Problems in real analysis advanced calculus on the real axis

R˘adulescu Titu Andreescu Problems in Real Analysis Advanced Calculus on the Real Axis... This carefully written book presents an extremely motivating and original approach,by means of p

Problems in Real Analysis

Teodora-Liliana T R˘adulescu

Vicen¸tiu D R˘adulescu

Titu Andreescu

Problems in Real Analysis

Advanced Calculus on the Real Axis

Titu Andreescu

School of Natural Sciences and Mathematics

University of Texas at Dallas

Library of Congress Control Number: 2009926486

Mathematics Subject Classification (2000): 00A07, 26-01, 28-01, 40-01

© Springer Science +Business Media, LLC 2009

All rights reserved.

permission of the publisher (Springer Science +Business Media, LLC, 233 Spring Street, New York, NY

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The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject

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This work may not be translated or copied in whole or in part without the writtenSpringer Dordrecht Heidelberg London New York

Springer is part of Springer Science+Business Media (www.springer.com)

to do mathematics And what does it mean doing mathematics? In the first place it means to be able to solve mathematical problems.

—George P´olya (1887–1985)

We come nearest to the great when we are great in humility.

—Rabindranath Tagore (1861–1941)

This carefully written book presents an extremely motivating and original approach,

by means of problem-solving, to calculus on the real line, and as such, serves as aperfect introduction to real analysis To achieve their goal, the authors have care-fully selected problems that cover an impressive range of topics, all at the core

of the subject Some problems are genuinely difficult, but solving them will behighly rewarding, since each problem opens a new vista in the understanding ofmathematics This book is also perfect for self-study, since solutions are provided

I like the care with which the authors intersperse their text with careful reviews

of the background material needed in each chapter, thought-provoking quotations,and highly interesting and well-documented historical notes In short, this book alsomakes very pleasant reading, and I am confident that each of its readers will enjoyreading it as much as I did The charm and never-ending beauty of mathematicspervade all its pages

In addition, this little gem illustrates the idea that one cannot learn mathematicswithout solving difficult problems It is a world apart from the “computer addiction”that we are unfortunately witnessing among the younger generations of would-bemathematicians, who use too much ready-made software instead or their brains, orwho stand in awe in front of computer-generated images, as if they had become theessence of mathematics As such, it carries a very useful message

One cannot help comparing this book to a “great ancestor,” the famed Problems

and Theorems in Analysis, by P´olya and Szeg˝o, a text that has strongly influenced

generations of analysts I am confident that this book will have a similar impact

Hong Kong, July 2008 Philippe G Ciarlet

vii

If I have seen further it is by standing on the shoulders of giants.

—Sir Isaac Newton (1642–1727), Letter to Robert Hooke, 1675

Mathematical analysis is central to mathematics, whether pure or applied Thisdiscipline arises in various mathematical models whose dependent variables varycontinuously and are functions of one or several variables Real analysis dates to themid-nineteenth century, and its roots go back to the pioneering papers by Cauchy,Riemann, and Weierstrass

In 1821, Cauchy established new requirements of rigor in his celebrated Cours

d’Analyse The questions he raised are the following:

– What is a derivative really? Answer: a limit

– What is an integral really? Answer: a limit

– What is an infinite series really? Answer: a limit

This leads to

– What is a limit? Answer: a number

And, finally, the last question:

– What is a number?

Weierstrass and his collaborators (Heine, Cantor) answered this question around1870–1872

Our treatment in this volume is strongly related to the pioneering contributions

in differential calculus by Newton, Leibniz, Descartes, and Euler in the seventeenthand eighteenth centuries, with mathematical rigor in the nineteenth century pro-moted by Cauchy, Weierstrass, and Peano

in the integral calculus developed by Riemann and Darboux

Due to the huge impact of mathematical analysis, we have intended in this book

to build a bridge between ordinary high-school or undergraduate exercises and moredifficult and abstract concepts or problems related to this field We present in thisvolume an unusual collection of creative problems in elementary mathematical anal-ysis We intend to develop some basic principles and solution techniques and to offer

a systematic illustration of how to organize the natural transition from solving activity toward exploring, investigating, and discovering new results andproperties

problem-ix

This presentation furthers modern directions

The aim of this volume in elementary mathematical analysis is to introduce,through problems-solving, fundamental ideas and methods without losing sight ofthe context in which they first developed and the role they play in science and partic-ularly in physics and other applied sciences This volume aims at rapidly developingdifferential and integral calculus for real-valued functions of one real variable,giving relevance to the discussion of some differential equations and maximum prin-ciples.

The book is mainly geared toward students studying the basic principles of ematical analysis However, given its selection of problems, organization, and level,

math-it would be an ideal choice for tutorial or problem-solving seminars, particularlythose geared toward the Putnam exam and other high-level mathematical contests

We also address this work to motivated high-school and undergraduate students.This volume is meant primarily for students in mathematics, physics, engineering,and computer science, but, not without authorial ambition, we believe it can be used

by anyone who wants to learn elementary mathematical analysis by solving lems The book is also a must-have for instructors wishing to enrich their teach-ing with some carefully chosen problems and for individuals who are interested insolving difficult problems in mathematical analysis on the real axis The volume isintended as a challenge to involve students as active participants in the course Tomake our work self-contained, all chapters include basic definitions and properties.The problems are clustered by topic into eight chapters, each of them containingboth sections of proposed problems with complete solutions and separate sectionsincluding auxiliary problems, their solutions being left to our readers Throughoutthe book, students are encouraged to express their own ideas, solutions, generaliza-tions, conjectures, and conclusions

prob-The volume contains a comprehensive collection of challenging problems, ourgoal being twofold: first, to encourage the readers to move away from routineexercises and memorized algorithms toward creative solutions and nonstandardproblem-solving techniques; and second, to help our readers to develop a host ofnew mathematical tools and strategies that will be useful beyond the classroom and

in a number of applied disciplines We include representative problems proposed atvarious national or international competitions, problems selected from prestigiousmathematical journals, but also some original problems published in leading publi-cations That is why most of the problems contained in this book are neither standardnor easy The readers will find both classical topics of mathematical analysis on thereal axis and modern ones Additionally, historical comments and developments arepresented throughout the book in order to stimulate further inquiry

Traditionally, a rigorous first course or problem book in elementary mathematicalanalysis progresses in the following order:

Sequences

Functions=⇒ Continuity =⇒ Differentiability =⇒ Integration

Limits

However, the historical development of these subjects occurred in reverse order:

Cauchy(1821) ⇐= Weierstrass (1872) ⇐= Newton (1665)

