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(1),. MATHEMATICAL ANALYSIS A STRAIGHTFORWARD. APPROACH. K.G. BINMORE Professor of Mathematics L.S.E.. -. .. CAMBRIDGE. .... UNIVERSITY. .... - ..~.:- .. PRESS. CAMBRIDGE LONDON. - NEW YORK. - MELBOURNE. http://4paye.ir.

(2) Published by the Syndics of the Cambridge University Press The Pitt Building, Trumpington Street, Cambridge CB2 1RP Bentley House, 200 Euston Road, London NW1 2DB 32 East 57th Street, New York, NY 10022, USA 296 Beaconsfield Parade, Middle Park, Melbourne 3206, Australia © Cambridge University Press 1977. First published 1977 Reprinted 1978 Printed in Great Britain at the University Press, Cambridge Library of Congress Cataloguing in Publication. Binmore, K G 1940Mathematical analysis. Includes index. 1. Mathematical analysis. I. Title. QA300.B536 515 76-28006 ISBN 0 521 214807 hard covers ISBN 0 521 291674 paperback. Data.

(3) CONTENTS. Preface. ix. 1. Real numbers. 1.1 1.2 1.3 1.4 1.9 1.10 1.13 1.14. Set notation The set of real numbers Arithmetic Inequalities Roots Quadratic equations Irrational numbers Modulus. 1 2 3 3 6 6 9 10. 2. Continuum Property. 12. 2.1 2.2 2.6 2.7 2.9 2.11. Achilles and the tortoise The continuum property Supremum and infimum Maximum and minimum Intervals Manipulations with sup and inf. 12 13 15 15 16 18. 3. Natural numbers. 3.1 3.2 3.7. Introduction Archimedean property Principle of induction. 20 20 22. 4. Convergent sequences. 26. 4.1 4.2 4.4 4.7 4.15 4.21 4.26. The bulldozers and the bee Sequences Definition of convergence Criteria for convergence Monotone sequences Some simple properties of convergent sequences Divergent sequences. 26 27 27 30 33 37 38. 5 5.1 5.8. Subsequences. 41. Subsequences Bo1zano- Weierstrass theorem. 41 46. 1. 20. V.

(4) vt. Contents. 5.12 5.16. Lim sup and lim inf Cauchy sequences. 47 49. 6. Series. 53. 6.1 6.4 6.7 6.12 6.20 6.23. Definitions Series of positive terms Elementary properties of series Series and Cauchy sequences Absolute and conditional convergence Manipulations with series. 53 54 57 60 61. 7 7.1 7.6 7.9 7.11 7.13. Functions. 64. Notation Polynomial and rational functions Combining functions Inverse functions Bounded functions. 64 66 68 68 70. 8. Limits of functions. 74. 8.1 8.2 8.3 8.6 8.8 8.11 8.16 8.18. Limits from the left Limits from the right [(x) -)- I as x -)- ~ Continuity at a point Connexion with convergent sequences Properties of limits Limits of composite functions Divergence. 74 74 75 77 78 79 81 82. 9. Continuity. 84. 9.1 9.7. Continuity on an interval Continuity property. 84 86. 10. Differentiation Derivatives Higher derivatives More notation Properties of differentiable functions Composite functions. 91 91 92 93 95 98. 10.1 10.2 10.4 10.5 10.12 11 11.1 11.3 11.5 11.9. 56. Mean value theorems. 100. Local maxima and minima Stationary points Mean value theorem Taylor's theorem. 100 101 102 105.

(5) Contents. vii. 12. Monotone functions. 108. 12.1 12.3 12.6 12.9 12.11 12.13. Definitions Limits of monotone functions Differentiable monotone functions Inverse functions Roots Convex functions. 108 108 110 110 112 114. 13 13.1 13.2 13.3 13.9 13.16 13.19 13.27 13.31. Integration. 119. Area The integral Some properties of the integral Differentiation and integration Riemann integral More properties of the integral Improper integrals Euler-Maclaurin summation formula. 119 120 121 124 127 129 132 134. 14. Exponential and logarithm. 137. 14.1 14.4 14.6. Logarithm Exponential Powers. 137 140 141. 15. Power series. 143. 15.1 15.4 15.7. Interval of convergence Taylor series Continuity and differentiation. 143 145 147. 16. Trigonometric functions. 150. 16.1 16.2 16.4. Introduction . Sine and cosine Periodicity. 17. The gamma function. 17.1 17.3 17.5. Stirling's formula The gamma function Properties of the gamma function. 18. Appendix. 150 151 153 156 156 157 158. 161 This contains the proofs of 'propositions' left unproved in the main body of the text.. Solutions to exercises. 172. Suggested further reading. 251. Notation. 253. Index. 255.

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(7) PREFACE. This book is intended as an easy and unfussy introduction to mathematical analysis. Little formal reliance is made on the reader's previous mathematical background, but those with no training at all in the elementary techniques of calculus would do better to turn to some other book. An effort has been made to lay bare the bones of the theory by eliminating as much unnecessary detail as is feasible. To achieve this end and to ensure that all results can be readily illustrated with concrete examples, the book deals only with 'bread and butter' analysis on the real line, the temptation to discuss generalisations in more abstract spaces having been reluctantly suppressed. However, the need to prepare the way for these generalisations has been kept well in mind. It is vital to adopt a ~stematic approach when studying mathematical analysis. In particular, one should always be aware at any stage of what may be assumed and what has to be proved. Otherwise confusion is inevitable. For this reason, the early chapters go rather slowly and contain a considerable amount of material with which many readers may already be familiar. To neglect these chapters would, however, be unwise. The exercises should be regarded as an integral part of the book. There is a great deal more to be learned from attempting the exercises than can be obtained from a passive reading of the text. This is particularly the case when, as may frequently happen, the attempt to solve a problem is unsuccessful and it is necessary to turn to the solutions provided at the end of the book. To help those with insufficient time at their disposal to attempt all the exercises, the less vital exercises nave been marked with the symbol t. (The same notation has been used to mark one or two passages in the text which can be omitted without great loss at a first reading.) The symbol * has been used to mark exercises which are more demanding than most but which are well worth attempting. The final few chapters contain very little theory compared with the number of exercises set. These exercises are intended to illustrate the power of the techniques introduced earlier in the book and to provide the opportunity of some revision of these ideas. This book arises from a course of lectures in analysis which is given at the London School of Economics. The students who attend this course are mostly not specialist mathematicians and there is little uniformity in their previous. ix.

(8) Preface. x. mathematical training. They are, however, quite well-motivated. The course is a 'one unit' course of approximately forty lectures supplemented by twenty informal problem classes. I have found it possible to cover the material of this book in some thirty lectures. Time is then left for some discussion of point set topology in simple spaces. The content of the book provides an ample source of examples for this purpose while the more general theorems serve as reinforcement for the theorems of the text. Other teachers may prefer to go through the material of the book at a more leisurely pace or else to move on to a different topic. An obvious candidate for further discussion is the algebraic foundation of the real number system and the proof of the Continuum Property. Other alternatives are partial differentiation, the complex number system or even Lebesgue measure on the line. Iwould like to express my gratitude to Elizabeth Boardman and Richard Holmes for reading the text for me so carefully. My thanks are also due to 'Buffy' Fennelly for her patience and accuracy in preparing the typescript. Finally, Iwould like to mention M.C. Austin and H. Kestelman from whom Ilearned so much of what Iknow. July 1976. K.G.B..

