p ≤ x

for x > 4.

A bound of divisor sums related to the Riemann hypothesis

233

2. For n ≥ 2, show that n2 ≤ φ(n)τ 2 (n); σ(n) ≤ nτ (n) − φ(n); φ(n) n+1 τ (n). σ(n) ≤ φ(n) + τ (n)(n − φ(n)); σ(n) ≤ 2

σ(n) ≤

3. When n is odd, show that σ(n) ≤ φ(n)τ 2 (n) can be improved to σ(n) ≤ φ(n)τ (n). Using this result prove that σ(n) ≤ 2φ(n)τ (n) for n ≥ 2 even. 4. (J. Sandor and L. Kovacs) Let P (n) be the largest prime factor of n. Show that σ(n) ≤

3 φ(n)P (n) for n ≥ 3 odd, and σ(n) ≤ 3 φ(n)P (n) for n even. 4

Comment. Since τ (n) and P (n) are not comparable, the inequalities in Problems 3 and 4 are independent of each other. 5. Let n = pa1 1 · · · pakk , where pi are distinct primes and ai ∈ N for 1 ≤ i ≤ k. Define k X σ ∗ (n) = pai i . i=1 2

∗

For example, 90 = 2 · 3 · 5, σ (90) = 2 + 32 + 5 = 16. Show that (a) σ ∗ (n) ≤ n and the equality holds only if k = 1, i.e., n = pa . (b) Let n be an odd integer with k ≥ 2. If n > 15, then σ ∗ (n) ≤

n−1 . 2

(c) If k ≥ 2 and piai ≥ k for all 1 ≤ i ≤ k, then σ ∗ (n) ≤

n . k k−2

6. Recall that Dedekind’s totient function ψ(n) is defined by ψ(n) = n

k  Y i=1

1 1+ pi

 .

Show that, for n ≥ 1, σ(n) < ζ(2)ψ(n). 7. (A. Grytczuk) If n is odd and n > 39 /2, show that σ(2n)
0. Another path is to apply Flett’s mean value theorem [42]: Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). If f 0 (a) = f 0 (b), then there 235

236

Monthly Miniatures

exists a number c ∈ (a, b) such that f 0 (c) =

f (c) − f (a) . c−a

Rb This enables us to show that if f : [0, b] → R is continuous and 0 f (x) dx = 0, Ra then there exists a ∈ (0, b) such that 0 xf (x) dx = 0. Thus, the auxiliary function Z x H(x) = (t2 + (1 − x)t)f (t) dt 0

fulfills the assumptions of Flett’s mean value theorem, so that H(c) − H(0) = cH 0 (c) leads to the desired equality. Based on the discussion above, we give two solutions. Solution I. If f is identically zero, there is nothing to prove, so assume that f (x) is not R1 identically zero. Since f is continuous on [0, 1] such that 0 f (x)dx = 0, there are distinct a, b ∈ [0, 1] such that f (a) = max f (x) > 0, f (b) = min f (x) < 0. x∈[0,1]

(5.1)

x∈[0,1]

Define

Z

2

F (x) = x f (x) −

x

(t + t2 )f (t)dt.

0

Note that F (x) is continues on [0, 1]. Since (t + t2 )f (a) ≥ (t + t2 )f (t) for all t ∈ [0, 1],   Z a 1 a 2 2 2 − f (a) > 0. F (a) ≥ a f (a) − (t + t )f (a)dt = a 2 3 0 Similarly, (t + t2 )f (b) ≤ (t + t2 )f (t) for all t ∈ [0, 1], so   Z b 1 b F (b) ≤ b2 f (b) − (t + t2 )f (b)dt = b2 − f (b) < 0. 2 3 0 The intermediate value theorem implies that R c there is a number c between a and b such that F (c) = 0, so that c2 f (c) = 0 (x + x2 )f (x)dx.  Solution II. First, for x ∈ [0, 1], define Z

x

Z f (t) dt −

F (x) = x 0

x

tf (t) dt. 0

Rx Note that F (x) is differentiable on (0, 1) and F 0 (x) = 0 f (t) dt. Since F 0 (0) = R 1 0 and F 0 (1) = 0 f (t) dt = 0, by Flett’s mean value theorem, there exists a ∈ (0, 1) such that F (a) − F (0) F 0 (a) = . a−0

Value defined by an integral

237

This implies that

a

Z

tf (t) dt = 0.

(5.2)

0

Rx Next, let G(x) = e−x 0 tf (t) dt. Then G(0) = G(a) = 0. By Rolle’s theorem, there exists b ∈ (0, a) such that G0 (b) = −e−b

Z

b

tf (t) dt + e−b bf (b) = 0.

0

Thus, Z

b

tf (t) dt = bf (b).

(5.3)

0

Finally, let H(x) =

Rx 0

(t2 + (1 − x)t)f (t) dt. Then H 0 (x) = xf (x) −

Z

x

tf (t) dt. 0

Thus, H 0 (0) = 0 and from (5.3) H 0 (b) = 0. Once again by Flett’s mean value theorem, there exists c ∈ (0, b) such that H 0 (c) = This is equivalent to c2 f (c) = confirmed.

H(c) − H(0) . c−0

Rc

(x + x2 )f (x) dx and the desired equality is  R1 Remark. Based on Solution I, the hypothesis 0 f (x) dx = 0 can be replaced by the hypothesis that f changes sign in (0, 1). Similarly, by changing the R1 R1 R1 hypothesis 0 f (x) dx = 0 to 0 f (x) dx = 0 xf (x) dx, the argument in Solution II yields the same result (see Problem 5 below). It is interesting to see that we can prove H(x) fulfills the assumptions of Flett’s mean value theorem along the same lines as in Solution I. In fact, since clearly H 0 (0) = 0, it suffices to Rshow that there is c ∈ (0, 1) such that x H 0 (c) = 0. Note that H 0 (x) = xf (x)− 0 tf (t) dt is continuous on [0, 1]. Using (5.1), we have Z a Z a H 0 (a) = af (a) − tf (t) dt ≥ af (a) − f (a) t dt = af (a)(1 − a/2) > 0 0

0

0

and H 0 (b) = bf (b) −

Z

b

Z tf (t) dt ≤ bf (b) − f (b)

0

b

t dt = bf (b)(1 − b/2) < 0. 0

The intermediate value theorem implies that there is a number c between a and b such that H 0 (c) = 0.

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Monthly Miniatures

Mean value theorems play an essential role in analysis. Rolle’s and Lagrange’s mean value theorems are two common forms. Flett’s mean value theorem provides a variant of Lagrange’s mean value theorem with a Rolle’s type condition. In general, Flett’s mean value theorem can be extended as: Let f be n times differentiable on (a, b) and f (n) (a) = f (n) (b). Then there exists ξ ∈ (a, b) such that f (ξ) − f (a) =

n X (−1)k+1 (ξ − a)k f (k) (ξ). k!

(5.4)

k=1

Recall the nth Taylor polynomial of f which is given by Tn (f, x0 )(x) = f (x0 ) +

f 0 (x0 ) f (n) (x0 ) (x − x0 ) + · · · + (x − x0 )n . 1! 1!

(5.4) has the following nice compact form: f (a) = Tn (f, ξ)(a). Now we end this section with some problems for additional practice. 1. Let f be continuous on [a, b] and differentiable on (a, b). Prove that there exists c ∈ (a, b) such that f 0 (c) =

f (c) − f (a) 1 f 0 (b) − f 0 (a) + (c − a). c−a 2 b−a

2. (A variant of Cauchy’s mean value theorem) Let f, g : [a, b] → R be continuous on [a, b] and differentiable on (a, b). If g 0 (x) 6= 0 on [a, b] and f 0 (a) f 0 (b) = , g 0 (a) g 0 (b) then there exists a number c ∈ (a, b) such that f (c) − f (a) f 0 (c) = . 0 g (c) g(c) − g(a) 3. Let f be n-times differentiable on (a, b). Prove that there exists c ∈ (a, b) such that f (a) =

n X f (k) (c) (a − c)n+1 f (n) (b) − f (n) (a) (a − c)k + . k! (n + 1)! b−a

k=0

R1 R1 4. Let f be continuous on [0, 1] such that 0 f (x)dx = 0 xf (x)dx. Prove that there exists c ∈ (0, 1) such that Z c f (x) dx = 0. 0

Another mean value theorem

239

R1 5. Let f be continuous on [0, 1] such that 0 f (x)dx = 0. Prove that there exists c ∈ (0, 1) such that Rc (a) f (c) = f 0 (c) 0 f (x) dx, if f is differentiable on (0, 1). Rc (b) (1 − c)f (c) = c 0 f (x) dx. R c (c) c2 f (c) = 2 0 xf (x) dx, if in addition f (0) = 0. 6. Problem 11581 (Proposed by D. V. Thong, 118(6), 2011). Let f be a continuous, nonconstant function from [0, 1] to R such R1 that 0 f (x) dx = 0. Also, let m = min0≤x≤1 f (x) and M = max0≤x≤1 f (x). Prove that Z 1 −1 mM ≤ xf (x) dx . 2 M −m 0 Comment. This offers R xa nice example of problems leading to publications. Let F (x) = 0 f (t) dt. Integration by parts yields Z 1 Z 1 Z 1 xf (x) dx = xF (x)|10 − F (x) dx = − F (x) dx. 0

0

0

Kouba generalized this problem and obtained the following optimal bound: Z 1 1/p mM −1 , for any p > 0, |F (x)|p dx ≤ √ p 1+pM −m 0 which is published in [60]. 7. Problem 11814 (Proposed by C. Lupu, 122(1), 2015). Let φ be a continuously differentiable function from [0, 1] into R, with φ(0) = 0 and φ(1) = 1, and suppose that φ0 (x) 6= 0 for 0 ≤ x ≤ 1. Let f be R1 a continuous function from [0, 1] into R such that 0 f (x) dx = R1 φ(x)f (x) dx. Show that there exists c with 0 < c < 1 such that R0c φ(x)f (x) dx = 0. 0

5.2

Another mean value theorem

Problem 11872 (Proposed by P. C. Le Van, 122(9), 2015). Let f be a conR1 tinuous function from [0, 1] into R such that 0 f (x)dx = 0. Prove that for all positive integers n there exists c ∈ (0, 1) such that Z c cn+1 f (c) = n xn f (x) dx. 0

240

Monthly Miniatures

Discussion. We can proceed with this problemR in two different ways. First, we apply c (5.2) to xk f (x) recursively to get 0 n xn f (x) dx = 0 for some cn ∈ (0, 1). Rx Now, introduce the auxiliary function F (x) = x1n 0 tn f (t) dt, the equality to be proved follows from Rolle’s theorem. Second, we establish the proposed equality by the intermediate value theorem as we did in Solution I of Problem 11555. Solution I. Applying (5.2) to xf (x) yields Z c2 x2 f (x) dx = 0,

for c2 ∈ (0, 1).