Leibniz (1675) ⇐=

ArchimedesKepler (1615)Fermat (1638)This book brings to life the connections among different areas of mathematicalanalysis and explains how various subject areas flow from one another The vol-ume illustrates the richness of elementary mathematical analysis as one of the mostclassical fields in mathematics The topic is revisited from the higher viewpoint ofuniversity mathematics, presenting a deeper understanding of familiar subjects and

an introduction to new and exciting research fields, such as Ginzburg–Landau tions, the maximum principle, singular differential and integral inequalities, andnonlinear differential equations

equa-The volume is divided into four parts, ten chapters, and two appendices, asfollows:

Part I Sequences, Series, and Limits

Chapter 1 Sequences

Chapter 2 Series

Chapter 3 Limits of Functions

Part II Qualitative Properties of Continuous and Differentiable FunctionsChapter 4 Continuity

Chapter 5 Differentiability

Part III Applications to Convex Functions and Optimization

Chapter 6 Convex Functions

Chapter 7 Inequalities and Extremum Problems

Part IV Antiderivatives, Riemann Integrability, and Applications

Chapter 8 Antiderivatives

Chapter 9 Riemann Integrability

Chapter 10 Applications of the Integral Calculus

Appendix A Basic Elements of Set Theory

Appendix B Topology of the Real Line

Each chapter is divided into sections Exercises, formulas, and figures are bered consecutively in each section, and we also indicate both the chapter and thesection numbers We have included at the beginning of chapters and sections quo-tations from the literature They are intended to give the flavor of mathematics as

num-a science with num-a long history This book num-also contnum-ains num-a rich glossnum-ary num-and index, num-aswell as a list of abbreviations and notation

Key features of this volume:

– contains a collection of challenging problems in elementary mathematicalanalysis;

– includes incisive explanations of every important idea and develops illuminatingapplications of many theorems, along with detailed solutions, suitable cross-references, specific how-to hints, and suggestions;

– is self-contained and assumes only a basic knowledge but opens the path to petitive research in the field;

com-– uses competition-like problems as a platform for training typical inventive skills;– develops basic valuable techniques for solving problems in mathematical ana-lysis on the real axis;

– 38 carefully drawn figures support the understanding of analytic concepts;– includes interesting and valuable historical account of ideas and methods inanalysis;

– contains excellent bibliography, glossary, and index

The book has elementary prerequisites, and it is designed to be used for lecturecourses on methodology of mathematical research or discovery in mathematics Thiswork is a first step toward developing connections between analysis and other math-ematical disciplines, as well as physics and engineering

The background the student needs to read this book is quite modest Anyonewith elementary knowledge in calculus is well-prepared for almost everything to

be found here Taking into account the rich introductory blurbs provided with eachchapter, no particular prerequisites are necessary, even if a dose of mathematical so-phistication is needed The book develops many results that are rarely seen, and evenexperienced readers are likely to find material that is challenging and informative.Our vision throughout this volume is closely inspired by the following words ofGeorge P´olya [90] (1945) on the role of problems and discovery in mathematics:

Infallible rules of discovery leading to the solution of all possible mathematical problems would be more desirable than the philosopher’s stone, vainly sought by all alchemists The first rule of discovery is to have brains and good luck The second rule of discovery is to sit tight and wait till you get a bright idea Those of us who have little luck and less brain sometimes sit for decades The fact seems to be, as Poincar´e observed, it is the man, not the method, that solves the problem.

Despite our best intentions, errors are sure to have slipped by us Please let usknow of any you find

Vicent¸iu R˘adulescu Titu Andreescu

We acknowledge, with unreserved gratitude, the crucial role of Professors ine Bandle, Wladimir-Georges Boskoff, Louis Funar, Patrizia Pucci, Richard Stong,and Michel Willem, who encouraged us to write a problem book on this subject.Our colleague and friend Professor Dorin Andrica has been very interested in thisproject and suggested some appropriate problems for this volume We warmly thankProfessors Ioan S¸erdean and Marian Tetiva for their kind support and useful discus-sions.

Cather-This volume was completed while Vicent¸iu R˘adulescu was visiting the sity of Ljubljana during July and September 2008 with a research position funded

Univer-by the Slovenian Research Agency He would like to thank Professor Duˇsan Repovˇsfor the invitation and many constructive discussions

We thank Dr Nicolae Constantinescu and Dr Mirel Cos¸ulschi for the sional drawing of figures contained in this book

profes-We are greatly indebted to the anonymous referees for their careful reading ofthe manuscript and for numerous comments and suggestions These precious con-structive remarks were very useful to us in the elaboration of the final version of thisvolume

We are grateful to Ann Kostant, Springer editorial director for mathematics, forher efficient and enthusiastic help, as well as for numerous suggestions related toprevious versions of this book Our special thanks go also to Laura Held and to theother members of the editorial technical staff of Springer New York for the excellentquality of their work

We are particularly grateful to copyeditor David Kramer for his guidance, oughness and attention to detail

thor-V R˘adulescu acknowledges the support received from the Romanian ResearchCouncil CNCSIS under Grant 55/2008 “Sisteme diferent¸iale ˆın analiza neliniar˘a s¸iaplicat¸ii.”

xiii

Foreword vii

Preface ix

Acknowledgments xiii

Abbreviations and Notation xix

Part I Sequences, Series, and Limits 1 Sequences 3

1.1 Main Definitions and Basic Results 3

1.2 Introductory Problems 7

1.3 Recurrent Sequences 18

1.4 Qualitative Results 30

1.5 Hardy’s and Carleman’s Inequalities 45

1.6 Independent Study Problems 51

2 Series 59

2.1 Main Definitions and Basic Results 59

2.2 Elementary Problems 66

2.3 Convergent and Divergent Series 73

2.4 Infinite Products 86

2.5 Qualitative Results 89

2.6 Independent Study Problems 110

3 Limits of Functions 115

3.1 Main Definitions and Basic Results 115

3.2 Computing Limits 118

3.3 Qualitative Results 124

3.4 Independent Study Problems 133

xv

Part II Qualitative Properties of Continuous and Differentiable Functions

4 Continuity 139

4.1 The Concept of Continuity and Basic Properties 139

4.2 Elementary Problems 144

4.3 The Intermediate Value Property 147

4.4 Types of Discontinuities 151

4.5 Fixed Points 154

4.6 Functional Equations and Inequalities 163

4.7 Qualitative Properties of Continuous Functions 169

4.8 Independent Study Problems 177

5 Differentiability 183

5.1 The Concept of Derivative and Basic Properties 183

5.2 Introductory Problems 198

5.3 The Main Theorems 218

5.4 The Maximum Principle 235

5.5 Differential Equations and Inequalities 238

5.6 Independent Study Problems 252

Part III Applications to Convex Functions and Optimization 6 Convex Functions 263

6.1 Main Definitions and Basic Results 263

6.2 Basic Properties of Convex Functions and Applications 265

6.3 Convexity versus Continuity and Differentiability 273

6.4 Qualitative Results 278

6.5 Independent Study Problems 285

7 Inequalities and Extremum Problems 289

7.1 Basic Tools 289

7.2 Elementary Examples 290

7.3 Jensen, Young, H¨older, Minkowski, and Beyond 294

7.4 Optimization Problems 300

7.5 Qualitative Results 305

7.6 Independent Study Problems 308

Part IV Antiderivatives, Riemann Integrability, and Applications 8 Antiderivatives 313