(9) I. REAL NUMBERS. 1.1. Set notation. A set is a collection of objects which are called its elements. If x is an element of the set S, we say that x belongs to S and write xES.. If y does not belong to S, we write y t/=. S. The simplest way of specifying a set is by listing its elements. We use the notation A =. U, 1,y2, e, 1T}. to denote the set whose elements are the real numbers ~, I, Similarly. y2, e and. 1T.. B = {Romeo, Juliet} denotes the set whose elements are Romeo and Juliet. This notation is, of course, no use in specifying a set which has an infinite number of elements. Such sets may be specified by naming the property which distinguishes elements of the set from objects which are not in the set. For example, the notation. c. =. {x: x. > O}. (which should be read 'the set of all x such that positive real numbers. Similarly. x> 0') denotes the set of all. D = {y: y loves Romeo} denotes the set of all people who love Romeo. It is convenient to have a notation for the empty set 0. This is the set which has no elements. For example, if x denotes a variable which ranges over the set of all real numbers, then. {x :x2 + 1 = O} = 0. This is because there are no real numbers x such that x2 = -1. If Sand T are two sets, we say that S is a subset of T and write. 1.

(10) 2. Real numbers SC T. -, if every element of S is also an element of T. As an example, consider the setsP = {I, 2, 3, 4} and Q = {2, 4}. Then Q C P. Note that this is not the same thing as writing Q E P, which means that Q is an element of P. The elements of P are simply 1,2,3 and 4. But Q is not one of. these. The sets A, B, C and D given above also provide some examples. We have Ace and (presumably) BCD.. 1.2. The set of real numbers. It will be adequate for these notes to think of the real numbers as being points along a straight line which extends indefinitely in both directions. The line may then be regarded as an ideal ruler with which we may measure the lengths of line segments in Euclidean geometry.. 1. -2. I. .1. -I. o. 2. I. ,j2 2. 4. 3. The set of all real numbers will be denoted by IR. The table below distinguishes three important subsets of IR.. Subset. Notation. Elements. Natural numbers (or whole numbers). N. 1,2,3,4,5,. 71... .. -2,-1,0,1,2,3,. Integers. ... .... Rational numbers (or fractions). Not all real numbers are rational. Some examples of irrational numbers are. y2, e and. 1T.. While we do not go back to first principles in these notes, the treatment will be rigorous in so far as it goes. It is therefore important to be clear, at every stage, about what our assumptions are. We shall then know what has to be proved and what may be taken for granted. Our most vital assumptions are concerned with the properties of the real number system. The rest of this chapter and the following two chapters are consequently devoted to a description of the.

(11) Real numbers. 3. properties of the real number system which we propose to assume and to some of their immediate consequences.. 1.3. Arithmetic. The first assumption is that the real numbers satisfy all the usual laws of addition, subtraction, multiplication and division. The rules of arithmetic, of course, include the proviso that division by zero is not allowed. Thus, for example, the expression 2. o makes no sense at all. In particular, it is not true that. -o2 = .. -,. 00. We shall have a great deal of use for the symbol but it must clearly be does not represent a real number. Nor can it be treated as understood that such except in very special circumstances. 00,. 00. 1.4. Inequalities. The next assumptions concern inequalities between real numbers and their manipulation. We assume that, given any two real numbers a and b, there are three mutually exclusive possibilities:. (i). a> b. (ii) a = b. (a is greater than b) (a equals b). a. We:have,. for example, the. following inequalities. 1>0; 3>2; 2<3; -1 <0; -3<-2. There is often some confusion about the statements (iv) a;;> b. (a is greater than or equal to b). (v) a ~ b. (a is less than or equal to b).. To clear up this confusion, we note that the following are all true statements. 1>0; 3>2; 1>1; 2~3; -1~0;. -3~-?. We assume four basic rules for the manipulation of inequalities. From these.

(12) 4. Real numbers. the other rules may be deduced. (I) If a > band b > e, then a > e.. I. I. c. b. I a .. (II) If a> band c is any real number, then a+e>b+e. (III) If a> band e > 0, then ae through by a positive factor). (IV) If a> band e < 0, then ae reverses the inequality).. 1.5. Example. > be (i.e. inequalities < be. can be multiplied. (i.e. multiplication by a negative factor. If a> 0, prove that a-I> O.. Proof We argue by contradiction. Suppose that a > 0 but that a-I ~ O. ~t cannot be true that a-I = 0 (since then 0 = O~a = 1). Hence. a-I. < O.. By rule III we can multiply this inequality through by a (since a 1 = a-l.a<O.a But 1 < 0 is a contradiction.. > 0). Hence. = O.. Therefore the assumption a-I ~ 0 was false. Hence. a-I> O.. 1.6. If x andy are positive, then x <y if and only ifx2 <y2. Proof We have to show two things. First, that x < y implies x2 < y2,. Example. and secondly, that x2 < y2 implies x < y. (i) We begin by assuming that x < y and try to deduce that x2 the inequality x < y through by x > 0 (rule III). We obtain. < y2.. Multiply. x2 <xy. Similarly xy <y2. But now x2 < y2 follows from rule I. (ii) We now assume that x2 < y2 and try and deduce thatx to both sides of x2 < y2 (rule II), we obtain y2. i.e.. -x2. (y -x)(y. < y.. Adding -x2. >0 + x) > O.. (1).

(13) I'. t. Real numbers. I. t. I. 5. Since x + y > 0, (x + yfl through inequality (1) by (x. > 0 (example 1.6). We can therefore + y rl to obtain. multiply. y-x>O. x <yo. i.e.. (Alternatively, we could prove (ii) as follows. Assume that x2 < y2 but that x ~ y. From .c ~ y it follows (as in (i)) that x2 ~ y2, which is a contradiction.). 1.7. Suppose that, for any e > 0, a. Example. a<b. b. Then a - b > O. But, for any e> 0, particular case when e = b - a. Thus. + e. Hence a b must be false. Therefore a e.b. (Note: The symbol e in this example is the Greek letter epsilon. It should be carefully distinguished from the 'belongs to' symbol E and also from the symbol ~ which is the Greek letter xi.) ~:,c>. 1.8. I t i.~:. Exercise. (1) If x is any real number, prove that x2 prove that (i) 0 (2) If b. a. <a <a < 1 2. (ii) b2. ~. O. If 0. < a < 1 and b > 1,. > b > 1.. > 0 and B > 0 and A. b<B' prove. that aB. < bA.. a. A. a+A. Deduce that. b<b+B<Ji' (3) If a> band c > d, prove that a + c > b + d (Le.lnequalities can be added). If, also, b > 0 and d > O,prove that ac > bd (Le. inequalities between ~umbers can be m~ltiplied). (4) Show that each of the following i~s may fail to hold even though a > band c > d.. (i) a -c >b-d.