0

Repeating this process yields Z cn xn f (x) dx = 0,

for cn ∈ (0, 1).

0

Define F (x) =

 

1 xn



0,

Rx 0

tn f (t) dt, if x 6= 0, if x = 0.

Then F (x) is continuous on [0, cn ] and differentiable on (0, cn ) and F (0) = F (cn ) = 0. By Rolle’s theorem, these exists c ∈ (0, cn ) such that Z c n F 0 (c) = f (c) − n+1 xn f (x) dx = 0. c 0 Rc n n+1 This implies c f (c) = n 0 x f (x) dx.  Solution II. If f is identically zero, then there is nothing to prove. Assume that f (x) is R1 not identically zero. Since 0 f (x)dx = 0, there exist distinct a, b ∈ [0, 1] such that f (a) = max f (x) > 0, f (b) = min f (x) < 0. x∈[0,1]

Let

x∈[0,1]

 Rx n n   f (x) − xn+1 0 t f (t) dt, if x > 0, F (x) =     f (0) 1 − n , if x = 0. n+1

Since limx→0+ F (x) = F (0), the function F (x) is continuous on [0, 1]. We claim F (a) > 0. This is clear when a = 0. For a > 0,   Z a n n n F (a) ≥ f (a) − n+1 t f (a)dt = 1 − f (a) > 0. a n+1 0

Another mean value theorem

241

Similarly, we have F (b) < 0. The intermediate value theorem implies that n+1 there f (c) = R c isn a number c between a and b such that F (c) = 0. That is, c n 0 x f (x) dx.  Remark. What if we replace xn by some function w(x)? Here we give one generalization: Let w(x) ∈ C 1 [0, 1] with w(0) = 0 and w0 (x) > 0 for x ∈ (0, 1). Then there exists c ∈ (0, 1) such that Z c w0 (c) w(x)f (x) dx. f (c) = 2 w (c) 0 An editorial comment following our featured solution noted that Omajee and Tauraso also obtained the above extension. To this end, define Z x 1 G(x) = w(t)f (t) dt. w(x) 0 Notice that w(x) is increasing. By the mean value theorem for integrals, we have Z w(ξ)|f (ξ)|x 1 x = ≤ |f (ξ)|x, for some ξ ∈ (0, x). |G(x)| = w(t)f (t) dt w(x) 0 w(x) This implies that limx→0+ G(x) = 0. Since Z x w0 (x) 0 G (x) = − 2 w(t)f (t) dt + f (x), w (x) 0 by Rolle’s theorem, it suffices to show that G(x0 ) = 0 for some x0 ∈ (0, 1). We prove this by contradiction. Without R x loss of generality, we assume that G(x) > 0 for all x ∈ (0, 1). Let F1 (x) = 0 f (t) dt. Then integration by parts yields Z x 1 1 x (w(t)F1 (t))|0 − w0 (t)F1 (t) dt = F1 (x) − F2 (x), G(t) = w(x) w(x) 0 Rx 0 1 where F2 (x) := w(x) w (t)F1 (t) dt. The function F2 (x) is differentiable in 0 (0, 1) and Rx w(x)w0 (x)F1 (x) − w0 (x) 0 w0 (t)F1 (t) dt 0 F2 (x) = w2 (x) 0 w (x)(F1 (x) − F2 (x)) w0 (x)G(x) = = > 0. w(x) w(x) Since limx→0+ F2 (x) = limx→0+ (G(x) − F1 (x)) = 0, it follows that limx→1− F2 (x) > 0. But, on the other hand, the assumption F1 (1) = 0 and G(x) > 0 imply that lim F2 (x) = lim (F1 (x) − G(x)) ≤ 0,

x→1−

x→1−

242

Monthly Miniatures

which is a contradiction. Therefore, there exists x0 ∈ (0, 1) such that G(x0 ) = 0. Here are more problems for your practice. 1. Problem 10739 (Proposed by O. Ciaurri, 106(6), 1999). Suppose that f : [0, 1] → R has a continuous second derivative with f 00 (x) > 0 on (0, 1), and suppose that f (0) = 0. Choose a ∈ (0, 1) such that f 0 (a) < f (1). Show that there is a unique b ∈ (a, 1) such that f 0 (a) = f (b)/b. 2. Monthly Problem 11892 (Proposed by F. Perdomo and A. Plaza, 123(2), 2016). Let f be a real-valued continuous and differentiable function on [a, b] with positive derivative on (a, b). Prove that, for all pairs (x1 , x2 ) with a ≤ x1 < x2 ≤ b and f (x1 )f (x2 ) > 0, there exists t ∈ (x1 , x2 ) such that f (t) x1 f (x2 ) − x2 f (x1 ) =t− 0 . f (x2 ) − f (x1 ) f (t) 3. Problem 11290 (Proposed by C. Lupu and T. Lupu, 114(4), 2007). Let f and g be continuous real functions on [0, 1]. Prove that there exists c ∈ (0, 1) such that Z 1 Z c Z 1 Z c f (x) dx xg(x) dx = g(x) dx xf (x) dx. 0

0

0

0

Comment: The x in the above equality can be replaced by any function w(x) ∈ C 1 [0, 1] with w0 (x) ≥ 0 on [0, 1]. 4. Problem 11313 (Proposed by C. Lupu and T. Lupu, 114(8), 2007). Let f be a four-times differentiable function on R with f (4) continuous on [0, 1] such that Z

1

Z

3/4

f (x) dx + 3f (1/2) = 8 0

f (x) dx. 1/4

Prove that there is some c between 0 and 1 such that f (4) (c) = 0. 5. Problem 11429 (Proposed by C. Lupu and T. Lupu, 116(4), 2009). For a continuous real-valued function φ on [0, 1], let T φ be the Rt function mapping C[0, 1] → R given by T φ(t) = φ(t) − 0 φ(u) du, Rt and similarly define S by Sφ(t) = tφ(t) − 0 uφ(u) du. Show that if f and g are continuous real-valued functions on [0, 1], then there exist numbers a, b, and c ∈ (0, 1) such that each of the following is true: T f (a) = Sf (a). Z 1 Z 1 T g(b) f (u) du = T f (b) g(u) du. 0

0

The product of derivatives by Darboux’s theorem Z 1 Z Sg(c) f (u) du = Sf (c) 0

243 1

g(u) du.

0

Comment: These equalities can be viewed as some intermediate value variants. 6. Problem 11517 (Proposed by C. Lupu and T. Lupu, 117(6), 2010). Let f be a three-times differentiable real-valued function on [a, b] with f (a) = f (b). Prove that Z Z b (b − a)4 (a+b)/2 f (x) dx − f (x) dx ≤ sup |f 000 (x)|. a 192 x∈[a,b] (a+b)/2 7. Problem 11981 (Proposed by C. Lupu, 124(5), 2017). Suppose f : [0, 1] → R is a differentiable function with continuous derivative R1 R1 and with 0 f (x) dx = 0 xf (x) dx = 1. Prove that Z

1

|f 0 (x)|3 dx ≥



0

128 3π

2 .

8. Problem 12046 (Proposed by M. Omarjee, 125(5), 2018). Suppose that f : [0, 1] → R has a continuous and nonnegative third R1 derivative, and suppose that 0 f (x) dx = 0. Prove that Z 10

1

x3 f (x) dx + 6

0

5.3

Z

1

Z xf (x) dx ≥ 15

0

1

x2 f (x) dx.

0

The product of derivatives by Darboux’s theorem

Problem 11753 (Proposed by P. Pongsriiam, 121(1), 2014). Let f be a continuous map from [0, 1] to R that is differentiable on (0, 1), with f (0) = 0 and f (1) = 1. Show that for each positive integer n there exist distinct numbers c1 , . . . , cn ∈ (0, 1) such that n Y

f 0 (ck ) = 1.

k=1

Discussion. For n = 1, we wish to find c1 ∈ (0, 1) such that f 0 (c1 ) = 1. This follows from the mean value theorem immediately. For n = 2, consider the subintervals

244

Monthly Miniatures

[0, x] and [x, 1] where x ∈ (0, 1) is to be determined. By the mean value theorem, there is a c1 ∈ (0, x) and c2 ∈ (x, 1) such that f 0 (c1 ) =

f (x) − f (0) f (x) = x−0 x

and

f 0 (c2 ) =

f (1) − f (x) 1 − f (x) = . 1−x 1−x

Thus, f 0 (c1 )f 0 (c2 ) = 1 if and only if f (x)(1 − f (x)) = x(1 − x). This does not hold unless f (x) = x or f (x) = 1 − x. However, we see that if f (x) > x, then f 0 (c1 ) > 1 and f 0 (c2 ) < 1. Similarly, if f (x) < x, then f 0 (c1 ) < 1 and f 0 (c2 ) > 1. Let D = {f 0 (t) : t ∈ (0, 1)}. Then there is a δ > 0 such that (1 − δ, 1 + δ) ⊂ D. By Darboux’s theorem, f 0 (x) has the intermediate value property. Thus, we can select distinct c1 and c2 such that f 0 (c1 ) = y ∈ (1, 1 + δ)

and

f 0 (c2 ) =

1 ∈ (1/(1 + δ), 1) ⊂ (1 − δ, 1). y

For example, let f (x) = x2 . We can choose δ = 1/2, y = 4/3, c1 = 2/3, c2 = 3/8. With the above analysis we can carry out the case for an arbitrary positive integer n. Solution. If f (x) ≡ x, the required equality holds trivially, so assume that f (x) is not identically equal to x. Based on the analysis in the discussion above, we see that the required equality holds for n = 2. For any n > 2, let m = bn/2c. Choose m distinct numbers yk ∈ (1, 1 + δ) (1 ≤ k ≤ m). Darboux’s theorem assures that there are distinct ck (1 ≤ i ≤ 2m) such that f 0 (ck ) = yk , f 0 (cm+k ) =

1 , (1 ≤ k ≤ m). yk

If n is odd, just take yn = 1. Thus, n Y k=1

f 0 (ck ) =

m Y k=1

f 0 (ck ) · f 0 (cm+k ) =

m  Y k=1

yk ·

1 yk

 = 1. 