8.1 Main Definitions and Properties 313

8.2 Elementary Examples 315

8.3 Existence or Nonexistence of Antiderivatives 317

8.4 Qualitative Results 319

8.5 Independent Study Problems 324

9 Riemann Integrability 325

9.1 Main Definitions and Properties 325

9.2 Elementary Examples 329

9.3 Classes of Riemann Integrable Functions 337

9.4 Basic Rules for Computing Integrals 339

9.5 Riemann Iintegrals and Limits 341

9.6 Qualitative Results 351

9.7 Independent Study Problems 367

10 Applications of the Integral Calculus 373

10.1 Overview 373

10.2 Integral Inequalities 374

10.3 Improper Integrals 390

10.4 Integrals and Series 402

10.5 Applications to Geometry 406

10.6 Independent Study Problems 409

Part V Appendix A Basic Elements of Set Theory 417

A.1 Direct and Inverse Image of a Set 417

A.2 Finite, Countable, and Uncountable Sets 418

B Topology of the Real Line 419

B.1 Open and Closed Sets 419

B.2 Some Distinguished Points 420

Glossary 421

References 437

Index 443

We have tried to avoid using nonstandard abbreviations as much as possible Otherabbreviations include:

AMM American Mathematical Monthly

GMA Mathematics Gazette, Series A

MM Mathematics Magazine

IMO International Mathematical Olympiad

IMCUS International Mathematics Competition for University StudentsMSC Mikl´os Schweitzer Competitions

Putnam The William Lowell Putnam Mathematical Competition

SEEMOUS South Eastern European Mathematical Olympiad for University

Students

Notation

We assume familiarity with standard elementary notation of set theory, logic,algebra, analysis, number theory, and combinatorics The following is notation thatdeserves additional clarification

N the set of nonnegative integers(N = {0,1,2,3, })

N∗ the set of positive integers(N∗ = {1,2,3, })

Z the set of integer real numbers(Z = { ,−3,−2,−1,0,1,2,3, })

Z∗ the set of nonzero integer real numbers(Z∗ = Z \ {0})

Q the set of rational real numbers

Q =m

n ; m ∈ Z, n ∈ N ∗ , m and n are relatively prime

R the set of real numbers

R∗ the set of nonzero real numbers(R∗ = R \ {0})

R+ the set of nonnegative real numbers(R+= [0,+∞))

R the completed real line

R = R ∪ {−∞,+∞}

xix

C the set of complex numbers

sup A the least upper bound of the set A ⊂ R

inf A the greatest lower bound of the set A ⊂ R

x+ the positive part of the real number x (x+= max{x,0})

x − the negative part of the real number x (x − = max{−x,0})

|x| the modulus (absolute value) of the real number x ( |x| = x++ x −)

{x} the fractional part of the real number x (x = [x] + {x})

Card(A) cardinality of the finite set A

dist(x,A) the distance from x ∈ R to the set A ⊂ R (dist(x,A) = inf{|x − a|; a ∈ A})

IntA the set of interior points of A ⊂ R

f (A) the image of the set A under a mapping f

f −1 (B) the inverse image of the set B under a mapping f

f ◦ g the composition of functions f and g: ( f ◦ g)(x) = f (g(x))

k ≥n x k

f (n) (x) nth derivative of the function f at x

C n (a,b) the set of n-times differentiable functions f : (a,b)→R such that f (n)is

+∂2f

∂x2 2

+ ··· +∂2f

∂x2

N

Landau’s notation f (x) = o(g(x)) as x→x0if f (x)/g(x)→0 as x→x0

f (x) = O(g(x)) as x→x0if f (x)/g(x) is bounded in a neighborhood of x0

f ∼ g as x→x0if f (x)/g(x)→1 as x→x0

Hardy’s notation f ≺≺ g as x→x0if f (x)/g(x)→0 as x→x0

f  g as x→x0if f (x)/g(x) is bounded in a neighborhood of x0

As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.

—Albert Einstein (1879–1955)

Abstract In this chapter we study real sequences, a special class of functions

whose domain is the setN of natural numbers and range a set of real numbers

1.1 Main Definitions and Basic Results

Hypotheses non fingo [“I frame no hypotheses.”]

Sir Isaac Newton (1642–1727)

Sequences describe wide classes of discrete processes arising in various tions The theory of sequences is also viewed as a preliminary step in the attempt tomodel continuous phenomena in nature Since ancient times, mathematicians haverealized that it is difficult to reconcile the discrete with the continuous We under-stand counting 1, 2, 3, up to arbitrarily large numbers, but do we also under-

applica-stand moving from 0 to 1 through the continuum of points between them? Around450

essential way As he put it in his paradox of dichotomy:

course) before it arrives at the end.

Aristotle, Physics, Book VI, Ch 9

A sequence of real numbers is a function f : N→R (or f : N ∗ →R) We usually

write a n (or b n , x n , etc.) instead of f (n) If (a n)n≥1 is a sequence of real numbers

and if n1< n2< ··· < n k < ··· is an increasing sequence of positive integers, then

the sequence(a n k)k≥1 is called a subsequence of (a n)n ≥1.

n n ≥1 is said to be nondecreasing (resp.,

increas-ing) if a n ≤ a n+1(resp., a n < a n+1), for all n ≥ 1 The sequence (a n)n ≥1 is called

A sequence of real numbers(a )

Springer Science + Business Media, LLC 2009

Problems in Real Analysis: Advanced Calculus on the Real Axis,

There is no motion because that which is moved must arrive at the middle (of its

T.-L Rădulescu et al.,

Fig 1.1 Adapted from G.H Hardy, Pure Mathematics, Cambridge University Press, 1952.

We recall in what follows some basic definitions and properties related to

sequences One of the main notions we need in the sequel is that of convergence.

Let(a n)n≥1 be a sequence of real numbers We say that(a n)n≥1 is a convergent

sequence if there exists  ∈ R (which is called the limit of (a n)n≥1) such that foreach neighborhoodN of , we have a n ∈ N for all n ≥ N, where N is a positive

integer depending onN In other words, (a n)n≥1converges to if and only if for

eachε> 0 there exists a natural number N = N(ε) such that |a n − | <ε, for all

n ≥ N In this case we write lim n →∞a n =  or a n → as n→∞ The concentration of

a n’s in the strip( −ε, +ε) for n ≥ N is depicted in Figure 1.1.

seq-e (seq-e= 2.71828 ) We will prove in the next chapter that e is an irrational number.