(14) 6. Real numbers. .. a (11) ~. >d. (iii) ac. > bd.. b. What happens if we impose the extra condition that b > 0 and d > O? (5) Suppose that, for any E > 0, a - E < b < a + E. Prove that a = b. (6) Suppose that a < b. Show that there exists a real number x satisfying a c. x c.b,. 1.9. Roots. Let n be a natural number. The reader will be familiar with the notation y =xn. For example.x" =x.x and x? =x.x.x. Our next assumption about the real number system is the following. Given any y ~ 0 there is exactly one value of x ~ 0 such that y =. x".. (Later on we shall see how this property may be deduced from the theory of continuous functions.) If y ~ 0, the value of x ~ 0 which satisfies the equation y = x" is called the nth root of y and is denoted by x = yl/n.. vy. = yll2. Note that, with this conWhen n = 2, we also use the notation vention, it is always true that O. If Y > 0, there are, of course, two numbers whose square is y. The positive one is and the negative one is - Vy. The notation ± -Jv means 'Vy or - Vy'. If r = mln is a positive rational number. and y ~ 0, we define. -J» ~. yr. =. -J». (ym)lln.. If r is a negative rational, then r is a positive rational and hence y-r is defined. If y > 0 we can therefore define yr by +. yr. =. _1_ y-r'. We also write yO = 1. With these conventions it follows that, if y > 0, then yr is defined for all rational numbers r. (The definition of yX when x is an irrational real number must wait until a later chapter.). 1.10. Quadratic equations. If y > 0, the equation x2 = y has two solutions. We denote the positive solution by Vy. The negative solution is therefore -Vy. We note again that.

(15) Real numbers. 7. there is no ambiguity about these symbols and that ± -Vy'. The general quadratic equation has the form ax". + bx + c. Vy simply means. 'Vy or. = 0. where a =1= O. Multiply through by 4a. We obtain 4a2x2. + b)2 -b2 + 4ac + b)2 = b2 -4ac.. (2ax (2ax 'I. ~. + 4abx + 4ac =. +b x. =. ± V(b2 -4ac). -b. 2a. + bx + c = 0 are therefore -4ac) b + V(bl -4ac) ~ = -_....:....->-_--'-. The roots of the equation ax2. I. a =. - b - V(b2. +. and. 2a. II. jl. 0. = ± V(b2 - 4ac). (-. l. =. It follows that the quadratic equation has no real solutions if b2 - 4ac < 0, one real solution if b2 - 4ac = 0 and two real solutions if iJ2.- 4ac > O. If b2 -4ac ~ 0, 2ax. I. 0. 2a. It is a simple matter to check that, for all values of x, ax2. + bx + c. = a(x - a)(x - ~).. With the help of this formula, we can sketch the graph of the equation y = ax" + bx + c.. y = ax' + bx. +c. a>O. 1.11 Example A nice application of the work on quadratic equations described above is the proof of the important Cauchy-Schwarz inequality. This asserts that, if ai, a2, ... ,an and b 1, b2, ... .b'; are any real numbers, then. (a1b1 +a2b2. + ... + anbn)2. :s:;;;;. (ar. + a~ + ... + a~)(br + b~ + .. , + b~)..

(16) Real numbers. 8. Proof For any x,. o ~ (alx + bl)2 + (a2X + b2)2 + =. Since y. + (anx + bn)2. (ar + ... +a~)x2 + 2(alb, +. +anbn)x + (br + ... + b~). = Ax2 + 2Bx + C. = Ax2 + 2Bx + C ~ 0 for all values of x,. it follows that the equation. AX2 + 2Bx + C = 0 cannot have two (distinct) roots. Hence (2B)2 - 4AC ~ 0 B2 ~ AC. i.e.. which is what we had to prove.. Exercise. 1.12. (1) Suppose that n is an even natural number. Prove that the equation xn = y has no solutions if y < 0, one solution if y = 0 and two solutions if y > O. Suppose that n is an odd natural number. Prove that the equation xn = y always has one and only one solution. Draw graphs of y = x2 and y = x3 to illustrate these results. (2) Simplify the following expressions: (i) 82/3 (3) If y ing:. (ii) 27-4/3. > 0, Z > 0 and rand. (iii) 326/5•. s are any rational numbers, prove the follow(iii) (yz. Y = yr zr .. (4) Suppose that a > 0 and that 0: and {3 are the roots of the quadratic equation ax? + bx + c = 0 (in which b? - 4ac > 0). Prove that y = ax? + bx + c is negative when 0: < x < {3 and positive when x < 0: or x > {3. Show also (without the use of calculus) that y = ax2 + bx + c achieves a minimum value of c - b'214awhen x = - sn«. (5) Let a I, az, ... ,an be positive real numbers. Their arithmetic mean An and harmonic mean Hn are defined by. An. =. al+a2+. ... +an n. 1(11. 1]. H;; I = -. - + - + ... + - . n al. a2. an. Deduce from the Cauchy-Schwarz inequality that Hn ~ An. (6) Let ai, a'2,' .. ,an and bl, b'2," . .b ; be any real numbers. Prove Minkowski's inequality, i.e.. 1). L~, )'!'" .; It (a. + bk. ]"'+. a1. (. n ]"2 Lb~. k ::1.

(17) Real numbers. 9. For the case n = 2 (or n = 3) this inequality amounts to the assertion that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides. Explain this.. 1.13. Irrational numbers. In § 1.2 we mentioned the existence of irrational real numbers. That such numbers exist is by no means obvious. For example, one may imagine the process of marking all the rational numbers on a straight line. First one would mark the integers. Then one would move on to the multiples of ~ and then to the multiples of! and so on. Assuming that this program could ever be completed, one might very well be forgiven for supposing that there would be no room left for any more points on the line.. -----r'.!!I1!1!N1.!."HM~ ••. "'!!!!lUI!!~!I!!I!!II1!'''!~!!Itn!II!1''"'I!!I!1I!I!''!!ltpi''"l!"M""!'''I!!!1!~I''lIIttrlMt!!'''''!!!1. -. 1. --t. 0. t 1- 1. I. -1 -t 1-. 2. But our assumption about the existence of nth roots renders this view untenable. This assumption requires us to accept the existence of a positive real number x (namely y2) which satisfies x2 = 2. If x were a rational number it would be expressible in the form. x=-. m n. where m and n are natural numbers with no common divisor (other than 1). It follows that. and so m? is even. This implies that m is even. (If m were odd, we should have m = 2k + 1. But then m2 = 4k2 + 4k + 1 which is odd.) We may therefore write m = 2k. Hence 4k2 = 2n2 n2 = 2k2•. Thus n is even. We have therefore shown that both m and n are divisible by 2. This is a contradiction and it follows that x cannot be rational, i.e. y2 must be an irrational real number. Of course, -J: is not the only irrational number and the ability to extract nth roots allows us to construct many others. But it should not be supposed that all irrational numbers can be obtained in this way. It is not even true that every irrational number is a root of an equation of the form.