Remark. The featured solution by John Hagood [53] used induction. By contradiction, he proved that if there exist distinct c1 , c2 , . . . , cn−1 with Qn−1 0 i=1 f (ci ) = 1, then there is a number cn 6∈ {c1 , c2 , . . . , cn−1 } such that f 0 (cn ) = 1. This problem pushes the mean value theorem in another direction: Once there exists c such that f 0 (c) = 1, so does the geometric mean of distinct derivatives. These kinds of extensions have appeared in recent journals and books. Here we collect a list of problems related to arithmetic, geometric and harmonic means for additional practice.

The product of derivatives by Darboux’s theorem

245

1. Let f (x) be differentiable on [0, 1] with f (0) = 0 and f (1) = 1. For Pn each positive integer n and arbitrary weights wk , (wk > 0, k=1 wk = 1), show that there exist distinct points c1 , c2 , . . . , cn ∈ [0, 1] such that n X k=1

wk 0 f (ck )

= 1.

2. Let f (x) be continuous on [a, b] and c ∈ (a, b) which is not an extreme point of f . For each positive integer n, show that (a)there exist distinct points c1 , c2 , . . . , cn ∈ [a, b] such that f (c) =

n 1 X f (ck ); n k=1

(b)there exist distinct points c1 , c2 , . . . , cn ∈ [a, b] such that v u n uY n f (c) = t f (ck ); k=1

(c)if f (c) 6= 0, then there exist distinct points c1 , c2 , . . . , cn ∈ [a, b] such that n f (c) = Pn . k=1 1/f (ck ) 3. Let f (x) and g(x) be differentiable on [a, b] and g 0 (x) 6= 0 for every x ∈ (a, b). For each positive integer n, show that (a)there exist distinct points c1 , c2 , . . . , cn ∈ (a, b) such that n f (b) − f (a) 1 X f 0 (ck ) = ; g(b) − g(a) n g 0 (ck ) k=1

(b)there exist distinct points c1 , c2 , . . . , cn ∈ (a, b) such that v u n u Y f 0 (ck ) f (b) − f (a) n = t ; g(b) − g(a) g 0 (ck ) k=1

(c)if f (b) 6= f (a), then there exist distinct points c1 , c2 , . . . , cn ∈ (a, b) such that f (b) − f (a) n = Pn g0 (c ) . k g(b) − g(a) 0 k=1 f (ck )

246

Monthly Miniatures 4. Let f (x) and g(x) be continuous on [a, b] and g(x) 6= 0 for every x ∈ (a, b). For each positive integer n, show that (a)there exist distinct points c1 , c2 , . . . , cn ∈ (a, b) such that Rb a Rb a

f (x) dx

=

g(x) dx

n 1 X f (ck ) ; n g(ck ) k=1

(b)there exist distinct points c1 , c2 , . . . , cn ∈ (a, b) such that v Rb u n u Y f (ck ) f (x) dx a n = t ; Rb g(ck ) g(x) dx k=1 a Rb (c)if a f (x) dx 6= 0, then there exist distinct points c1 , c2 , . . . , cn ∈ (a, b) such that Rb a

Rb a

f (x) dx g(x) dx

= Pn

n

g(ck ) k=1 f (ck )

.

Comments. Math. Magazine Problem 1867, proposed by A. Plaza and C. Rodriguez, is a special case of parts (a) and (c) with R1 [a, b] = [0, 1], 0 f (x) dx = 1, g(x) ≡ 1. CMJ Problem 956, proposed by D. V. Thong, is a special case of (b) with n = 3, [a, b] = R1 [0, 1], 0 f (x) dx = 1, g(x) ≡ 1.

5.4

An integral-derivative inequality

Problem 11417 (Proposed by C. Lupu and T. Lupu, 116(2), 2009). Let f be a R 2/3 continuous differentiable real-valued function on [0, 1] such that 1/3 f (x)dx = 0. Show that Z 1 2 Z 1 (f 0 (x))2 dx ≥ 27 f (x) dx . (5.5) 0

0

Discussion. A natural approach is to try to apply the Cauchy-Schwarz inequality. In view of (5.5), the Cauchy-Schwarz inequality yields Z 0

1

(f 0 (x))2 dx

Z 0

1

g 2 (x) dx ≥

Z 0

1

f 0 (x)g(x) dx

2 .

An integral-derivative inequality

247

Thus, it suffices to find a function g(x) for which Z

1

g 2 (x) dx =

0

1

Z

1 27

f 0 (x)g(x) dx =

and 0

Solution. Integration by parts yields Z Z 1 f 0 (x)g(x) dx = f (x)g(x)|10 −

1

Z

f (x) dx. 0

1

f (x)g 0 (x) dx

0

0

1/3

Z

= f (x)g(x)|10 −

f (x)g 0 (x) dx +

0

Z

2/3

Z

f (x)g 0 (x) dx

1/3

!

1 0

+

f (x)g (x) dx . 2/3

The expected equality 1

Z

f 0 (x)g(x) dx =

0

Z

1

f (x) dx 0

suggests that we choose g(x) such that g(0) = g(1) = 0 and  x ∈ [0, 1/3];  −1, constant, x ∈ [1/3, 2/3]; g 0 (x) =  −1, ∈ [2/3, 1]. In the simplest form, we have  x ∈ [0, 1/3];  −x, 2x − 1, x ∈ [1/3, 2/3]; g(x) =  1 − x, x ∈ [2/3, 1].

(5.6)

Notice that g is continuous and its derivative is piecewise continuous on [0, 1]. Moreover, 1

Z

1/3

Z

2

2

g (x) dx = 0

Z

2/3 2

1

(2x − 1) dx +

x dx + 0

Z

1/3

(1 − x)2 dx =

2/3

1 . 27

The inequality to be proved then follows from the Cauchy-Schwarz inequality Z

1 0

2

Z

2

Z

g (x) dx ≥

(f (x)) dx 0

1

0

0

1

2 Z f (x)g(x) dx = 0

1

2 f (x) dx .

0



248

Monthly Miniatures

Remark. The selection of g(x) is not unique. In fact, for any continuous R1 R1 function φ(x) with 0 φ(x) dx = 1 and 0 f (x)φ(x) dx = 0, define Z x g(x) = φ(t) dt − x. 0

With the essentially same argument above, regardless of the condition that R 2/3 f (x)dx = 0, we have 1/3 Z 0

1

1 (f (x)) dx ≥ C 0

2

2 f (x) dx ,

1

Z 0

R1

where C = 0 g 2 (x) dx. The equality Z

1

1

Z

0

f (x)g(x) dx = 0

f (x) dx 0

has played a key role to establish (5.5). It is interesting to see what Rhappens if x we replace f 0 (x) by f 00 (x): Let g(x) be given by (5.6). Define G(x) = 0 g(t) dt. Then  x ∈ [0, 1/3];  −x2 /2, x2 − x + 1/6, x ∈ [1/3, 2/3]; G(x) =  −x2 /2 + x − 1/2, x ∈ [2/3, 1]. Notice that G is continuously differentiable with G(0) = G(1) = G0 (0) = G0 (1) = 0 and its second derivative is piecewise continuous. Integrating by parts twice gives Z Z 1

1

f 00 (x)G(x) dx =

0

f (x) dx. 0

Since Z

1

11 , 4860 0 the Cauchy-Schwarz inequality then yields a solution to the following Problem 11946 (Proposed by M. Omarjee, 123(10), 2016). Let f be a twice differentiable function from [0, 1] to R with f 00 continuous on [0, 1] and R 2/3 f (x)dx = 0. Prove 1/3 Z 4860

G2 (x) dx =

2

1

f (x) dx 0

Z ≤ 11

1

(f 00 (x))2 dx.

0

To end this section, we compile a few more problems for your practice. 1. Let f be twice continuously differentiable on [0, π] P∞ Pnwith f (0) = f (π) = 0. Let f (x) = k=1 ak sin kx and Sn (x) = k=1 ak sin kx. Show that, for every n ∈ N, Z π Z π 1 (f (x) − Sn (x))2 dx ≤ 2 |f 00 (x)|2 dx. 3n 0 0

An integral-derivative inequality

249

2. Problem 11133 (Proposed by P. Bracken, 112(2), 2005). Let f be a nonnegative, continuous, concave function on [0, 1] with f (0) = 1. Prove that 2 Z 1 Z 1 1 2 2 x f (x) dx + f (x) dx . ≤ 12 0 0 Comment. Seiffert offered the following extension: For any p > 0, 1

Z

x2p f (x) dx +

(p + 1) 0

2p − 1 ≤ 8p + 4

1

Z

2 f (x) dx .

0

3. Problem 11548 (Proposed by C. Lupu and T. Lupu, 118(1), 2011). Let f be a twice-differentiable real-valued function with continuous second derivative, and suppose that f (0) = 0. Show that 1

Z

00

1

Z

2

(f (x)) dx ≥ 10 −1

2 f (x) dx .

−1

4. Problem 11756 (Proposed by P. Perfetti, 121(2), 2014). Let f be a function from [−1, 1] to R with continuous derivatives of all orders up to 2n + 2. Given f (0) = f 00 (0) = · · · = f (2n+2) (0) = 0, prove (4n + 5)((2n + 2)!)2 2

Z

1

2 Z f (x) dx ≤

−1

1

(f (2n+2) (x))2 dx.

−1

5. Problem 11812 (Proposed by C. Chiser, 122(1), 2015). Let f be a twice continuously differentiable function from [0, 1] to R. Let p be an integer greater that 1. Given that p−1 X k=1

1 f (k/p) = − (f (0) + f (1)), 2

prove that Z

1

2 f (x) dx ≤

0

1 5! p4

Z

1

(f 00 (x))2 dx.

0

6. Problem 11861 (Proposed by P. C. Le Van, 122(8), 2015). Let n be a natural number and let f be a continuous function from [0, 1] R1 to R such that 0 f 2n+1 (x) dx = 0. Prove that (2n + 1)2n+1 (2n)2n

Z 0

1

4n Z f (x) dx ≤ 0

1

(f (x))4n dx.