Another important example is given by the following formula due to Stirling, which

asserts that, asymptotically, n! behaves like n ne−n √

Stirling’s formula is important in calculating limits, because without this asymptotic

property it is difficult to estimate the size of n! for large n In this capacity, it plays

an important role in probability theory, when it is used in computing the probableoutcome of an event after a very large number of trials We refer to Chapter 9 for acomplete proof of Stirling’s formula

We say that the sequence of real numbers(a n)n≥1 has limit+∞ (resp., −∞) if

to eachα∈ R there corresponds some positive integer N0such that n ≥ N0implies

a n >α(resp., a n <α) We do not say that (a n)n≥1converges in these cases

An elementα∈ R is called an accumulation point of the sequence (a n)n≥1 ⊂ R

if there exists a subsequence(a n k)k≥1 such that a n k →α as n k →∞.

Any convergent sequence is bounded The converse is not true (find a example!), but we still have a partial positive answer, which is contained in thefollowing result, which is due to Bernard Bolzano (1781–1848) and

counter-Karl Weierstrass (1815–1897) It seems that this theorem was revealed for the firsttime in Weierstrass’s lecture of 1874.

convergent subsequence

However, some additional assumptions can guarantee that a bounded sequenceconverges An important criterion for convergence is stated in what follows

monotone Then(a n)n≥1is a convergent sequence If increasing, thenlimn →∞a n=supn a n, and if decreasing, thenlimn →∞a n= infn a n

In many arguments a central role is played by the following principle due toCantor

Nested Intervals Theorem.Suppose thatI n = [a n ,b n]are closed intervals suchthatI n+1⊂ I n, for alln ≥ 1 Iflimn →∞(b n − a n) = 0, then there is a unique realnumberx0that belongs to everyI n

A sequence(a n)n≥1 of real numbers is called a Cauchy sequence [Augustin Louis

Cauchy (1789–1857)] if for every ε> 0 there is a natural number Nε such that

|a m − a n | <ε, for all m ,n ≥ Nε A useful result is that any Cauchy sequence is abounded sequence The following strong convergence criterion reduces the study ofconvergent sequences to that of Cauchy sequences

Cauchy’s Criterion.A sequence of real numbers is convergent if and only if it

is a Cauchy sequence

Let (a n)n≥1 be an arbitrary sequence of real numbers The limit inferior of (a n)n≥1 (denoted by lim infn →∞a n ) is the supremum of the set X of x ∈ R such

that there are at most a finite numbers of n ∈ N ∗ for which a

n < x The limit perior of (a n)n≥1(denoted by lim supn →∞a n ) is the infimum of the set Y of y ∈ R

su-such that there are finitely many positive integers n for which a n > y Equivalent

characterizations of these notions are the following:

(i)  = limsup n →∞a n( ∈ R) if and only if wheneverα<  the set {n ≥ N ∗ ; a

n >

α} is infinite, and whenever  <β the set{n ≥ N ∗ ; a

n >β} is finite;

(ii)  = limsup n →∞a nif and only if = inf m≥1supn ≥m a n

Exercise Formulate the above characterizations for lim infn →∞a n

The limit inferior and the limit superior of a sequence always exist, possibly in

R Moreover, the following relations hold:

The existence of the limit of a sequence is closely related to “lim inf” and

“lim sup.” More precisely, the sequence (a n)n≥1 has a limit if and only iflim infn →∞a n= limsupn →∞a n

A very useful result in applications concerns the following link between the ratio

and the nth root of the terms of a sequence of positive numbers.

Theorem.Let(a n)n ≥1be a sequence of positive numbers Then

can give an elementary proof to the fact that a n > a n+1, for all n ≥ 3 Indeed, define

b n = (a n /a n+1)n (n+1) , for all n ≥ 2 Then b n > b n−1 for n ≥ 2 is equivalent to n2=

a 2n n > a n+1

n+1a n n−1 −1 = n2− 1 It remains to show that b n > 1 for all n ≥ 3 This follows

from the fact that(b n)n≥2 is increasing, combined with b3= 34/43= 81/64 > 1.

A sequence that is not convergent is called a divergent sequence An important example of a divergent sequence is given by a n = 1 + 1/2 + 1/3 + ···+ 1/n, n ≥ 1.

Indeed, since(a n)n ≥1is increasing, it has a limit ∈ R ∪ {+∞} Assuming that  is

finite, it follows that

which is impossible because 1/1 > 1/2, 1/3 > 1/4, 1/5 > 1/6, and so on

Con-sequently, a n → + ∞ as n→∞ The associated series 1 + 1/2 + 1/3 + 1/4 + ··· is

usually called the harmonic series Furthermore, if p < 1 we have 1/n p > 1/n, for

all n ≥ 2 It follows that the sequence (a n)n ≥1 defined by a n= ∑n

2p+ 1

4p + ··· + (2n)1 p

+

1

2p+ 1

4p + ··· + (2n)1 p

+

1

because a n < a 2n+1 Thus(1 − 21−p )a 2n+1 < 1 Since p > 1, we have 1 − 21−p > 0.

Hence a 2n+1 < (1 − 21−p)−1 for all n ≥ 1 So, the sequence (a n)n≥1 (for p > 1)

is increasing and bounded above by(1 − 21−p)−1, which means that it converges.

The corresponding limit limn →∞∑n

k=11/k p= ∑∞n=11/n pis denoted byζ(p) (p > 1) and is called the Riemann zeta function [Georg Friedrich Bernhard Riemann

(1826–1866)] This function is related to a celebrated mathematical conjecture(Hilbert’s eighth problem) The Riemann hypothesis, formulated in 1859, is closelyrelated to the frequency of prime numbers and is one of the seven grand challenges

of mathematics (“Millennium Problems,” as designated by the Clay MathematicsInstitute; see http://www.claymath.org/millennium/).

The following result (whose “continuous” variant is l’Hˆopital’s rule for tiable functions) provides us a method to compute limits of the indeterminate form

differen-0/0 or ∞/∞ This property is due to Otto Stolz (1859–1906) and Ernesto Ces`aro

Then there existslimn →∞a n /b nand, moreover,limn →∞a n /b n = 

(ii) Assume thatb n →+ ∞asn→∞and that(b n)n ≥1is increasing for all sufficiently

largen Suppose that there exists

We first prove with elementary arguments the following basic result

num-bersa1,a2, ,a nwe have

a1+ a2+ ··· + a n

a1a2···a n

Replacing a k with a k √ n

a1···a n (for 1≤ k ≤ n), we have a1a2···a n= 1, so it

is enough to prove that a1+ a2+ ··· + a n ≥ n We argue by induction For n = 1

the property is obvious Passing from n to n + 1, we can assume that a1≤ 1 ≤ a2.Hence(1−a1)(a2−1) ≥ 1, that is, a1+a2≥ 1+a1a2 Next, we set b1= a1a2and,

for any k = 2, ,n, we set b k = a k+1 Hence b1b2···b n= 1 So, by the inductionhypothesis,

a a + a + ··· + a ≥ n.