(18) 10. Real numbers .ao + aJx. + a2x2 + ... + a.x". = 0. in which the coefficients ao, aI, ... .a.; are rational numbers. Real numbers which are not the roots of such an equation are called transcendental, notable examples being e and 1T. This fascinating topic, however, lies outside the scope of this book.. 1.14. Modulus. Suppose that x is a real number. Its modulus (or absolute value) [x I is defined by. (x~o) (x. < 0).. y = Ixl. x. o. Thus 131= 3,1-61 = 6 and 101 = O. Obviously [x] ~ It is sometimes useful to note that Ix I = .jx2. 1.15. 0 for all values ofx.. Theorem For any real number x, -Ixl. ~ x ~ [x].. ,. Proof Either x ~ 0 or x < O. In the first case, -Ixl the second case, -Ixl = x < 0 < [x],. ~ 0 ~ x = [x], In. I. i /. 1.16. Theorem. I. For any real numbers a and b. lab I = lal.lb I Proof The most elegant proof is the following. labl. =. .j«ab)2). =. .j(a2b2). =. .j(a2) • .j(b2). =. lal.lbI.. The next theorem is of great importance and is usually called the triangle. inequality, This nomenclature may be justified by observing that the theorem is the special case of Minkowski's inequality (exercise 1.12(6)) with n = 1. We give a separate proof.. 1.17. Theorem (triangle inequality) la + bl ~ lal. +. Proof We have. Ibl.. For any real numbers a and b,. I.

(19) Real numbers la+bl2. =. 11. (a+b)2. It follows (from example. = a2+2ab+b2 = lal2 + 2ab + Ibl2 ~ lal2 + 21abl + Ibl2. (theorem l.IS). =. lal2 + 2lal.lbl + Ibl2. =. (Ial + Ibl)2.. (theorem. 1.16). 1.6) that. la + bl ~ lal + Ibl. 1.18. Theorem For any real numbers c and d,. Ic -dl. ~ Icl-Idl.. Proof Take a a + b = c and so lei 1.19. =. la + bl ~ lal + Ibl. =. d in the triangle inequality. Then. Idl + Ie -dl.. Example (i) 2 = (ii) 7 =. 1.20. = d and b = c -. 1-1 + 31 ~ 1-11 + 131 = 1 + 3 = 4. 16 - (-1) I ~ 161- 1- I I = 6 - I = 5.. Exercise (1) Prove that lal < b if and only if =b <a < b. [Recall that you have two things to prove as in example 1.6.} (2) Prove that. Ic -dl. ~ IIcl-ldli.. (3) The distance d(x,y) between two real numbers x andy is defined by d(x,y) = Ix -yl. Show that, for any X,y and z, (i) d(x ,y) ~ 0 (iii) d(x ,y) = 0 if and only if x = y (ii) d(x,y) = d(y ,x) (iv) d(x,y) ~ d(x, z) + d(z ,y). t (4) If rand s -are rational numbers, prove that. r+sy2 is irrational unless s = O. t (5) Suppose that the coefficients a :f= 0, b and c of the quadratic equation. ax? + bx + c = 0 are all rational numbers and that Q = r + sy2 is a root of this equation, where rand s are also rational numbers. Prove that (3 = r - sy2 is also a root. t (6) Prove that 3112 and 2113 are irrational numbers..

(20) 2. CONTINUUM. 2.1. Achilles and the tortoise. PROPERTY. The following is one of the famous paradoxes of Zeno. Achilles is to race a tortoise. Since Achilles runs faster than the tortoise, the tortoise is given a start of Xo feet. When Achilles reaches the point where the tortoise started, the tortoise will have advanced a bit, say XI feet. Achilles soon reaches the tortoise's new position, but, by then, the tortoise will have advanced a little bit more, say X2 feet. This argument may be continued indefinitely and so Achilles can never catch the tortoise.. ~------~~~-----------x,,------;>;,.. ....• "';:----.t, ~. The simplest way to resolve this paradox is to say that Achilles catches the tortoise after he has run a distance of X feet, where X is 'the smallest real number larger than all of the numbers Xo, Xo+ Xl, Xo+ Xl + X2 ••• .'. Zeno's argument then simply reduces to subdividing a line segment of length X into an infinite number of smaller line segments of respective lengths Xo, X I, X 2, .... ~.=-=---=-_x=-o-=-~_x='tx'~ Formulated in this way, the paradox loses its sting. This solution, of course, depends very strongly on the existence of the real number X, i.e. the smallest real number larger than all of the numbers X 0, Xo +XI,XO. 12. +XI. +X2,'". ,.

(21) 13. Continuum property A similar lesson is to be learned from Archimedes' method of exhaustion. This was invented by Archimedes as a means of evaluating the area of regions with curved boundaries. In the case of a circle, the method involves inscribing a sequence of larger and larger regular polygons inside the circle and then calculating their areas •• A1,A2,A3,·. The area A of the circle is then identified with the 'smallest real number larger than each of the numbers A 1, A 2, A 3, .. .'. By taking the radius of the circle to be 1 and calculating the value of An for a large enough value of n, Archimedes was able to obtain as close an approximation to n as he chose. Note that this method again depends strongly on the existence of a smallest real number larger than all of the real numbers A 1, A 2, A 3, •... 2.2. The Continuum Property. The discussion of the previous section indicates the necessity of making a further assumption about the real number system. It would, after all, be most unsatisfactory if Achilles never caught the tortoise or if a circle had no area. We shall call our new assumption the Continuum Property of the real numbers. The Continuum Property is not usually mentioned explicitly in a school algebra course, but it is very important for what follows in these notes. The rest of this chapter is therefore devoted to its consideration. We begin with some terminology . A set S of real numbers is bounded above if there exists a real number H which is greater than or equal to every element of the set, i.e. if, for some H, H. «<n. --~I---. ~·+IAAII1I41~1~11I-+1~1. . s. for any xES. The number H (if such a number exists) is called an upper bound of the set S. A set S of real numbers is bounded below if there exists a real number h.

(22) 14. Continuum property. which is less than or equal to every element of the set, i.e. if, for some h,. h. ~~~II~I+I~I~II~I#II-----'-----v----"'. xe n. s. for any xES. The number h (if such a number exists) is called a lower bound of the set S. A set which is both bounded above and bounded below is just said to be bounded. 2.3 Proposition A set S of real numbers is bounded if and only if there exists a real number K such that [x]. for any xES.. 2.4. <;. K. (This is easily proved with the help of exercise 1.20(1).). I. ~xamples. (i) The set {I, 2, 3} is bounded above. Some upper bounds are 100,10, 4 and 3. The set is also bounded below. Some lower bounds are - 27, 0 and 1. (ii) The set {x : 1 <; x < 2} is bounded above. Some upper bounds are 100, and 10,4 and 2. The set is also bounded below. Some lower bounds are -27, l.. °. (iii) The set {x: x > O] is unbounded above. If H> 0 is proposed as an upper bound, one has only to point to H + 1 to obtain an element of the set larger than the supposed upper bound. However, the set {x :x> o} is bounded below. Some lower bounds are -27 and 0.. We now state the Continuum Property which will be fundamental for the remainder of these notes.. Continuum Property Every non-empty set of real numbers which is bounded above has a smallest upper bound. Every non-empty set of real numbers which is bounded below has a largest lower bound.. Thus, if S is a non-empty set which is bounded above, then S has an upper bound B such that, given any other upper bound H of S,.