250

Monthly Miniatures 7. Problem 11884 (Proposed by C. Lupu and T. Lupu, 123(1), 2016). Let f be a real-valued function on [0, 1] such that f and its first two derivatives are continuous. Prove that if f (1/2) = 0, then 1

Z

(f 00 (x))2 dx ≥ 320

1

Z

2 f (x) dx .

0

0

8. Problem 11918 (Proposed by P. C. Le Van, 123(6), 2016). Let f be continuously differentiable on [0, 1], with f (1/2) = 0 and f (i) (0) = 0 when i is even and less than n. Prove Z 1 2 Z 1 1 f (x) dx ≤ (f (n) (x))2 dx. n (n!)2 (2n + 1)4 0 0 9. Monthly Problem 11981 (Proposed by C. Lupu, 124(5), 2017). Suppose f : [0, 1] → R is a differentiable function with continuous R1 R1 derivative and with 0 f (x) dx = 0 xf (x) dx = 1. Prove that Z

1 0

3



|f (x)| dx ≥ 0

128 3π

2 .

10. Problem 12088 (Proposed by F. Stanescu, 126(1), 2019). Let k be a positive integer with k ≥ 2, and let f : [0, 1] → R be a function with continuous k-th derivative. Suppose f (k) (x) ≥ 0 for all x ∈ [0, 1], and suppose f (i) (0) = 0 for all i ∈ {0, 1, . . . , k − 2}. Prove Z

1

x 0

5.5

k−1

(k − 1)!k! f (1 − x) dx ≤ (2k − 1)!

Z

1

f (x) dx. 0

Eigenvalues of a (0, 1)-matrix

Problem 10958 (Proposed by R. Chapman, 124(2), 2017). Let An be the n × n (0, 1)-matrix with 1s in exactly those positions (j, k) such that n ≤ j + k ≤ n + 1. Find the eigenvalues of An . Discussion. The published solution is in a verification manner. Unfortunately, it is hard to see where the claimed eigenvalues and corresponding eigenfunctions come from. Let the characteristic polynomial of An be Pn (λ). Since the zeros of Pn (λ) are the eigenvalues of An , to get a feel for Pn (λ) = det(λIn − An ),

Eigenvalues of a (0, 1)-matrix

251

we compute Pn (λ) directly by cofactor expansions along the last row of the matrix for n = 1, 2, 3, 4: = λ − 1; λ − 1 −1 P2 (λ) = −1 λ λ −1 P3 (λ) = −1 λ − 1 −1 0 P1 (λ)

= λ(λ − 1) − 1 = λP1 (λ) − 1; −1 λ −1 −1 0 = λ − −1 λ − 1 λ − 1 λ

= λP2 (λ) − P1 (λ); λ 0 −1 0 λ − 1 −1 P4 (λ) = −1 λ −1 −1 0 0

−1 0 0 λ

−1 0

= λP3 (λ) − P2 (λ).

In general, these facts suggest that Pn (λ) satisfies the recurrence relation: Pn (λ) = λPn−1 (λ) − Pn−2 (λ)

for n ≥ 2

(5.7)

with P0 (λ) = 1, P1 (λ) = λ − 1. Thus, we can find Pn by solving the recurrence (5.7), which can be carried out either by a special function or by a generating function. Solution I. By induction, for n ≥ 2, we can verify that Pn (λ) = det(λIn − An ) satisfies (5.7) with P0 (λ) = 1, P1 (λ) = λ − 1. Now, we determine Pn (λ) in terms of the well-known Chebyshev polynomials of the second kind Un (λ). Recall that Un (λ) is defined by Un (λ) = 2λUn−1 (λ) − Un−2 (λ),

with

U0 (λ) = 1, U1 (λ) = 2λ.

(5.8)

We claim that Pn (λ) := Un (λ/2) − Un−1 (λ/2) satisfies (5.7) for n = 2, 3, . . .. In fact, this is verified directly for n = 2. For n ≥ 3, using (5.8) twice yields Pn (λ)

=

Un (λ/2) − Un−1 (λ/2)

=

λUn−1 (λ/2) − Un−2 (λ/2) − (λUn−2 (λ/2) − Un−3 (λ/2))

= λ(Un−1 (λ/2) − Un−2 (λ/2)) − (Un−2 (λ/2) − Un−3 (λ/2)) = λPn−1 (λ) − Pn−2 (λ). Since

sin(n + 1)θ , sin θ by the sum to product and double angle formulas, we find that Un (cos θ) =

Pn (2 cos θ) = Un (cos θ) − Un−1 (cos θ) =

sin(n + 1)θ − sin nθ cos[(2n + 1)θ/2] = . sin θ cos(θ/2)

252

Monthly Miniatures

Hence, the zeros of Pn (i.e., the eigenvalues of An ) are   2k − 1 λk = 2 cos θk = 2 cos π , 2n + 1 where k = 1, 2, . . . , n.



Solution II. We provide another approach to solve (5.7) by using the generating function method. Let ∞ X G(x) = Pn (λ)xn . n=0

Using (5.7) and the initial conditions, we obtain G(x)

=

P0 (λ) + P1 (λ)x +

∞ X

Pn (λ)xn

n=2 ∞ X

∞ X

Pn−1 (λ)xn −

Pn−2 (λ)xn

=

1 + (λ − 1)x + λ

=

1 + (λ − 1)x + λx(G(x) − 1) − x2 G(x).

n=2

Therefore, G(x) =

n=2

1−x . 1 − λx + x2

Let λ = 2 cos θ. Then 1 − λx + x2 = (1 − xeiθ )(1 − xe−iθ ). Using partial fractions, we have A B G(x) = + , 1 − xeiθ 1 − xe−iθ where 1 eiθ , B= . A= 1 + eiθ 1 + eiθ Invoking the geometric series, we find that Pn (λ) = Aeinθ + Be−inθ . To find the zeros of Pn (λ), rewrite Pn (λ) = Be−inθ (ei(2n+1)θ + 1). For Pn (λ) = 0, we must have ei(2n+1)θ + 1 = 0. This implies that θ = (2k − 1)π/(2n + 1) for k = 1, 2 . . . , n. Hence, the eigenvalues of An are   2k − 1 λk = 2 cos θk = 2 cos π , 2n + 1

Eigenvalues of a (0, 1)-matrix

253

for k = 1, 2 . . . , n.



Remark. The key insight of using the substitution λ = 2 cos θ can be revealed by the structure of Pn itself. Let z = eiθ . Then λ = 2 cos θ = z + z −1 . It is interesting to see that P1 (z + z −1 ) P2 (z + z −1 )

z − 1 + z −1 ; z 2 − z + 1 − z −1 + z −2 ;

= = ······ Pn (z + z −1 ) =

z n − z n−1 + · · · + z −n .

Hence,
1 + z 2n+1 . Pn (z + z −1 ) = z −n z 2n − z 2n−1 + · · · + 1 = z −n 1+z In this form, we can easily find all zeros of Pn . The (0, 1)-matrices have played a fundamental role in a wide variety of combinatorial problems. For example, let X = {x1 , x2 , . . . , xn } and let X1 , X2 , . . . , Xm be subsets of X. We can define a (0, 1)-matrix A = (aij )m×n as follows: aij = 1 if xj ∈ Xi , otherwise, aij = 0. Here the 1’s in the ith row of A indicate the elements that belong to Xi and the 1’s in the jth column of A specify the sets that contain xj . Thus, the matrix A characterizes the m subsets X1 , . . . , Xm of X. Moreover, let ri be the sum of the ith row and sj be the sum of the jth column. Define vectors R = (r1 , r2 , . . . , rm ) ∈ Rm

S = (s1 , s2 , . . . , sn ) ∈ Rn . Pn Pm Let U(R, S) be all m × n (0, 1)-matrices with j=1 aij = ri and i=1 aij = sj . As an illustration, let R = (4, 3, 3, 3, 3) and S = (5, 3, 3, 3, 2). Then     1 0 1 1 1 1 1 1 0 1  1 1 0 1 0   1 1 0 1 0         A =  1 1 0 0 1 , B =   1 0 1 0 1  ∈ U(R, S).  1 1 1 0 0   1 0 1 1 0  1 1 0 1 0 1 0 1 1 0 and

In graph theory, the matrices in U(R, S) are the incidence matrices of the bipartite simple graphs of BG(R, S). For more details, we direct the interested reader to Brualdi’s expository paper [26]. Here are a few additional problems for your practice. 1. Let

 A=

be invertible, and let x = √a+d 2

a c

det(A)

b d



. Show that

An = det(A)(n−1)/2 Un−1 (x)A − det(A)n/2 Un−2 (x)I, where Un (x) is the Chebyshev polynomial of the second kind.

254

Monthly Miniatures 2. Problem 12025 (Proposed by A. Dzhumadil’daev, 125(2), 2018). The Chebyshev polynomials of the second kind are defined by the recurrence relation U0 (x) = 1, U1 (x) = 2x, and Un (x) = 2xUn−1 (x)−Un−2 (x) for n ≥ 2. For an integer n with n ≥ 2, prove  0 1 1 ··· 1  x 0 1 ··· 1   x2 x 0 ··· 1  det  . . . . .. .. ..  .. . ..   xn−2 xn−3 xn−4 · · · 0 xn−1 xn−2 xn−3 · · · x

1 1 1 .. .



   √   = (−1)n−1 xn/2 Un−2 ( x).   1  0

3. Let 

a b 0 .. .

    A=    0 0 

b a b .. . 0 0

0 ··· b ··· a ··· .. . . . . 0 ··· 0 ···

0 0 0 .. .

0 0 0 .. . a b 0 0 0 .. .