It follows that

a1+ a2+ ··· + a n+1≥ a1a2+ 1 + a3+ ··· + a n+1≥ n + 1.

The AM–GM inequality also implies a relationship between the harmonic mean

and the geometric mean: for any positive numbers a1, a2, ,a nwe have

which shows that the limit sought is A1:= min{Aj; 1≤ j ≤ k}  

Comments We easily observe that the above result does not remain true if we

do not assume that the numbers A1, ,A k are nonnegative Moreover, in such a

case it is possible that the sequence defined by B n:=A n + ··· + A n

k

1/n

is not evenconvergent (give an example!)

We have already provided some basic examples of convergent and divergentsequences The next exercise shows how, by means of monotony principles, wecan construct further examples of such sequences

1.2.2 Prove thatlimn →∞√ n

n!/n = e −1 An alternative argument is based on the Stolz–Ces`aro lemma

applied to a n = log(n!/n n ) and b n = n and using again lim n →∞

1+ n −11/n= e



A nonobvious generalization of the above property is stated below Our proofapplies subtle properties of real-valued functions (see Chapter 5) but we stronglysuggest that the reader refine the monotony arguments developed above.

apply the Lagrange mean value theorem to the function f (x) = (x3ln x)/3−x3/9 on

the interval[k,k + 1], 1 ≤ k ≤ n.

We have seen above that √ n

n!→∞ as n→∞ A natural question is to study the

asymptotic behavior of the difference of two consecutive terms of this sequence.The next problem was published in 1901 and is due to the Romanian mathemati-cian Traian Lalescu (1882–1929), who wrote one of the first treatises on integralequations

1.2.3 Find the limit of the sequence(a n)n≥2defined bya n= n+1 (n + 1)!− √ n

So, by (1.1) and (1.2), we obtain that a n →e −1 as n →∞  

then(b n − 1)/lnb n →1 as n→∞ This follows directly either by applying the Stolz–

Ces`aro lemma or after observing that

ln e= 1.

of the sequence(x n)n≥1defined by

We already know that the sequence(s n)n ≥1 defined by s n = 1 + 1/2 + ···+ 1/n

diverges to+∞ In what follows we establish the asymptotic behavior for s n /n as

n →∞ As a consequence, the result below implies that lim n →∞n (1 − √ n

n) = ∞

1.2.4 Consider the sequence(s n)n≥1defined bys n = 1 + 1/2 + ···+ 1/n Provethat

(a) n (n + 1)1/n < n + s n, for all integersn > 1;

(b) (n − 1)n −1/(n−1) < n − s n, for all integersn > 2

Solution (a) Using the AM–GM inequality we obtain

n + s n

n =(1 + 1) +



1+1 2

1/(n−1)

= n −1/(n−1)



The following exercise involves a second-order linear recurrence

1.2.5 Letα∈ (0,2) Consider the sequence defined by

x n+1=αx n + (1 −α)x n−1 , for alln ≥ 1.

Find the limit of the sequence in terms ofα,x0, andx1

Solution We have x n − x n−1= (α− 1)(x n − x n−1 ) It follows that x n − x n−1=(α− 1) n−1 (x1− x0) Therefore

Find a necessary and sufficient condition such that the sequence is convergent.

Solution If limn →∞x n =  then  = a + 2, that is,

1.2.7 Let f (x) = 1/4 + x − x2 For anyx ∈ R, define the sequence(x n)n≥0 by

x0= xandx n+1= f (x n) If this sequence is convergent, letx∞be its limit

(a) Show that ifx= 0, then the sequence is bounded and increasing, and computeits limitx∞= 

(b) Find all possible values of and the corresponding real numbersxsuch that

Since all the terms of the sequence are positive, we deduce that = 1/2.

(b) By the definition of f it follows that

We now prove that if x ∈ (−1/2,3/2), then the sequence converges and its limit

equals 1/2 Indeed, in this case we have



f(x)−12

The next exercise gives an example of a convergent sequence defined by means

of an integer-valued function We invite the reader to establish more properties ofthe function· : [0,∞)→N.

1.2.8 For any integern ≥ 1, letnbe the closest integer to√

Solution Since(k − 1/2)2= k2− k + 1/4 and (k + 1/2)2= k2+ k + 1/4, it

fol-lows thatn = k if and only if k2− k + 1 ≤ n ≤ k2+ k Hence

prop-1.2.9 (i) Let(a n)n≥1be a sequence of real numbers such thata1= 1anda n+1>

3a n /2for alln ≥ 1 Prove that the sequence(b n)n≥1defined by

(3/2) n−1

either has a finite limit or tends to infinity

(ii) Prove that for allα> 1there exists a sequence(a n)n≥1with the same propertiessuch that

lim

n →∞

a n

(3/2) n−1 =α.

International Mathematical Competition for University Students, 2003

Solution (i) Our hypothesis a n+1> 3a n /2 is equivalent to b n+1> b n, and theconclusion follows immediately

(ii) For anyα> 1 there exists a sequence 1 = b1< b2< ··· that converges toα.

Choosing a n = (3/2) n−1 b

n, we obtain the required sequence(a n)n ≥1. 

Qualitative properties of a sequence of positive integers are established in thenext example

1.2.10 Consider the sequence(a n)n≥1defined bya n = n2+ 2

(i) Find a subsequence

(ii) Consider the subsequence(b n)n≥1 defined by b1= 3, b2= b2

1+ 2, and, for any

integer n ≥ 3, b n = (b1···b n−1)2+ 2  

The Stolz–Ces`aro lemma is a powerful instrument for computing limits of

seq-uences (always keep in mind that it gives only a sufficient condition for the

exis-tence of the limit!) A simple illustration is given in what follows

1.2.11 The sequence of real numbers(x n)n≥1 satisfieslimn →∞(x 2n + x 2n+1) =

315andlimn →∞(x 2n + x 2n −1) = 2003 Evaluatelimn →∞(x 2n /x 2n+1)

Harvard–MIT Mathematics Tournament, 2003

Solution Set a n = x 2n and b n = x 2n+1and observe that

as n →∞ Thus, by the Stolz–Ces`aro lemma, the required limit equals −1  

Remark We observe that the value of limn →∞(x 2n /x 2n+1) does not depend on

the values of limn →∞(x 2n + x 2n+1) and limn →∞(x 2n + x 2n −1) but only on the vergence of these two sequences

con-We refine below the asymptotic behavior of a sequence converging to zero Theproof relies again on the Stolz–Ces`aro lemma, and the method can be extended tolarge classes of recurrent sequences

1.2.12 Let (a n)n≥1 be a sequence of real numbers such that limn →∞a n∑n

k=1a2k Then the condition a n s n →1 implies that s n →∞ and

a n →0 as n→∞ Hence we also have that a n s n−1 →1 as n→∞ Therefore

Thus, by the Stolz–Ces`aro lemma, s3/n→3 as n→∞, or equivalently, lim n →∞n −2/3

s2= 32/3 So, by (1.3), we deduce that lim

n →∞(3n)1/3 a

n = 1  

We study in what follows a sequence whose terms are related to the coefficients

of certain polynomials

1.2.13 Fix a real numberx = −1 ± √2 Consider the sequence(s n)n≥1defined

bys n= ∑n

k=0a k x ksuch thatlimn →∞s n = 1/(1−2x −x2) Prove that for any integer

n ≥ 0, there exists an integermsuch thata2+ a2

erty related to bounded sequences of real numbers Does the property given below

remain true if b nare not real numbers?