(23) 15. Continuum property. _~-----1B. B~H.. ~ S. Similarly, if S is a non-empty set which is bounded below, then S has a lower bound b such that, given any other lower bound h,. . hJ b.-L-U..I-.J..-U I----I~--. s>».. S. 2.5. Examples. (i) The smallest upper bound of the set {1, 2, 3} is 3. The largest lower bound of the set {1, 2, 3} is 1. (ii) The smallest upper bound of the set {x : 1 ~ x < 2} is 2. The largest lower bound is 1. (iii) The set {x :x> o) has no upper bounds at all. The largest lower bound of the set {x :x > O}is O.. 2.6. Supremum and infimum. If a non-empty set S is bounded above, then, by the Continuum Property, it has a smallest upper bound B. This smallest upper bound B is sometimes called the supremum of the set S. We write B = sup S or. B. =. sup x. xES. Similarly, a set which is bounded below has a largest lower bound b. We call b the infimumoi the set S and write b = inf S or b = inf x. xES. Sometimes you may encounter the notation sup S = + This simply means that S is unbounded above. Similarly, inf S = means that S is unbounded below. 00.. 00. 2.7. Maximum and minimum. If a set S has a largest element M, we call M the maximum of the set S and write M = max S. If S has a smallest element m, we call m the minimum of S and write m = min S..

(24) 16. Continuum property. It is fairly obvious that, if a set S has a maximum M, then it is bounded above and its smallest upper bound isM. Thus, in this case, sup S = max S. A common error is to suppose that the smallest upper bound of a set S is always the maximum of the set.' However, some sets which are bounded above (and hence have a smallest upper bound) do not have a maximum. (See example 2.8(ii).) Similar remarks, of course, apply to largest lower bounds and minima of sets.. 2.8. Examples. (i) The set {I, 2, 3} has a maximum 3 and this is equalto its smallest upper bound. The set {I, 2, 3} has minimum 1 and this is equal to its largest lower bound. (ii) The set {x : 1 ~ x < 2} has no maximum. The number 2 cannot be the largest element of the set because it does not belong to the set. On the other hand, any x in the set satisfies 1 ~ x < 2. But then. x+2. Y - --. ~l. 2. is an element of the set which is larger than x. Hence x cannot be the largest element of the set. However, {x : 1 ~ x < 2} has smallest upper bound 2. The set {x : 1 ~ x < 2} has minimum 1 (why?). This is equal to its largest lower bound. (iii) The set {x :x> O} has no maximum, nor does it have any upper bounds. The set {x :x > O} has no minimum (why not?). Its largest lower bound is O.. 2.9. Intervals. An interval I is a set of real numbers with the property that, if x E I and y E I and x ~ z ~ y, then z E I, i.e. if two numbers belong to I, then so does every number between them. In describing intervals we use the following notation:. (a,b) [a,b]. = =. {x:a<x<b} ~. {x:a~x~b}. [a,b) = {x:a~x<b}. (a, b] = {x: a < x ~ b}.. (a,b). a. b. ---I~d~l--'-----v---' [a,b J.

(25) 17. Continuum property These are the bounded intervals (classified by whether or not they have a maximum and whether or not they have a minimum). We also need to consider the unbounded intervals. For these we use the notation below. a. (a,oo) = {x:x>a} [a,oc) (-oo,b) (-oo,b]. = = =. ~~~ y. {x :x~a}. (a, 00). {x:x<b} {x:x~b}. (_00, b). (Do not be rnisled by this notation into supposing that 00or - 00are real numbers or that they can be treated as such.) All intervals (with the exception of the empty set and the set of all real numbers) fall into one of the categories described above. We call the intervals (a, b), (a, 00)and (- 00, b) open and the intervals [a, b], [a, 00) and (- 00, b] closed. (The intervals [a, b) and (b, a] are sometimes called half-open.) Of particular importance are the closed, bounded intervals [a, b] . We shall call such an interval compact.. 2.10. Exercise (1) Which of the following statements are true? (i)3E(1,2) (ii)2E(-00;3] (iii) 1 E [1,2) (iv)2E(0,2) (v)3E [1,3]. (2) If ~ is a real number and 0 > 0, prove that. {x:I~-xl<o}. =. (~-o,~+o).. (3) Decide in each of the following cases whether or not the given set is bounded above. For those sets which are bounded above, write down three different upper bounds including the smallest upper bound. Decide which of the sets have a maximum and, where this exists, write down its value. (i)(O,l) (ii)(-00,2] (iii){-1,0,2,5} (iv)(3,00) (v) [0,1]. (4) Repeat the above question but with the words 'above, upper and maximum' replaced by 'below, lower and minimum'. (5) Give an example of a set which has smallest upper bound 4 but contains no element x satisfying 3 < x < 4. (6) Show that, given any element x of the set (0,00), there is another element y E (0, 00) with the property that y < x. Deduce that (0,00) has no minimum..

(26) 18. Continuum property. 2.11. Manipulations with sup and inf We prove one result and list some others as exercises.. 2.12 Theorem Suppose that S is a non-empty set of real numbers which is bounded above and ~ > 0. Then sup ~x = ~ sup x. xES. xES. Proof Let B = sup S. Then B is the smallest number such that, for any xES,. »«:s. Let T= {~x: xES}.. Since ~ > 0,. ~x ~ ~B for any xES. Hence T is bounded above by ~B. By the Continuum Property, Thas a smallest upper bound (or supremum) C. We have to prove that C = ~B. Since ~B is an upper bound for T and C is the smallest upper bound for T, C~ ~B. Now repeat the argument with the roles of Sand T reversed. We know that C is the smallest number such that, for any yET,. y~c. Since ~ > 0, it follows that ~-ly ~. ~-1. C. for any yET. But S= {~-1Y :yET}. Hence ~-1 C is an upper bound for S. But B is the smallest upper bound for S. Thus B ~ ~-IC ~B ~ C. We have shown that C ~ ~Band also that ~B~ C. Hence ~B = C.. 2.13. Exercise (1) Suppose that S is bounded above and that So C S. Prove that sup So ~ sup S, (2) Suppose that S is bounded above and that ~ is any real number. Prove that sup (x xES. + ~). sup x xES. +~..

(27) Continuum property. 19. (3) Suppose that S is bounded above. Prove that inf (-x). = -. xES. sup x. xES. Hence obtain analogies of theorem 2.12 and exercises 2.13(1 and 2) in which inf replaces sup. (4) The distance d(t S) between a real number ~ and a non-empty set S of real numbers is defined by d(~,S). = inf. I~-xl. xES. (see exercise 1.20(3». Find the distance between the real number ~ = 3 and the sets (i) S = {O, 1, 2} (ii) S = (0,1) (iii) S = [1,2] (iv) S = (2, 3). (5) (i) If ~ E S, prove that d(t S) = O. Give an example of a real number ~ and a non-empty set S for which d(t S) = 0 but ~ f/. S. (ii) If S is bounded above and ~ = sup S, prove that d(t S) = o. Deduce that the same is true if S is bounded below and ~ = inf S. t (iii) If I is a closed interval, prove that d(~, I) = 0 implies that ~ E I. If I is an open interval (other than IR or f/J), show that a ~ q:. I can always be found for which d(t I) = O. * (6) Suppose that every element of an interval I belongs to one or the other of two non-empty subsets Sand T. Show that one of the two sets S or T contains an element at zero distance from the other. (Do not assume that S and Tare necessarily intervals. The set S, for example, could consist of all rational numbers in I and the set T of all irrational numbers in I.) [Hint: Suppose that s < t, where s E Sand t E T. Let To = {x : x E T and x> s}. Show that To =1= f/J and is bounded below. Write b = inf To and consider the two cases bET and b q:. T.].