0 0 0 .. .



       b  a

    B=    (−1)n−2 b (−1)n−2 a (−1)n−1 a (−1)n−1 b

··· ··· ··· .. . ··· ···

and

 0 b a −b −a −b   a b 0   .. ..  . .. . . .   0 0 0  0 0 0

Determine the eigenvalues of A and B. 4. SIAM Problem 61-5 (Proposed by C. Sealander). Let A = (aij )n−1 i,j=1 with aij = sin

jπ (i − j)π iπ sin cos . n n n

Find the eigenvalues of A. 5. Problem 11415 (Proposed by F. Holland, 116(2), 2009). Let (A1 , . . . , An ) be a list of n positive-definite 2×2 matrices of complex numbers. Let G be the group of all unitary 2 × 2 complex matrices, and define the function F on the Cartesian product Gn by ! n X ∗ F (U ) = F (U1 , U2 , . . . , Un ) = det Uk Ak Uk . k=1

A matrix of secants

255

Show that minn F (U ) =

U ∈G

n X

σ1 (Ak ) ·

k=1

n X

σ2 (Ak ),

k=1

where σ1 (Aj ) and σ2 (Aj ) denote the greatest and the least eigenvalues of Aj , respectively. 6. Problem 12100 (Proposed by F. Holland, T. Laffey, and R. Smyth, 126(3), 2019). For a positive integer n, let An be the n-by-n tridiagonal matrix whose i, j-entry is given by   −2j(n − j + 1) if j = i; j(n − j + 1) if j = i ± 1; aij =  0 if |i − j| > 1. Determine the eigenvalues of An .

5.6

A matrix of secants

Problem 11969 (Proposed by A. Dzhumadil’daev, 124(3), 2017). Let x1 , . . . , xn be indeterminates, and let A be the n × n matrix with (i, j) entry sec(xi − xj ). Prove Y n det(A) = (−1)( 2 ) tan2 (xi − xj ). 1≤i 1, show that ∞ X

 s2 (n)

n=1

1 1 − p n (n + 1)p

 =

1 − 21−p ζ(p). 1 − 2−p

5. Show that the finite generating function of s2 (n) can be expressed as the following finite Lambert series: n 2X −1

k=1

s2 (k)xk =

n n−1 i 1 − x2 X x2 . 1 − x i=0 1 + x2i

6. Let u(n) be the number of digit blocks of 11 in the binary expansion of n. For example, u(14) = 2 since 14 = 11102 . Show that P∞ u(n) = 32 ln 2 − π4 . (i) n=1 n(n+1) √ Q∞  (2n+1)2 (−1)u(n) (ii) n=0 (n+1)(4n+1) = 22 . 7. Let sb (n) be the sum of all digits in the expansion of n in base b. Prove that P∞ sb (n) b (i) n=1 n(n+1) = b−1 ln b.  (−1)sb (n) √ Q∞ Qb−1 nb+k (ii) n=0 k=1,3,... nb+k+1 = bb (Sondow). Comment. The digit sum sequence sb (n) can be implemented by Mathematica as

278

Monthly Miniatures Digitsum[n_, b_: 10] := Total[IntegerDigits[n, b]] 8. Problem 11685 (Proposed by D. Knuth, 120(1), 2013). Prove that ∞  Y k=0

1 1 + 2k 2 −1

∞

 =

1 X 1 . + Qk−1 2j 2 j=0 (2 − 1) k=0

9. Problem 11762 (Proposed by R. Stanley, 121(3), 2014). Let f (n) be the least number of strokes needed to draw the Young diagrams of all the partitions of n. Let F (x) =

∞ X

f (n)xn = x + 2x2 + 5x3 + 12x4 + 21x5 + 40x6 + · · · .

n=1

Find the coefficients g(n) of the power series G(x) = satisfying G(x) F (x) = 1 + x + Q∞ . i i=1 (1 − x )

P∞

n=1

g(n)xn

10. Problem 11828 (Proposed by R. Tauraso, 122(3), 2015). Let n be a positive integer, and let z be a complex number that is not a kth root of unity for any k with 1 ≤ k ≤ n. Let S be the Pnset of al lists (a1 , . . . , an ) of n nonnegative integers such that k=1 kak = n. Prove that n XY a∈S k=1

n Y 1 1 . = a k a k k ak k (1 − z ) 1 − zk k=1

For example, for n = 3 we have 1 1 1 1 + + = . 6(1 − z)3 2(1 − z)(1 − z 2 ) 3(1 − z 3 ) (1 − z)(1 − z 2 )(1 − z 3 ) 11. Problem 12113 (Proposed by R. Stanley, 126(5), 2019). Define f (n) and g(n) for n ≥ 0 by X

f (n)xn =

n≥0

X

x2

j

j≥0

n≥0

g(n)xn =

k

k

(1 + x2 + x3·2 )

k=0

and X

j−1 Y

Y

j

j

(1 + x2 + x3·2 ).

i≥0

Find all values of n for which f (n) = g(n), and find f (n) for these values.

An infinite sum-product identity

279

12. Independent Study. Recall that an L-function is defined as P χ(n) n≥1 ns , where χ is a Dirichlet character and satisfies χ(nk+r) = χ(r), which resembles sb (nb + r) = sb (n) + r. Thus, the methods used to study the L-function could potentially lead to the study of Dirichlet series of sb (n). (a)Let the Hurwitz function be defined by ζ(s, x) :=

∞ X n=0

1 . (x + n)s

For s > 0, s 6= 1 and x ≥ 0, prove that n bX −1

 sb (n)

k=1

=

n−1 X

1 1 − s (x + n) (x + n + 1)s



 1  ζ(s, 1 + x/bi ) − ζ(s, 1 + (x + bn )/bi ) bsi

i=0 n X

−

i=1

 b  ζ(s, 1 + x/bi ) − ζ(s, 1 + (x + bn )/bi ) . si b

(b)Let the Barnes zeta function be defined by ζ2 (s, x; (a, b)) :=

X n1 ,n2 ≥0

1 (x + an1 + bn2 )s

.

Let sb (n) be the sum of all digits in the expansion of n in base b. Prove that n bX −1

k=1

=

sb (k) (n + x)s n−1 X

  ζ2 (s, x + bi ; (1, bi )) − ζ2 (s, x + bi + bn ; (1, bi ))

i=0

−b

n X   ζ2 (s, x + bi ; (1, bi )) − ζ2 (s, x + bi + bn ; (1, bi )) . i=1

280

5.10

Monthly Miniatures

A polynomial zero identity

Problem 12077 (Proposed by M. A. Alekseyev, 125(10), 2018). Let f (x) be a monic polynomial of degree n with distinct zeros a1 , . . . , an . Prove n X ain−1 = 1. f 0 (ai ) i=1

Discussion. By the assumption, let f (x) =

n Y

(x − ai ).

i=1

Then f 0 (x) =

n Y X (x − aj ). i=1 j6=i

0

In particular, f (ai ) = nodal points a1 , . . . , an Li (x) =

Q

j6=i (ai

− aj ). Recall the Lagrange polynomials with

Y x − aj 1 Y (x − aj ), = 0 ai − aj f (ai )

(i = 1, 2, . . . , n).

j6=i

j6=i

In view of the left-hand side of the proposed identity, this suggests that we apply the Lagrange interpolation for some selected polynomial. Solution. Let f (x) and Li (x) (1 ≤ i ≤ n) be as defined in the discussion above. Let P (x) = xn − f (x), which is a polynomial with degree not exceeding n−1. Applying the Lagrange interpolation formula yields P (x) =

n X

Li (x)P (ai ) =

i=1

n X

Li (x)ani .

(5.23)

i=1

We now consider two cases. (I) If all ai 6= 0, letting x = 0 in (5.23) gives P (0) = (−1)n+1 a1 a2 . . . an =

n X

Li (0)ani .

i=1

Since Li (0) = (−1)n−1

1 f 0 (ai )

Y i6=j

aj ,

(5.24)

A polynomial zero identity

281

from (5.24) it follows that a1 a2 . . . an = a1 a2 . . . an

n X an−1 i 0 (a ) f i i=1

! .

This proves the desired identity by cancelling the common factor a1 . . . an . (II) If one of ai = 0, without loss of generality, we assume that a1 = 0. Removing the common factor x in (5.23) then applying the above process yields ! n X an−1 i , a2 . . . an = a2 . . . an 0 (a ) f i i=2 which is equivalent to the proposed identity. Remark. Based on the idea of the proof above, we can shorten the proof by directly establishing n X an−1 f (x) i xn−1 = . 0 f (ai ) x − ai i=1 The proposed equality follows from matching the coefficients of xn−1 . Similarly, the reader can try to prove the following generalization:  n if 0 ≤ k < n − 1;  0, k X ai 1, if k = n − 1; = f 0 (ai )  Pn i=1 a , if k = n. i=1 i We now end this section by offering a few more problems for additional practice. 1. Problem 10697 (Proposed by J. L. Diaz-Barrero, 105(10), 1998). Given n distinct nonzero complex numbers z1 , z2 , . . . , zn , show that n X 1 zk

k=1

n Y j=1,j6=k

1 (−1)n+1 = . zk − zj z1 z2 · · · zn

2. Problem 11008 (Proposed J. L. Diaz-Barrero and J. Eqozcue, Pby n 110(4), 2003). Let A(z) = k=0 ak z k be a monic polynomial with complex coefficients and with zeros z1 , z2 , . . . , zn . show that n 1 X |zk |2 < 1 + max |an−k |2 . 1≤k≤n n k=1

3. Problem 11012 (Proposed by C. Popescu, 110(5), 2008). Given a positive integer n, find the minimum value of x31 + · · · + x3n x1 + · · · + xn subject to the condition that x1 , . . . , xn be distinct positive integers.