1.2.14 Suppose that (a n)n≥1 is a sequence of real numbers such that

limn →∞a n= 1and(b n)n≥1is a bounded sequence of real numbers Ifkis a positiveinteger such thatlimn →∞(b n − a n b n +k ) = , prove that = 0

Solution (C˘alin Popescu) Let b= liminfn →∞b n and B= limsupn →∞b n Since

(b n)n ≥1 is bounded, both b and B are finite Now there are two subsequences (b p r)and(b q r ) of (b n)n ≥1 such that b p r → b and b q r → B as r → ∞ Since a n → 1 and

b n − a n b n +k →  as n → ∞, it follows that the subsequences (b p r +k ) and (b q r +k) of

(b n)n≥1 tend to b −  and B − , respectively, as r → ∞ Consequently, b −  ≥ b and

B −  ≤ B, hence  = 0  

Elementary trigonometry formulas enable us to show in what follows that a verysimple sequence diverges A deeper property of this sequence will be proved inProblem 1.4.26

1.2.15 Seta n = sinn, for anyn ≥ 1 Prove that the sequence(a n)n ≥1is divergent.

Solution Arguing by contradiction, we assume that the sequence(a n)n≥1is

con-vergent Let a= limn →∞sin n Using the identity

sin(n + 1) = sinncos1 + cosnsin1

we deduce that the sequence(cosn) converges and, moreover,

a = acos1 + bsin1,

where b= limn →∞cos n Using now

cos(n + 1) = cosncos1 − sinnsin1

we deduce that

b = bcos1 − asin1.

These two relations imply a = b = 0, a contradiction, since a2+ b2= 1  

Under what assumption on the function f one can deduce that a sequence (a n)n≥1

converges, provided the sequence( f (a n))n≥1is convergent? The next exercise offers

a sufficient condition such that this happens

1.2.16 Let(a n)n≥1be a sequence of real numbers such thata n ≥ 1for allnandthe sequence



n≥1is bounded (since it is

conver-gent)

Arguing by contradiction, assume that A > a Choose subsequences (a n k)k≥1and

(a m k)k ≥1 such that a n k →A as k→∞ and a m k →a as k→∞ Therefore a n k +1/a n k →A+

1/A as k→∞ and a m k + 1/a m k →a + 1/a as k→∞ But the sequencea n + a −1

Solution We first observe that for all x > −1,

hence the desired conclusion 

1.2.19 Let(x n)n ≥1be a sequence and sety n = x n −1 + 2x nfor alln ≥ 2 Supposethat the sequence(y n)n ≥2converges Show that the sequence(x n)n ≥1converges.

Solution Let ˜y= limn →∞y nand set ˜x = ˜y/3 We show that ˜x = lim n →∞x n For

ε> 0 there is a positive integer n0such that for all n ≥ n0,|y n − ˜y| <ε/2 Hence

2+ 2−(m+1) |x n−1 − ˜x|.

By taking m large enough, 2 −(m+1) |x n −1 − ˜x| <ε/2 Thus for all sufficiently large

k, |x k − ˜x| <ε 

1.2.20 (i) Let(a n)n≥1be a bounded sequence of real numbers such thata n = 0

for alln ≥ 1 Show that there is a subsequence(b n)n≥1 of(a n)n≥1 such that thesequence(b n+1/b n)n≥1converges

(ii) Let (a n)n≥1 be a sequence of real numbers such that for every subsequence

(b n)n≥1of(a n)n≥1,limn →∞|b n+1/b n | ≤ 1 Prove that(a n)n≥1has at most twolimit points Moreover, if these limit points are not equal and if one of them is

t, then the other is−t

Solution (i) We distinguish two cases.

CASE1: there existsε> 0 such that for infinitely many k, |a k | ≥ε Let(b n)n≥1bethe subsequence of(a n)n≥1 consisting of those a kwith|a k | ≥ε Then|b n+1/b n | ≤

(supk |a k |)/εfor all n.

CASE2: for everyε> 0 there is an integer Nε such that for all k ≥ Nε,|a k | <ε.Then there is a subsequence(b n)n≥1of(a n)n≥1such that|b n+1| < |b n |.

In both cases, the sequence of ratios(b n+1/b n)n≥1is bounded, and thus by theBolzano–Weierstrass theorem there exists a convergent subsequence

(ii) By hypothesis it follows that the sequence(a n)n≥1 is bounded Now supposethat(a n)n≥1 has at least two distinct limit points s and t, where s = −t Then

there are subsequences(b p)p≥1and(c q)q≥1 of(a n)n≥1 converging to s and t

respectively Let(d j)j≥1be the following subsequence of(a n)n≥1:

d j=



b p if j is odd (that is, d1= b1, d3= b2, ),

c q if j is even (that is, d2= c1, d4= c2, ).

Then limj →∞|d j+1/d j | does not exist This contradiction concludes the proof  

1.2.21 Let(a n)n≥1 be a sequence of real numbers such thatlimn →∞(2a n+1−

a n ) =  Prove thatlimn →∞a n = 

Solution We first show that the sequence(a n)n≥1is bounded Since the sequence

(2a n+1−a n)n≥1 is bounded (as a convergent sequence), there exists M > 0 such that

|a1| ≤ M and for all n ≥ 1, |2a n+1− a n | ≤ M We prove by induction that |a n | ≤ M

for all n Indeed, suppose that |a n | ≤ M Then

This concludes the induction and shows that(a n)n ≥1is bounded.

Taking lim sup in

This yields lim supn →∞a n ≤  Similarly, we deduce that liminf n →∞a n ≥  We

conclude that limn →∞a n =   

1.2.22 Prove that a countably infinite set of positive real numbers with a finitenonzero limit point can be arranged in a sequence(a n)n≥1 such that (a1/n

n )n≥1 isconvergent

P Orno, Math Magazine, Problem 1021

Solution Let(x n)n ≥1 denote the real numbers of concern and let a be a nonzero

finite limit point Choose A such that 1 /A < a < A and let a1denote the x nof est subscript that lies in the interval(1/A,A) Assuming a1, a2, , a k −1 to have

small-been chosen, let a k be the x n of smallest subscript different from the n − 1 already

chosen lying in the interval(A − n ,A √ n ) Since a is a limit point, such an x n can

be found, and since A √ n →∞ and A − n →0, we deduce that every x neventually isincluded in an interval of the form(A − n ,A √ n) and will therefore eventually be-come part of the sequence(a n)n≥1 For each n we have 1 /A √ n < a n < A √ n andtherefore(1/A)1/ √ n < a1/n

n < A1/ √ n But limn →∞A1/ √ n= limn →∞(1/A)1/ √ n= 1and therefore limn →∞a1n /n = 1  

1.3 Recurrent Sequences

Mathematics is trivial, but I can’t do

my work without it.