(28) 3. NATURAL. 3.1. Introduction. NUMBERS. In this chapter we explore some implications of the Continuum Property for sets 01 natural numbers. As always, arithmetical properties are taken for granted. But where a proof hinges on such a property, the property will be explicitly stated.. 3.2. Archirnedean property The Archimedean property is the assertion of the following theorem.. 3.3. Theorem The set N of natural numbers is unbounded above.. Proof For this proof we need to assume that, if n is a natural number, then so is n + 1. Suppose that the theorem is false. Then N is bounded above. By the Continuum Property it therefore has a smallest upper bound B. Since B is the smallest upper bound for N, B-1 is not an upper bound for N. Thus, for some nEN,. n >B-l n + 1 > B. But then n + 1 is an element of the set N which is greater than B. But B is an upper bound of the set N. This is a contradiction. Hence N is unbounded above.. 3.4. Example Prove that the set S = {n-1: n EN}. is bounded below with largest lower bound. o.. Proof Since the natural numbers are all positive,. -1-0. -+1++1. t-+t--+t---~ s. 20.

(29) 21. Natural numbers. for all n EN. Hence the set S is bounded below by O. We have to prove that 0 is the largest lower bound. Suppose that h > 0 is a lower bound for S. Then, for each n E N,. Hence, for each n EN, n<'h-I. But then h -I is an upper bound for the set N which we know is unbounded above. It follows from this contradiction that no h > 0 can be a lower bound for S. Thus 0 is the largest lower bound for S. Note that 0 r$.S and hence S has no minimum.. 3.5 mum.. Theorem Every set of natural numbers which is not empty has a mini-. Proof For this proof we need to assume that the distance between distinct natural numbers is at least 1. Let S be a non-empty set of natural numbers. Given that all natural numbers are positive, 0 is a lower bound for S. By the Continuum Property, it follows that S has a largest lower bound b. Since b is the largest lower bound, b + 1 is not a lower bound. Therefore, for some n E S, n<b+1. If n is the minimum of S, there is nothing to prove. If not, then, for some m E S, m < n. We obtain the inequality b<.m<n<b+1 from which follows the contradiction 0. 3.6. <n-. m. < 1.. Exercise (1) Prove that the set n -1 S = ( -n-:nEN. ). is bounded above with smallest upper bound 1. Does S have a maximum? (2) Let X > 1. Prove that the set S. = {Xn:. n EN}. is unbounded above. [Hint: if B were the smallest upper bound, then BX-1 could not be an upper bound.] Show that, if 0 < x < 1, then the.

(30) 22. Natural numbers set T = {x'": n EN} (3). (4). (5). (6). 3.7. is bounded below with largest lower bound O. Prove that any non-empty set of integers which is bounded above has a maximum and that any non-empty set of integers which is bounded below has a minimum. Let a < b. Prove that there exists a rational number r satisfying a < r < b. [Hint: justify the existence of a natural number n satisfying n > (b -art and consider the rational number r = mln where m is the smallest integer satisfying m > an.] Let S be the set of all rational numbers r which satisfy 0 < r < 1. Show that S has no maximum and no minimum. Prove that S is bounded and has largest lower bound 0 and smallest upper bound 1. [Hint: for the last part, use the previous question.] Let a < b. Prove that there exists an irrational number ~ satisfying a < ~< b. [Hint: exercise 1.20(4).]. Principle of induction. Suppose that a line of dominoes is arranged so that, if the nth one falls, it will knock over the (n + 1)th. If the first domino is pushed over, most people would agree that a1l the dominoes would then fall down. The principle of induction is an idealisation of this simple notion. We show below how it may be deduced from theorem 3.5. 3.8. Theorem (principle a/induction). Suppose that, for each n EN, Pen) is a statement about the natural number n. Suppose also that (i) P(1) is true (ii) if P(n) is true, then Pen + 1) is true. ThenP(n) is true for every n EN.. Proof For this proof we need to assume that, if n is a natural number other than 1, then n - 1 is a natural number as well. Let S = {n :Pen) is false}. We want to prove that S is empty. Suppose that S is not empty. Then from theorem 3.5 it follows that S has a minimum m. Since P(1) is true, 1 S. Hence m =1= 1. Therefore m - 1 is a natural number. But m is the minimum of S. Thus m -1 S and so P(m -1) is true. But then P(m) is true (because of hypothesis (ii) of the theorem). Hence m S which is a contradiction.. tI.. tI.. tI..

(31) Natural numbers Example. 3.9. 1+ 2. 23. For each n EN. + 3 + ... + n. =. tn(n + 1).. Proof Let Sn = 1 + 2 + 3 + ... + n = 'Lh=1 k. Let pen) be the statement Sn = tn(n + 1). We have S1 = 1, and ~ .1(1 + 1) = 1. Hence P(1) is true. We now wish to show that 'If pen) is true, thenP(n + 1) is true'. To do this, we assume that pen) is true and try and deduce that pen + 1) is true. Thus we assume that S; = ~n(n + 1) and try and deduce that Sn+1 = ten + 1)(n + 2). But Sn+1 = 1 + 2 + 3 + ... + n + (n + 1) = Sn. + (n + 1). ~n(n =. + 1) + (n + 1). Hn + 1)(n + 2). as required. The result now follows by induction.. 3.10. Example. The real number. is the arithmetic mean of the numbers a1, a2, ... , an. If a1, a2, ... , an are positive, their geometric mean G; is defined by Gn =. (a1a2 ... an)lIn.. The 'inequality of the arithmetic and geometric means' asserts that. Proof This inequality has an entertaining proof using 'backwards induction' (see exercise 3.11 (5)). Let P(n) be the statement 'For any positive numbers a 1, a2, ... , an, Gn'::;;An'. We shall show that (i) p(2n) is true for each n EN. (ii) If P(n) is true, then pen - 1) is true. The result then follows by backwards induction. (i) We begin by provingP(2). NoV{, O'::;;(va1-va2)2. =. a1-2v(a1a2)+a2. (a 1a2)1/2 .::;;Ha 1 + a2). This is simply P(2). We now show that p(2n) implies p(2n + 1). Let m = 2n. Then 2n+1 = 2m. We have to assume P(m) and try and deduce P(2m). Since P(m) is true,.

(32) 24. Natural numbers. But we know that P(2) is true. Hence. {(a 1 a2 ... am)lIm(am+l am+2 ... a2m)l/mpl2 ~ ~~al + ... +am + am+1 + ... +a2m}. 2 ~ m m Thus. which is P(2m). We have shown that P(21) is true and that, if p(2n) is true, then p(2n+l) is true. Hence p(2n) is true for all n E N by straightforward induction. (ii) We now show that, if pen) is true, then pen -1) is true. Suppose that P(n) is true. Then. {a 1a2 ... an -1Gn-1 }ltn. + a2 + al --=--_..::.-. ~. ... + an-l + qn-l :.:.....!._...o.:.::....:. n. 'l-1 G }lIn ~ (n - I)An-l + Gn-1 {Gn-l n-l ...••.. .. nGn-1. ~. n. (n - I)An-1. + Gn-1. Thus pen - 1) is true. This completes the proof.. 3.11. Exercise (1) Prove by induction !hat n. (i). L. k:l. k2. =. 12 + 22. + ... + n2 =. !n(n. k3. =. 13 + 23. + ... + n3 =. !n2(n. n. (ii). L k:l. (2) Prove by induction that, if x n. L k:o. Xk. = 1 +x. =1=. + 1)(2n + 1) + 1)2.. 1,. + x2 + ... + x" =. 1 - xn+1. I-x.