282

Monthly Miniatures 4. Problem 11098 (Proposed by C. Hillar and D. Rhea, 111(7), 2004). Let   n X (−1)i+1 n f (n) = . 2i − 1 i i=1 Prove that there are constants c and c0 such that c ≤ f (n)/ ln n ≤ c0 for sufficiently large n (that is, f (n) = Θ(ln n)). 5. Problem 11403 (Proposed by Y. Yu, 115(10), 2008). Let n be an integer greater than 1, and let fn be the polynomial given by i−1 n   X Y n (−x)n−i (x + j). i i=0 j=0

Find the degree of fn . 6. Problem 11354 (Proposed by M. Beck and A. Berkovich, 115(3), 2008). Find a polynomial f in two variables such that for all pairs (s, t) of relatively prime positive integers, s−1 X t−1 X

|mt − ns| = f (s, t).

m=1 n=1

7. Problem 11577 (Proposed by P. P. Dalyay, 120(5), 2013). Let n be a positive even integer and Pn let p be prime. Show that the polynomial f given by f (z) = p + k=1 z k is irreducible over Q. 8. Problem 11798 (Proposed by F. Holland, 121(10), 2014). For positive integer n, let fn be the polynomial given by fn (x) =

n   X n bk/2c x . k

k=0

(a)Prove that if n + 1 is prime, then fn is irreducible over Q. (b)Prove for all n, bn/2c

fn (1 + x) =

X n − k  2n−2k xk . k

k=0

9. Problem 11736 (Proposed by M. Merca, 120(9), 2013). For n ≥ 1, let f be the symmetric polynomial in variables x1 , . . . , xn , given by f (x1 , . . . , xn ) =

n−1 X k=0

(−1)k+1 ek (x − 1 + x21 , x2 + x22 , . . . , xn + x2n ),

A polynomial zero identity

283

where ek is the kth elementary polynomial in n variables. Also, let ω be a primitive nth root of unity. Prove that f (1, ω, ω 2 , . . . , ω n−1 ) = Ln − L0 , where Lk is the kth Lucas number. 10. Problem 11720 (Proposed by I. Gessel, 120(7), 2013). Let En (t) be the Eulerian polynomial defined by ∞ X

(k + 1)n tk =

k=0

En (t) , (1 − t)n+1

and let Bn be the nth Bernoulli number. Show that (En+1 (t) − (1 − t)n )Bn is a polynomial with integer coefficients. Hint. Use that !   n k X X i n+1 n (−1) (k + 1 − i) tk . En (t) = i i=0 k=0

11. Problem 11947 (Proposed by G. Stoica, 123(10), 2016). Let n be a positive integer, and let z1 , . . . , zn be the zeros of z n +1. For a > 0, prove Pn−1 2k n a 1 X 1 = k=0 n 2 . 2 n |zk − a| (1 + a ) k=1

12. Problem 12022 (Proposed by M. Merca, 125(2), 2018). Let n be a positive integer, and let x be a real number not equal to −1 or 1. Prove n−1 X (1 − xn )(1 − xn−1 ) · · · (1 − xn−k ) =n 1 − xk+1 k=0

and n−1 X

(−1)k

k=0

n (1 − xn )(1 − xn−1 ) · · · (1 − xn−k ) (n−k−1 x 2 ) = nx( 2 ) . 1 − xk+1

A List of Problems

In this list, bold face denotes the Monthly problem featured solution in each section. P = Putnam Problem, M = Mathematics Magazine, C = The College Mathematics Journal. Problems from The American Mathematical Monthly, Mathematics Magazine, and The College Mathematics Journal Mathematical Association of America, 2020. All rights reserved.

©

11559; 118(3), 2011 E3356; 98(4), 1991 11528; 117(9), 2010 11659; 119(7), 2012 11995; 124(7), 2017 E3034; 91(4), 1984 1966-A3 (P) 11773; 121(4), 2014 E1557; 70(9), 1963 6376; 89(1), 1982 12079; 125(10), 2018 2087(M); 93(1), 2020 11976; 124(4), 2017 11941; 123(9), 2016 E1245; 63(10), 1956 4828; 66(2), 1959 11225; 113(5), 2006 11611; 118(10), 2011 12120; 126(6), 2019 12153; 127(1), 2020 2097 (M); 93(3), 2020 11535; 117(9), 2010 11438; 116(5), 2009 11853; 122(7), 2015 11930; 123(8), 2016 12090; 126(2), 2019 12101; 126(3), 2019 12118; 126(6), 2019 11505; 117(5), 2010

11367; 115(5), 2008 12063; 125(8), 2018 1966-A6 (P) 11967; 124(3), 2017 11592; 118(8), 2011 11494; 117(3), 2010 11612; 118(10), 2011 12029; 125(3), 2018 11677; 119(10), 2012 11821; 122(2) , 2015 11206; 113(2), 2006 11637; 119(4), 2012 12031; 125(3), 2018 12181; 127(5), 2020 11837; 124(1), 2017 2012-B4 (P) 11068; 111(3), 2004 4552; 60(7), 1953 4305; 55(7), 1948 4946; 68(1), 1961 11885; 123(1), 2016 11810; 122(1), 2015 4431; 58(2), 1951 4564; 62(2), 1955 10635; 105(1), 1998 10754; 106(8), 1999 12102; 126(3), 2019 11765; 121(3), 2014 11509; 117(6), 2010

11164; 112(6), 2005 11499; 117(2), 2010 11873; 122(10), 2015 12189; 127(6), 2020 12060; 125(7), 2018 11302; 114(6), 2007 11633; 119(3), 2012 11802; 121(8), 2014 11921; 123(6), 2016 11400; 115(10), 2008 11333; 114(10), 2007 11793; 121(7), 2014 11755; 121(2), 2014 11519; 117(7), 2010 11682; 119(10), 2012 12134; 126(8), 2019 12194; 127(6), 2020 11829; 122(3), 2015 11865; 122(9), 2015 12084; 126(1), 2019 12012; 124(10), 2017 11739; 120(9), 2013 11685; 120(1), 2013 11883; 123(1), 2016 11423; 116(3), 2009 2020 (M), 90(2), 2017 5529; 74(8), 1967 11329; 114(10), 2007 11426; 116(4), 2009

285

286 11935; 123(8), 2016 11771; 121(4), 2014 11338; 115(1), 2008 11724; 120(7), 2013 11808; 121(10), 2014 11875; 122(10), 2015 11811; 122(1), 2015 12057; 125(7), 2018 10973; 109(9), 2002 12129; 126(7), 2019 1953-A6 (P) E2835; 87(6), 1980 11277; 114(3), 2007 11564; 118(4), 2011 5035; 69(6), 1962 12184; 127(5), 2020 11937; 123(9), 2016 11322; 114(9), 2007 11331; 114(10), 2007 11650; 119(6), 2012 11322; 114(9), 2007 11275; 114(2), 2007 11953; 124(1), 2017 12070; 125(9), 2018 12054; 125(6), 2018 10884; 108(6), 2001 11152; 112(5), 2005 11961; 124(2), 2017 4826; 66(1), 1959 4865; 66(8), 1959 12158; 127(1), 2020 11850; 122(6), 2015 10777; 107(1), 2000 4212; 53(4), 1946 11933; 123(8), 2016 E2483; 81(6), 1974 11770; 121(3), 2014 11009; 100(3), 2003 11308; 114(7), 2007 11869; 122(9), 2015 11369; 115(6), 2008 11139; 112(3), 2005 11397; 115(10), 2008 12013; 124(10), 2017 12065; 125(8), 2018 10611; 104(5), 1997 11219; 113(4), 2006

List of Problems 11716; 120(6), 2013 11832; 122(4), 2015 11343; 115(2), 2008 11356; 115(4), 2008 11406; 116(1), 2009 11757; 121(2), 2014 11862; 122(8), 2015 11897; 123(3), 2016 11274; 114(2), 2007 11940; 123(9), 2016 12049; 125(6), 2018 11026; 110(7), 2003 11769; 121(4), 2014 11746; 120(10), 2013 11819; 122(2), 2015 11840; 122(5), 2015 11925; 123(7), 2016 11989; 124(6), 2017 11692; 120(2), 2013 1119 (C); 49(1), 2018 11504; 117(5), 2010 11923; 123(7), 2016 11024; 110(5), 2003 11137; 112(2), 2005 11135; 112(2), 2005 11847; 122(6), 2015 E3432: 98(3), 1991 11954; 124(1), 2017 11985; 124(6), 2017 10949; 109(6), 2002 E2611; 83(8), 1976 6015; 82(2), 1975 11555; 118(2), 2011 11581; 118(5), 2011 11814: 122(1), 2015 11872; 122(9), 2015 10739; 106(6), 1999 11892; 123(2), 2016 11290; 114(4), 2007 11313; 114(8), 2007 11429; 116(4), 2009 11517; 117(6), 2010 11981; 124(5), 2017 12046; 125(5), 2018 11753; 121(1), 2014 11417; 116(2), 2009 11946; 123(10), 2016

11966; 124(3), 2017 2005-A5 (P) 10884; 108(6), 2001 11101; 111(7), 2004 11113; 111(9), 2004 11639; 119(4), 2012 11418; 116(3), 2009 11629; 119(3), 2012 12051; 125(6), 2018 11993; 124(7), 2017 11234; 113(6), 2006 1968-B4 (P) 11981; 124(5), 2017 12088; 126(1), 2019 10958; 124(2), 2017 12025; 125(2), 2018 11415; 116(2), 2009 12100; 126(3), 2019 11969; 124(3), 2017 10387; 101(5), 1994 11475; 117(1), 2010 11463; 116(9), 2009 11471; 116(10), 2009 11293; 114(5), 2007 12066; 125(8), 2018 12003; 124(8), 2017 12099; 126(3), 2019 12114; 126(5), 2019 10829; 107(8), 2000 10797; 107(4), 2000 6615; 96(9), 1989 11883; 123(1), 2016 E2692; 85(1), 1978 1977-A4 (P) 11685; 120(1), 2013 11762; 121(3), 2014 11828; 122(3), 2015 12113; 126(5), 2019 12077; 125(10), 2018 10697; 105(10), 1998 11008; 110(4), 2003 11012; 110(5), 2003 11098; 111(7), 2004 11403; 115(10), 2008 11354; 115(3), 2008 11577; 120(5), 2013 11798; 121(10), 2014

287 11145; 112(4), 2005 1964-A5 (P) 6663; 98(6), 1991 11202; 113(2), 2006 12004; 124(8), 2017 11680; 119(10), 2012 E1682; 71(3), 1964

11133; 11548; 11756; 11812; 11861; 11884; 11918;

112(2), 118(1), 121(2), 122(1), 122(8), 123(1), 123(6),

2005 2011 2014 2015 2015 2016 2016

11736; 11720; 11947; 12022;

120(9), 2013 120(7), 2013 123(10), 2016 125(2), 2018

B Glossary

P∞ Abel’s Limit Theorem. Let f (x) = n=0 an xn be a power series which P∞ converges for |x| < R. If n=0 an Rn converges, then lim− f (x) =

x→R

∞ X

an Rn .

n=0

Abel’s Summation Formula. PkLet a1 , a2 , . . . , an and b1 , b2 , . . . , bn be real or complex numbers with Ak = i=1 ai for 1 ≤ k ≤ n. Then n X

ai bi = An bn+1 +

i=1

n X

Ai (bi − bi+1 ).

i=1

The limit version of this formula is ∞ X

ai bi = lim (An bn+1 ) + n→∞

i=1

∞ X

Ai (bi − bi+1 ).

i=1

Arithmetic-Geometric-Harmonic Mean Inequality (AM-GM-HM Inequality). Let a1 , a2 , . . . , an be positive real numbers. Then 1 a1

+

1 a2

n + ··· +

1 an

≤

√ n

a1 a2 · · · an ≤

a1 + a2 + · · · + an , n

with equality if and only if a1 = a2 = · · · = an . PnThe weighted version assumes that, for any pi > 0 (1 ≤ i ≤ n) with i=1 pi = 1, p1 a1

+

p2 a2

1 + ··· +

pn an

≤ ap11 ap22 · · · apnn ≤ p1 a1 + p2 a2 + · · · + pn an .