Richard Feynman (1918–1988)

Recurrent sequences are widely encountered in nature, and they should be seen as a

major step toward the discretization of various continuous models One of the most

famous recurrent sequences goes back to Leonardo Fibonacci (1170–1250) About

1202, Fibonacci formulated his famous rabbit problem, which led to the Fibonaccisequence 1,1,2,3,5,8,13, The terms of this sequence have beautiful properties,

mainly related to the golden ratio At the same time, it appears in several

applica-tions in biology, including leaves and petal arrangements, branching plants, rabbitcolonies, and bees’ ancestors (see Figure 1.2) It seems that the Fibonacci recurrence

F n = F n−1 +F n−2was first written down by Albert Girard around 1634 and solved by

de Moivre in 1730 Bombelli studied the equation y n = 2 + 1/y n−1in 1572, which

is similar to the equation z n = 1 + 1/z n−1satisfied by ratios of Fibonacci numbers,

in order to approximate√

2 Fibonacci also gave a rough definition for the concept

of continued fractions that is intimately associated with difference equations, which

are now intensively applied in the modeling of continuous phenomena A more cise definition was formulated by Cataldi around 1613 The method of recursionwas significantly advanced with the invention of mathematical induction by Mau-rolico in the sixteenth century and with its development by Fermat and Pascal in

pre-Number of ancestors 1 2 3 5 8 13

Fig 1.2 Bees have Fibonacci-number ancestors.

the seventeenth century Sir Thomas Harriet (1560–1621) invented the calculus offinite differences, and Henry Briggs (1556–1630) applied it to the calculation oflogarithms It was rediscovered by Leibniz around 1672 Sir Isaac Newton, Leon-hard Euler (1707–1783), Joseph-Louis Lagrange (1736–1813), Carl Friedrich Gauss(1777–1855), and many others used this calculus to study interpolation theory Thetheory of finite differences was developed largely by James Stirling (1692–1770)

in the early eighteenth century Recurrence relations were extended to the study

of several sequences A celebrated example is related to the cooperative recurrences

x n = (x n−1 +y n−1 )/2 and y n = √x n−1 y n−1, which were associated by Lagrange withthe evaluation of elliptic integrals

In this section we are interested in the study of recurrent sequences that are notnecessarily linear

The first example gives an interesting connection with Euler’s indicator function

1.3.1 Consider the sequence(a n)n≥1defined bya1= 1,a2= 2,a3= 24, and

i=1(2i − 1) To prove that a n is a multiple of n, let n= 2k m, where

m is odd Then k ≤ n ≤ n(n−1)/2 and there exists i ≤ m−1 such that m is a divisor

of 2i − 1 [for i =ϕ(m), whereϕ denotes Euler’s function] Consequently, a nis a

multiple of n 

In certain cases a sequence defined by a nonlinear recurrence relation may bedescribed in a linear way We give below an example

1.3.2 Define the sequence (a n)n ≥1 by a1= a2= 1 and a n = a2

n−1 + 2/a n −2,

for anyn ≥ 3 Prove that for alln,a nis an integer

Solution The recurrence relation is of second order, so we try to find a n of

the form a n=αq n1+βq n2 From a1= a2= 1 we obtainα = q2− 1/q1(q2− q1),

β= 1 − q1/q2(q2− q1) Substituting in the recurrence relation, we obtain

so q1q2= 1 and q1+ q2= 4 It follows that q1and q2are the roots of the

second-order equation q2− 4q + 1 = 0 This is the characteristic (secular) equation of the

sequence defined by

a n = 4a n−1 − a n−2 , a1= a2= 1.

Thus we obtain that if a n−1 ∈ N and a n−2 ∈ N then a n ∈ N  

Recurrent sequences may be also applied to finding coefficients in various sions, as illustrated in the following exercise

expan-1.3.3 Consider the expression

Find the coefficient ofx2

Solution Let a n be the coefficient of x2, and b n the coefficient of x We observe that for any n, the term not containing x in the above development is 4 It follows

that

a n = 4a n−1 + b2

n−1 and b n = 4b n−1 ,

where a1= 1 and b1= −4 We first obtain that b n = −4 n Substituting in the

rec-urrence relation corresponding to a n , we obtain a n = 4a n−1+ 42n −2, which implies

Solution Let(F n)n≥1 be the Fibonacci sequence Then F1= F2= 1 and F n+2=

F n+1+ F n , for all n ≥ 1 Let us assume that F n ≤ N < F n+1 So, 0≤ N − F n <

F n −1 It follows that there exists s < n − 1 such that F s ≤ N − F n < F s+1 Hence

0≤ N − F n − F s < F s−1 and s − 1 < n − 2 We thus obtain that N may be written

as N = F n + F s + F p + ··· + F r , where the consecutive subscripts n ,s, p, ,r are

nonconsecutive numbers 

We need in what follows elementary differentiability properties of polynomials

1.3.5 For any real numberaand for any positive integerndefine the sequence

c = n − 1 − 2r, for r = 0,1, ,n − 1 In this case, x k is the coefficient of t k−1in the

polynomial f (t) = (1−t) r (1+t) n−1−r This property follows after observing that f

satisfies the identity

(k + 1)x k+2− (k − 1)x k = (n − 1 − 2r)x k+1− (n − 1)x k

In particular, the largest c is n − 1, and x k = C k−1

n −1 , for k = 1,2, ,n  

The next problem is devoted to the study of the normalized logistic equation,

which is a successful model of many phenomena arising in genetics and cal biology In its simplest form, the logistic equation is a formula for approximating

mathemati-the evolution of an animal population over time The unknown a nin the following

recurrent sequence represents the number of animals after the nth year It is easy

to observe that the sequence(a n)n≥1 converges to zero The interesting part of the

problem is to deduce the first-and second-order decay terms of this sequence

1.3.6 Consider the sequence(a n)n≥1satisfyinga1∈ (0,1)anda n+1= a n (1−a n),for alln ≥ 1 Show thatlimn →∞na n= 1andlimn →∞n (1 − na n )/lnn = 1

Solution We first prove that a n < 1/(n+1), for all n ≥ 2 For this purpose we use

the fact that the mapping f (x) = x(1−x) is increasing on (0,1/2] and decreasing on [1/2,1) Thus, a2= a1(1−a1) ≤ 1/4 < 1/3 Let us now assume that a n < 1/(n+1).