(33) 25. Natural numbers Deduce that, for any real numbers a and b,. + a"-2 b + a"-3. a" - b" = (a - b )(a"-I (3) A polynomial. =. P(x). b2. + ... + ab"-2 + b" -I).. P(x) is an expression of the form. a"x". + a"_lx"-1 + ... + alx + aQ.. *-. If a" 0, the polynomial is said to be of degree n. Suppose that P(x) is a polynomial of degree n and that, for some t P(~) = 0. Prove that P(x) = (x - nQ(x) where Q(x) is a polynomial of degree n - 1. [Hint: use the previous question.] If ~I' ~2" .. .t, are distinct real numbers and ?(~I)= P(~2) = ... = P(~Il) = 0, prove that. =. P(x). k(x - ~I)(X - ~2) ... (x - ~"). where k is a constant. (4) Assuming that n! = 1. 2.3 ... nand O! = 1, we define. C). = r!(nn~r)!. (r=0,1,2,. .... ,n).. Prove that ( nr). + (r. _n. 1). = (n +r. 1). (r= 1,2, ... ,n).. Using this result and the principle of induction, obtain the binomial theorem in the form (a. + b)". = a". + na'"" b + n(n;. 1) a"-2b2. + ... + b". (5) Let pen) be a statement about the natural number n and suppose that (i) P(2") is true for each n E N (ii) if P(n) is true, then Pen - 1) is true. Prove that P(n) is true for every n EN. [Hint: the set {2" : n EN} is unbounded above (exercise 3.6(2)).] (6) Suppose that 0< X ~ Y and that x and y belongto the interval [X, Y]. Prove that, for each n EN, Xy1/". 1. x - y. 1 ~. nXY. 1. [Hint: exercise 3.11(2).]. x'!" - yl/". 1 ~. YX1/". 1. x - y. I..

(34) 4. CONVERGENT. 4.1. The bulldozers and the bee. SEQUENCES. Two bulldozers are moving towards each other at a speed of one mile per hour on a collision course. When they started they were one mile apart and a bee was perched on the front of one of them. The bee began to fly back and forward between the bulldozers at a constant speed of two miles per hour, vainly seeking to avoid his. fate. How far from his starting point will the unfortunate insect be crushed?. ~~~--- ---~'-·~---1-----)~·-----t----~». This riddle is usually posed in the hope that the victim will embark on some complicated calculations involving the flight of the bee. But it is quite obvious that the bee will be crushed when the bulldozers collide. Since the bulldozers travel at the same speed, this will happen halfway between their starting points. The answer is therefore one half mile. Let us, however, take up the role of the riddler's victim and examine the flight of the bee. Let Xn denote the distance the bee is from his starting point when he makes his nth landing. Then Xl X2. = ~ = ~- ~. and, in general, X n. 26. = ~~ 3 9 +~ 27. -. • • .. + (-. 1) 11-12 ( 3"1) n.

(35) 27. Convergent sequences =. HI. + (-l) + (- ~)2 + ... + (-. ~ H\-=-(~li;l~. ~t-I}. J(l-(-l)n).. How is the answer! to be extracted from this formula? It seems clear that, as n gets larger and larger, (gets closer and closer to zero and so x., gets closer and closer to We say that 'xn tends to as n tends to infinity' - or, in symbols, 'xn ~ as n ~ 00'. This idea is of the greatest importance, but, before we can make proper use of it, it is necessary to give a precise formulation of what it means to say that xn ~ I as n ~ 00.. !. 4.2. ~t. !.. !. Sequences. A sequence may be regarded as a list of numbers Xl> X2, X3, X4, ... More precisely, we can say that a sequence is determined by a rule which assigns to each natural number n a unique real number Xn. We call Xn the nth term of the sequence. The notation. means the sequence whose nth The set {x n : n EN} is called sequence is bounded above (or Thus (xn) is bounded above by. 4.3. term is xn. the range of the sequence. We say that a below) if its range is bounded above (or below). H if and only if xn <:( H (n = 1,2,3, ... ).. Examples (i) «(- 1)") (iii) (1). =. 1, + 1, - 1,. = -. x.;,. =. (xn) defined. "21 (. Xn-I. +- 2. xn-I. The first few terms are 2,. Definition A sequence. following. criterion. inductively ). 1 ,2,3,4,.·· 1 ! 1. ..) (n -I) -(11. (iv) (2n). 1, 1, 1, 1, .... (v) The sequence. 4.4. + 1, .... by x:. =. 2,4,8,16,. = 2 and. (n = 2, 3,4, ... ).. t H, Wa, .... of convergence. (xn) is said to converge to the limit I if and only if the is satisfied.. ....

(36) 28. Convergent sequences. Given any e > 0, we can find an N such that, for any n Ixn-/I<€.. We write Xn. -+. I as n. -+. 00. > N,. or. lim Xn = I.. n->-oo. It may be helpful to think of x ••x 2, X 3, ..• as successive approximations to the number I. The distance IXn -II between Xn and I is then the error involved in approximating I by Xn. The definition of convergence then simply asserts that we can make this error as small as we choose by taking n large enough. The diagram is supposed to represent a sequence (xn) with the property that xn -+ I as n -+ 00.. 1. r------i,---i\~---;.:---~--~ L---/~'iL__ -Vl----~--_~_--x,. - -. ,lx9-11. ,. I I. o. '.,. .---L 2. I I. 3. I. I. I. I. 4. 5. 6. 7. 8. 9. 10. II. n>6. For the value of € indicated in the diagram, a suitable value for N in the definition is N = 6. For each value of n > 6, IXn -II. < €.. In particular, the value of IX9 -II has been noted in the diagram. Notice that the definition of convergence begins 'Given any € > 0, we can find an N ... ' Here the emphasis is on the very small values of e > 0. It is clear from the diagram above that, if we had chosen to look at a very much smaller value of € > 0, then we should have had to have picked out a very much larger value of N. In general, the smaller the value of f > 0, the bigger must be the corresponding value of N.. 4.5. l:xample. 1 1 + - -+ 1 as n. -+. 00.. n. Proof Let e > 0 be given. We must find a value of N such that, for any. n>N,.

(37) Convergent sequences. (1+;)-1 <e. (1+;) -1 =. But. I n. 29. =. n. and so we simply need to find an N such that, for any n. > N,. 1. -<e n. i.e.. I. n>-.. e. But this solves the problem. We simply chose N. = lie.. Then, for any n. > N,. I. n>N =e i.e.. 1. -<e n. (1+;) -1. i.e.. < •.. We have shown that, given any e > 0, we can find an N (namely N such that, for any n > N,. (1+~ -1. = lie). < •.. It is important to observe that N depends on e, i.e. for each € > 0 we use a different N. Some values of e and the corresponding values of N are entered in the table below.. e. N. 0'4. 2'5. 0'1. 10. 0'000 001. 1000 000. (Note: Some authors insist that N be a natural number. This makes the definition of convergence a little more elegant but renders examples like that above marginally more complicated. If we wanted N to be a natural number in the example above, we could not simply write N = lie. Instead we should have to choose N to be some natural number larger than lie.).