Bernoulli’s Inequality. Let α > 0. Then (1 + x)α ≥ 1 + αx

for all x ≥ −1.

Bernoulli Numbers. The Bernoulli numbers Bn are defined by the generating function ∞ ∞ X x Bn n x X B2n 2n = x =1− + x . ex − 1 n=0 n! 2 n=1 (2n)!

289

290

Glossary

The first few Bernoulli numbers are B1 = −1/2, B2 = 1/6, B4 = −1/30, and B6 = 1/42. They satisfy the recurrence relation B0 = 1, Bn = −

 n−1  1 X n+1 Bk , n+1 k

n ≥ 1.

k=0

This implies that all the Bernoulli numbers are rational. Asymptotically, as n → ∞, √  n 2n . B2n ∼ (−1)n+1 4 πn πe Bernoulli Polynomials. The Bernoulli polynomials Bn (x) are defined by the generating function ∞ X zexz Bn (x) n = z . ez − 1 n=0 n!

Here Bn (0) = Bn , the nth Bernoulli number for all n ≥ 1, and Bn (x) =

n   X n Bk xn−k . k

k=0

Beta Function. For p, q > 0, the beta function is defined by Z 1 B(p, q) = xp−1 (1 − x)q−1 dx. 0

A change of variables leads two equivalent forms: Z ∞ xp−1 dx B(p, q) = (1 + x)p+q 0 Z π/2 =2 sin2p−1 x cos2q−1 x dx. 0

In general, if a < b, p, q > 0, then Z b (x − a)p−1 (b − x)q−1 dx = (b − a)p+q−1 B(p, q). a

In terms of the gamma function, we have B(p, q) =

Γ(p)Γ(q) . Γ(p + q)

Asymptotically, as p → ∞ and q → ∞, B(p, q) ∼

√

2π

pp−1/2 q q−1/2 . (p + q)p+q−1/2

291 Catalan Numbers. The Catalan numbers are defined by   1 2n Cn = for n ≥ 0, n+1 n or alternatively by the recurrence relation Cn+1 =

n X

Ck Cn−k ,

C0 = 1.

k=0

The generating function of {Cn } is given by √ 1 − 1 − 4x C(x) := . 2x Carleman’s Inequality. Let a1 , a2 , . . . , an , . . . be a positive real number sequence. Then ∞ ∞ X X (a1 a2 · · · an )1/n ≤ e an . n=1

n=1

Cauchy-Schwarz Inequality (Discrete version). Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be real numbers. Then !2 ! n ! n n X X X 2 2 ai bi ≤ ai bi . i=1

i=1

i=1

Equality holds if and only if ai and bi are proportional for all 1 ≤ i ≤ n. Cauchy-Schwarz Inequality (Integral version). Let f, g : [a, b] → R be nonnegative and integrable. Then !2 ! Z ! Z Z b

b

f (x)g(x) dx a

b

f 2 (x) dx

≤ a

g 2 (x) dx .

a

If both f and g are continuous, equality holds if and only if f and g are proportional. Chebyshev Polynomials. The Chebyshev polynomials of the first kind are defined by Tn (cos θ) = cos nθ, or alternatively by the generating function: ∞ X n=0

Tn (x)tn =

1 − xt . 1 − 2xt + t2

The first few Chebyshev polynomials of the first kind are T0 (x) = 1, T1 (x) = x, T2 (x) = 2x2 −1, T3 (x) = 4x3 −3x, T4 (x) = 8x4 −8x2 +1.

292

Glossary

They also satisfy the recurrence relation Tn+1 (x) = 2xTn (x) − Tn−1 (x). In general, bn/2c

  n  X n Y (2k − 1)π n−2k 2 k n−1 Tn (x) = x (x − 1) = 2 x − cos . 2k 2n k=0

k=1

The Chebyshev polynomials of the second kind are defined by Un (cos θ) =

sin(n + 1)θ , sin θ

or alternatively by the generating function: ∞ X

Un (x)tn =

n=0

1 . 1 − 2xt + t2

The first few Chebyshev polynomials of the second kind are U0 (x) = 1, U1 (x) = 2x, U2 (x) = 4x2 − 1, U3 (x) = 8x3 − 4x, U4 (x) = 16x4 − 12x2 + 1. They also satisfy the recurrence relation Un+1 (x) = 2xUn (x) − Un−1 (x). In general, bn/2c

Un (x) =

  n  X n+1 Y kπ xn−2k (x2 − 1)k = 2n x − cos . 2k + 1 n+1

k=0

k=1

Digamma Function. The digamma function ψ(z) (also called psi function) is define by ∞

X Γ0 (z) ψ(z) = (ln Γ(z)) = = −γ + Γ(z) 0

k=1



1 1 − k z+k−1

It has a series representation ψ(z) =

∞ X 1 −γ+ (−1)n ζ(n)z n−1 , z n=2

where γ is Euler’s constant and ζ is the Riemann zeta function.

 .

293 Some useful formulas of ψ(z) with Re(z) > 0: n−1 X

1 + ψ(z) for n ≥ 1; z+k k=0  Z 1 1 tz−1 + dt; ψ(z) = − ln t 1 − t 0 Z ∞ 1 2t dt ψ(z) = ln z − − . 2 + t2 )(e2πt − 1) 2z (z 0

ψ(z + n) =

In particular, when z = p/q, 0 < p < q, a rational number, Gauss formula assumes that ψ

        bq/2c X 2npπ π pπ nπ p cos = −γ − cot − ln q + 2 ln 2 sin . q 2 q q q n=1

Asymptotically, as x → ∞, ψ(x) ∼ ln x −

1 1 1 1 − + − + O(x−8 ). 2x 12x2 120x4 256x6

Euler’s Constant. Euler’s constant γ is defined by   1 1 γ = lim 1 + + · · · + − ln n = 0.5772156649 . . . . n→∞ 2 n Euler-Maclaurin Summation Formula. Let Bn be nth Bernoulli number. Then Z n n X 1 f (k) = f (x) dx + (f (m) + f (n)) 2 m k=m

bp/2c

+

 X B2k  f (2k−1) (n) − f (2k−1) (m) + Rp . (2k)!

k=1

Euler’s Product Formula for the Riemann Zeta Function. Let ζ(s) be the Riemann zeta function. Then, for all s > 1, ζ(s) =

∞ X 1 = s n n=1

Y primes p

1 . 1 − p−s

Euler’s Product Formula for Sine:  ∞  Y sin πx x2 = 1− 2 . πx n n=1

294

Glossary

Logarithmic differentiation yields the following partial fraction expansions:  ∞  1 X 1 1 π cot πx = − + , x n=1 x + n x − n π sin πx

=

π tan πx =

π sec πx = 2

π sin2 πx

=

∞ X 1 (−1)n , + 2x x x2 − n2 n=1 ∞ X n=−∞ ∞ X n=−∞ ∞ X n=−∞

n X 1 1 = lim , n→∞ n − x + 1/2 k − x + 1/2

1 = lim n + x + 1/2 n→∞ 1 = lim n→∞ (n + x)2

k=−n n X

k=−n n X

k=−n

1 , k + x + 1/2

1 . (k + x)2

Fa` a di Bruno’s Formula. This formula gives an explicit expansion for the nth derivative of the composition function (f ◦ g)(x). Let both f and g be n times differentiable. Then  k  n k X n! Dg(x) 1 D g(x) n [(f ◦ g)(x)](n) = (Dk f )(g(x)) ··· , k1 ! · · · kn ! 1! n! where k = k1 + · · · + kn and the sum is over all partition of n, i.e., values of k1 , . . . , kn such that k1 + 2k2 + · · · + nkn = n. For example, when n = 3, [(f ◦ g)(x)](3) = 3g 0 (x)f 00 (g(x))g 00 (x) + (g 0 (x))3 f (3) (g(x)) + f 0 (g(x))g (3) (x). Fibonacci Numbers. The Fibonacci numbers Fn are defined recursively by F1 = F2 = 1 and Fn = Fn−1 + Fn−2

for all n ≥ 3.

The generating function of {Fn } is ∞ X n=1

Let

Fn x n =

x . 1 − x − x2

√ √ 1+ 5 1− 5 ˆ φ= and φ = , 2 2 where φ is called the golden ratio. The closed form of Fn is given by Binet’s formula 1 Fn = √ (φn − φˆn ). 5 Famous identities include

295 ˆ Ces` aro identity:

Pn

k=0

n k




Fk = F2n .

2 ˆ Catalan’s identity: Fn2 − Fn+m Fn−m = (−1)n−m Fm .

Frullani Integrals. If f is continuous on [0, ∞) and f (∞) = limx→∞ f (x) exists, then Z ∞ f (ax) − f (bx) b dx = (f (0) − f (∞)) ln . x a 0 Gamma Function. The gamma function is defined by Z ∞ Γ(x) = tx−1 e−t dt for all x > 0. 0

A product representation due to Weierstrass asserts that ∞ n Y 1 x  −x/n o = xeγx 1+ e , Γ(x) n n=1

where γ is Euler’s constant. The gamma function satisfies many interesting identities: ˆ Euler reflection formula: Γ(x)Γ(1 − x) =

π sin πx .