We observe that the sequence(a n)n ≥1 is decreasing, because a n+1−a n =−a2< 0.

Moreover,(a n)n ≥1 is bounded and limn →∞a n = 0 Let c n = 1/a n We computelimn →∞c n /n = lim n →∞1/(na n) We have

So, by the Stolz–Ces`aro lemma, limn →∞c n /n = 1 Therefore lim n →∞na n= 1

We provide the following alternative proof to this result Let us first observe that

(n + 1)a n+1= na n + a n − (n + 1)a2

n = na n + a n (1 − (n + 1)a n ). (1.4)

To see that(na n)n≥1is increasing, we need to show that 1−(n+1)a n ≥ 0 From the

graph of y = x(1−x) we note that a2≤ 1/4 and a n ≤ a ≤ 1/2 imply a n+1≤ a(1−a).

= 1 − 1

n2≤ 1.

Furthermore, na n < (n + 1)a n ≤ 1, and so (na n)n≥1 is bounded above by 1 Thus

na nconverges to a limit with 0 < na n <  ≤ 1 Now summing (1.4) from 2 to n,

we obtain

1≥ (n+1)a n+1= 2a2+a2(1−3a2)+a3(1−4a3)+···+a n (1−(n+1)a n ) (1.5)

If  = 1 then (1 − (n + 1)a n ) ≥ (1 − )/2 for all large n, and thus relation (1.5)

shows that the series∑∞n=1a n is convergent However, na n ≥ a1and so∑∞n=1a n ≥

a1∑∞n=11/n But ∑∞

n=11/n is divergent This contradiction shows that  = 1.

For the second part, set b n = 1/a n Then

b n+1= b2n

b n − 1 = b n+ 1 + 1

b n − 1 . (1.6)

Using this recurrence relation, a standard induction argument implies that b n ≥ n,

for all n ≥ 0 Thus, by (1.6),

where C is a real constant [we have used here the fact that 1 + 1/2 + ···+ 1/n − lnn

tends to a finite limit] This in turns proves the existence of a constant C0such that

for all integer n ≥ 1,

It follows from (1.7) and (1.8) that

This concludes the proof 

Independent Study.Find a sequence(x n)n≥1such that

where(F n)n≥1is the Fibonacci sequence.

USA, Proposed to the 33rd International Mathematical Olympiad, 1992

Solution Consider the function f (x) = (x + 1) −1 , and for any positive integer n,

define the mapping g n (x) = x+ f (x)+ f2(x)+···+ f n (x), where f n is the nth iterate

of f , that is, f n = f ◦ ··· ◦ f (n times) We easily check the following properties:

(i) for all 0≤ x < y ≤ 1 we have 0 < f (x) − f (y) < y − x;

(ii) g nis increasing in[0,1];

(iii) F1/F2= 1, f (F n /F n+1) = F n+1/F n+2and g n−1(1) = ∑n

k=1F k /F k+1, for all

pos-itive integers n.

We just remark that for proving (i) we use

g n (x) − g n (y) = (x − y) + [f (x) − f (y)] + ··· + [ f n (x) − f n (y)]

combined with the fact that any difference is less in modulus than and of oppositesign with respect to the previous one [see (i)]

If for some 2≤ k < n we have x k = 0, then by definition, x n= 0, and we conclude

the proof by induction with respect to the first n − 1 terms of the sequence If this

does not occur, then for all 2≤ k ≤ n, we may write x k−1 = (a k + x k)−1, where

a k > 0 is the integer part of x −1 k−1 Hence

We prove by induction with respect to k that for any fixed x n ∈ [0,1), the

right-hand side of the above equality is maximum if and only if a k = 1, for all k We first observe that a2must be 1, since a2appears only in the last expression Next,

let us assume that k > 2 and that for any values of x n , a n , a n−1 , ,a k+1, the above

expression attains its maximum for a k−1 = a k−2 = ··· = a2= 1 Observe that only

the last k − 1 terms on the right-hand side contain a k, and moreover, their sum is

This concludes our proof 

A simple linear recurrence generates a sequence of perfect squares, as shownbelow

1.3.8 A sequence of integers(a n)n≥1is given by the conditionsa1= 1,a2= 12,

a3= 20, anda n+3= 2a n+2+2a n+1−a nfor everyn ≥ 1 Prove that for every positiveintegern, the number1+ 4a n a n+1is a perfect square.

Problem M1174∗, Kvant

Solution Define the sequence(b n)n≥1 by b n = a n+2− a n+1− a n , for any n ≥ 1.

Observe that(b n)n≥1fulfills the same recurrence relation as the sequence(a n)n≥1

The idea is to prove that for all n, we have 1 + 4a n a n+1= b2 We argue by

induc-tion and we first observe that this equality holds for n= 1 Next, assuming that

1+ 4a n−1 a n = b2

n−1, we prove that 1+ 4a n a n+1= b2 Indeed, since

b n = (2a n+1+ 2a n − a n−1 ) − a n+1− a n = a n+1+ a n − a n−1

= (a n+1− a n − a n−1 ) + 2a n = b n−1 + 2a n ,

1.3.9 Prove that there exists a unique sequence(x n)n≥0of positive integers such

We assume, by induction, that x1, ,x n are integers Observing that x n −1 x n −3 − 1

and x n −1 are relatively prime, it follows that x n −1 is a divisor of x3+ 1 Therefore

W W Chao, Amer Math Monthly, Problem E 3356

a n a n+1= n and a n−1 a n = n−1, for all n ≥ 2 Subtracting these yields a n+1−a n−1=

Since a n+1= n/a n = a n −1 n /(n − 1), we can iterate to obtain

a 2n+1=22n (2n)! (n!)2 and a 2n+2=(2n + 1)!

22n (n!)2 .

By Stirling’s formula, we have a 2n+1 ∼ π/2 · √ 2n and a 2n+2 ∼ 2/π· √ 2n.

Hence the two contributions of the right-hand side of (1.9) approach 2/π and

and a straightforward computation shows that a1000< 45.1  

Bounded sequences are not necessarily convergent We give below a sufficient

condition that this happens Monotonicity arguments again play a central role

1.3.12 Let(a n)n≥0be a bounded sequence of real numbers satisfying

a n+2≤1

2(a n + a n+1), for alln ≥ 0.

con-We refine below the asymptotic behavior of a sequence converging to zero Theproof relies again on the Stolz–Ces`aro lemma, and the method... now intensively applied in the modeling of continuous phenomena A more cise definition was formulated by Cataldi around 1613 The method of recursionwas significantly advanced with the invention... withthe evaluation of elliptic integrals

In this section we are interested in the study of recurrent sequences that are notnecessarily linear

The first example gives an interesting