ˆ Legendre duplication formula: Γ(2x)Γ(1/2) = 22x−1 Γ(x)Γ x + 12 .
Qn−1 ˆ Gauss multiplication formula: k=0 Γ x + nk = (2π)(n−1)/2 n1/2−nx Γ(nx). The gamma function can be defined for all complex numbers except the nonpositive integers. It is related to the Riemann zeta function ζ(s) by Z ∞ s−1 t ζ(s)Γ(s) = dt; et − 1 0   s 1−s −s/2 Γ π ζ(s) = Γ π −(1−s)/2 ζ(1 − s). 2 2 Asymptotically, as x → ∞, Γ(x) ∼

√

2π xx−1/2 e−x .

Equivalently, ln Γ(x) ∼

  1 1 x− ln x − x + ln(2π) + O(1/x). 2 2

Hardy’s Inequality (Discrete version). Let an be a nonnegative real sequences. For p > 1, !p  p X ∞ n ∞ X 1 X p ak ≤ apn . n p − 1 i=1 i=1 k=1

296

Glossary

Equality  p holds if and only if an = 0 for all n ≥ 1. Moreover, the constant p is the best possible. p−1 Hardy’s Inequality (Integral version). Let f : [0, ∞] → [0, ∞) be continuous such that f ∈ Lp with p > 1. Then p  p Z ∞ Z ∞ Z x p 1 f (t) dt ≤ f p (x) dx. x 0 p−1 0 0  p p Equality holds if and only if f = 0 identically. Moreover, the constant p−1 is the best possible. Harmonic Numbers. The nth harmonic numbers, Hn , is defined by n

Hn = 1 +

X 1 1 1 + ··· + = , 2 n k k=1

which has the generating function: ∞ X

Hn xn = −

n=1

ln(1 − x) . 1−x

Hn satisfies the elementary inequality 1 1 < Hn − ln(n + 1/2) − γ < . 24(n + 1)2 24n2 Asymptotically, as n → ∞, ∞

Hn

∼ ln n + γ +

X B2k 1 − 2n 2kn2k k=1

=

1 1 1 1 − + − + .... ln n + γ + 2 4 2n 12n 120n 252n6

where B2k is the (2k)th Bernoulli numbers. Hermite-Hadamard Inequality. Let f : [a, b] → R be a convex function. Then   Z b a+b 1 f (a) + f (b) f ≤ f (x) dx ≤ . 2 b−a a 2 H¨ older’s Inequality (Discrete version). Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be positive numbers. If p, q > 1 and 1/p + 1/q = 1, then n X i=1

ai bi ≤

n X i=1

!1/p api

n X

!1/q bqi

.

i=1

Equality holds if and only if ai and bi are proportional for all 1 ≤ i ≤ n.

297 H¨ older’s Inequality (Integral version). Let f and g be nonnegative function. If f ∈ Lp [a, b], g ∈ Lq [a, b] with p, q > 1 and 1/p + 1/q = 1, then Z

b

Z f (x)g(x) dx ≤

!1/p

b

f p (x) dx

Z

g q (x) dx

.

a

a

a

!1/q

b

L’Hˆ opital’s Monotone Rule. Let f, g : [a, b] → R be continuous functions that are differentiable on (a, b) with g 0 (x) 6= 0 on (a, b). If f 0 (x)/g 0 (x) is increasing (decreasing) on (a, b), then the functions f (x) − f (a) f (x) − f (b) and g(x) − g(a) g(x) − g(b) are likewise increasing (decreasing) on (a, b). Hurwitz Zeta Function. Hurwitz zeta function is defined by ζ(x, a) =

∞ X n=0

1 . (n + a)x

For x > 1, the integral representation is Z ∞ x−1 −at 1 t e dt. ζ(x, a) = Γ(x) 0 1 − e−t In particular, if x = k ∈ N, a ∈ (0, 1), then ζ(k, a) − (−1)k−1 ζ(k, 1 − a) =

(−1)k−1 π dk−1 (cot(πt)) |t=a . (k − 1)! dtk−1

Jensen’s Inequality. Let f : [a, Pnb] → R be a convex function. For any pi > 0, xi ∈ [a, b] (1 ≤ i ≤ n) with i=1 pi = 1, f (p1 x1 + p2 x2 + · · · + pn xn ) ≤ p1 f (x1 ) + p2 f (x2 ) + · · · + pn f (xn ). Mathieu’s Inequality. For c 6= 0, ∞ X 1 2n 1 < < 2. c2 + 1/2 n=1 (n2 + c2 )2 c

Alzer refined this inequality as ∞ X 2n 1 1 < < 2 . c2 + 1/(2ζ(3)) n=1 (n2 + c2 )2 c + 1/6

Parserval’s Identity. Let f ∈ L2 [−π, π] and ∞

a0 X f (x) = + (an cos nx + bn sin nx). 2 n=1

298

Glossary

Then

∞

π

a20 X 2 + (an + b2n ). 2 0 n=1 P∞ For a Fourier sine series on [0, π], i.e., f (x) = n=1 bn sin nx, Parserval’s identity assumes the form 1 π

Z

[f (x)]2 dx =

2 π

π

Z

[f (x)]2 dx =

0

∞ X

b2n .

n=1

Polylogarithm Function. For |z| ≤ 1, the polylogarithm Lin is defined by Li1 (z) = − ln(1 − z) and Z z ∞ X zk Lin−1 (t) dt = Lin (z) := kn t 0

for n ≥ 2.

k=1

It has an integral representation: Lis (z) =

1 Γ(s)

∞

Z 0

ts−1 dt. −1

et /z

In particular, the dilogarithm Li2 (z) satisfies Z Li2 (z) = z

0

ln(1 − t) dt = − t

Z 0

1

ln(1 − zt) dt t

and the functional equations: Li2 (z) + Li2 (−z) =

1 Li2 (z 2 ); 2

Li2 (z) + Li2 (1 − z) = ζ(2) − ln z ln(1 − z). Riemann Zeta Function. The Riemann zeta function is defined by ζ(s) =

∞ X 1 , s n n=1

for all s > 1.

It has the integral representation ζ(s) =

1 Γ(s)

Z 0

∞

xs−1 dx. ex − 1

It is well-known that ζ(2) = π 2 /6, ζ(4) = π 4 /90. In general, ζ(2k) =

(−1)k−1 22k−1 B2k 2k π , (2k)!

299 where B2k is the (2k)th Bernoulli number. Since   1 lim ζ(s) − = γ, s→1 s−1 ζ(s) has the following Laurent series expansion: ∞

ζ(s) =

X (−1)n γn 1 + (s − 1)n , s − 1 n=0 n!

where γn are the so-called Stieltjes constants. These constants are recursively defined by γ0 = γ and ! m X lnn k lnn+1 m γn = lim − , for any n ≥ 1. m→∞ k n+1 k=1

The reciprocal of the ζ(s) can be represented by a Dirichlet series: ∞ X µ(n) 1 = , ζ(s) n=1 ns

where µ(n) is the M¨ obius function in the Number Theory. Series Tests: Guass’s Test. Let an be a positive sequence such that an+1 r = 1 − + O(1/np ) an n P∞ for some constant r and p > 1. Then the series n=1 an converges if r > 1 and diverges if r ≤ 1. Kummer’s Test. Let an be a positive sequences. P∞ P∞ (a) n=1 an converges iff there is a positive series n=1 bn and a constant c > 0 such that an cn = bn − bn+1 ≥ c for all n. an+1 P∞ P∞ (b) n=1 bn such that P∞ n=1 an diverges iff there is a positive series 1/b diverges and n n=1 cn =

an bn − bn+1 ≤ 0 for all n. an+1

Remark. Many other series tests, for example, Ratio test, Raabe’s test, Bertrand’s test and Guass’s test above are all special P∞ cases of Kummer’s test obtained by choosing specific parameter series n=1 bn . Stirling’s Formula. As n → ∞,  n n √ n! ∼ 2πn . e

300

Glossary

More accurately, we have   n n  √ 1 139 1 4 n! = 2πn 1+ − + O(1/n ) . + e 12n 288n2 51840n3 The corresponding logarithmic form becomes   1 1 1 1 ln n! = n + + O(1/n5 ). ln n − n + ln(2π) + − 2 2 12n 360n3 Stolz-Ces` aro Theorem. Let an and bn be two real sequences. (a) Assume that an → 0, bn → 0 as n → ∞ and bn is decreasing. If limn→∞ (an+1 − an )/(bn+1 − bn ) exists, then lim

n→∞

an an+1 − an = lim . n→∞ bn+1 − bn bn

(b) Assume that bn → ∞ as n → ∞ and bn is increasing. If limn→∞ (an+1 − an )/(bn+1 − bn ) exists, then lim

n→∞

an an+1 − an = lim . n→∞ bn bn+1 − bn

Wallis Product Formula. Let (2n)!! = 2n · (2n − 2) · · · 2; (2n − 1)!! = (2n − 1) · (2n − 3) · · · 1. Then lim

n→∞

 ∞  Y 1 2n [(2n)!!]2 2n π = · = . 2n + 1 [(2n − 1)!!]2 2n − 1 2n + 1 2 n=1

Some Useful Inequalities. 1. y 1/n − x1/n ≤ (y − a)1/n − (x − a)1/n for y ≥ x ≥ a ≥ 0, n ∈ N.  √ √ 2 √ x+ y 2. xy ≤ ln x−y ≤ x+y ≤ x−ln y 2 2 for x, y > 0. 3.

√

xy ≤

4.

2x 2+x

≤

5.

ln x x−1

≤

x1−α y α +xα y 1−α ≤ x+y 2 2 for x, y x √ ln(1 + x) ≤ x+1 for x ≥ 0. √1 x

> 0, α ∈ [0, 1].

for x > 0, x 6= 1.

6. x − 12 x3 ≤ x cos x ≤ x − 61 x3 ≤ sin x ≤ 13 (x cos x + 2x). n o 2 2 x 7. max π2 , ππ2 −x ≤ sinx x ≤ 1 + 13 x2 ≤ tan for 0 ≤ x ≤ π/2. 2 +x x
x π π x 8. π4 1−x 2 < tan 2 x < 2 1−x2 for x ∈ (0, 1).  

n 1 1 9. e 1 − 2n ≤ 1 + n1 < e 1 − 2(n+1) for n ∈ N.

301 √ √
n
n 2πn ne e1/(12n+1) ≤ n! ≤ 2πn 10. e ne ≤
n en ne .


n n 1 1 √4 11. √4nπ 1 − 8n ≤ 2n 1 − 9n . n ≤ nπ 12.

1 2n+2/5

< Hn − ln n − γ