(38) 30. Convergent sequences. 4.6. Exercise (1) Prove that. =. lim (~). n~+ 1. n-+oo. l.. (2) Let r be any positive rational number. Prove that. Let A be any real number. If x., n -+. (3). -+. I as n. -+. 00,. prove that Axn. -+. A.l as. 00.. 4.7. Criteria for convergence. In the example and exercises above it is fairly easy to decide what value of N is appropriate to any given value of e > O. But this is by no means always the case. It is therefore natural to look around for some shortcuts which will enable us to determine whether a sequence converges (and what its limit is) without our having to indulge in the painful process of appealing to the definition. In this and the next section we give some useful results of this sort. 4.8 n. -+. 00. Proposition (combination theorem) Let x.; and let A and J.J.be any real numbers. Then. (i) Axn (ii) XnYn. x. (iii) ~ Yn. + J.J.Yn -+ Al + J.J.m as n -+ -+. I. -+ -. lm as n as n. -+. -+. 00. -+. I as n. -+. 00. and Yn. -+. m as. 00. 00. (provided that m =1=0).. m. The proofs of (ii) and (iii) can be found in the appendix. These proofs are somewhat technical and we prefer not to hold up the discussion by presenting them at this stage. Readers who prefer to omit the proofs of (ii) and (iii) altogether will not suffer greatly as these results may be deduced from (i) once the theory of the exponential and logarithm functions has been developed. Proof of 4.8(i). After exercise 4.6(3) we need only show that, if Xn -+ I as n -+ 00 and v, -+ mas n -+ 00, then Xn + Yn -+ 1+ mas n -+ 00. Let > 0 be given. Then > O. Since xn -+ I as n -+ it follows that we can find an N) such that, for any n > Nt,. €. !€. IXn. -II < !€.. Similarly, we can find an N2 such that, for any n IYn. 00,. -ml < !€.. (1). > N2, (2).

(39) 31. Convergent sequences. Let N be whichever of N, and N2 is the larger, i.e. N = rnax {Nt. N2}. Then, if n > N, both inequalities (1) and (2) are true simultaneously. Thus, for any. «».«.. =. I(xn+Yn)-(l+m)1. l(xn-l)+(Yn-m)1. ~ IXn -II. + IYn -. <. =. !E +!E. ml. (triangle inequality). E.. Given any E > 0 we have found a value of N (namely N = max {Nt. N2}) such that, for any n > N, I(xn + Yn) - (I + m)1 < E. Hence xn + Yn ~ 1+ mas n ~ =.. Example Prove that. 4.9. nl~~. = ~. (S):~;~2). Proof We write 2n3 3. 3n. -. 2. = - as n ~ (Xl.. =. 2. Sn +4n -2. 5. The supporting argument is as follows. It is obvious from the definition of convergence that 2 ~ 2 as n ~ From exercise 4.6(2), n-2 ~ 0 as n -+ =. Hence, by proposition 4.8(i),. =.. 2-3n-2-+2-3.0. = 2asn-+(Xl.. Similarly, 5 + 4n-1. -. 2n-3. -+. 5. + 4.0. - 2.0. = 5 as n. -+. =.. The result then follows from proposition 4.8(iii).. 4.10. Theorem (the sandwich theorem) Suppose that y., that Zn -+ I as n -+ =. IfYn ~xn ~zn (n = 1,2, ... ), then. -+. I as n. -+ oo. and. (Here the sequence (xn> is 'sandwiched' between the two sequences (Yn>and (zn> - just as the bee of §4.1 was 'sandwiched' between the bulldozers.). Ix-II. Proof For this proof it is necessary to note that the inequality < E is true if and only if I - E < X 0 be. given. We have to find an N such that, for any n. > N,. < E.. Since Yn ~ / as n ~ n>N1,. we know that there exists an Nl such that, for any. 00,. (3). IYn -II < E. Similarly, since. zn -+ I as n ~. IZn -II. 00,. there exists an N2 such that, for any. n. < E.. > N2, (4). Let N be whichever of Nl and N2 is the larger, i.e. N = max {N}, N2}. Then, if n > N, both inequalities (3) and (4) are true simultaneously. Thus, for any n>N,. <Yn N,. <Yn <xn <zn 0, we have. 1-. E. Given that, for any n. > N, IXn -II < E.. Hence. 4.11. x., ~. I as n ~. found a value of N (namely N = max {N}, N2}) such. 00.. Corollary Supoose that Y n ~ 0 as n ~ IXn -II <Yn. Then Xn ~ / as n ~. 00. and that. (n = 1,2, ... ).. 00.. Proof The inequality IXn -II < Yn is equivalent to 1- Yn < x., < 1+ Yn' But, using proposition 4.8, 1- Yn ~ I as n ~ and I + Yn ~ I as n ~ Hence xn ~ I as n ~ by theorem 4.10. 00. 00.. 00,. 4.12. Example. Suppose that [x I < 1. Prove that. x" ~ 0 as n ~. 00.. Proof Write 1. n. [x]. Thus. = (1. + h)" =. 1 1 + nh. + ~n(n - 1)h2 + ... + h". <~ nh.

(41) Convergent sequences Ixn-OI=lxln<". 0 as n. 33. 1. (n=1,2,. nh. ... ).. But 1/n. -+. 4.13. Example In the problem of the bulldozers and the bee, we obtained. -+. 00. (by exercise 4.6(2)). Thus, by corollary 4.11,. the sequence. H1-(-it}·. Xn =. By example 4.12, (-. 00. and hence Xn --+ !as n. -+. 00.. Example Let x >0. Prove that. 4.14. 1=. it -+ 0 as n -+. al. (x-l)/n. Proof In the inequality of the arithmetic and geometric means, take =a2 = ... =an-1 and an =x. Then Gn =xlln and zl , = (n -1 +x)/n =. + 1.. Hence. Xlln~ x-I __. i.e.. Xlln -1 ~. + 1. n. l (x -. 1).. n. (Alternatively, take X = y = 1 in exercise 3.11(6).) Assume to begin with that x ;;. 1. Then. o~. X. l/n - 1 ~. l (x -. 1). n. and hence xl/n -+ 1 as n -+ 00 by the sandwich theorem. If 0 x = y-l where y > 1. But j ':" -+ 1 as n -+ 00 and hence .. xlln = -. 1. 1 -+ -. )'lln. 1. as n. . -+. 00. by proposition 4.8(iii).. 4.15. Monotone sequences A sequence (xn) is increasing if. xn+l ;;'xn. (n = 1,2, ... ).. < x < 1, then.

(42) 34. Convergent sequences. Similarly, (xn) is decreasing if. Xn+l~Xn. (n=1,2,. (Xn). ... ).. increases. (XII). decreases. A sequence (xn) is strictly increasing if xn + 1 > Xn (n = 1,2, ... ). It is strictly decreasing ii xc ., < Xn (n = 1,2, ... ). A sequence which is either increasing or else decreasing is called monotone.. 4.16. Example. (i) (ii) (iii) (iv) decreasing.. 4.17. The The The The. sequence (2n) is strictly increasing. sequences tn -1) and (- n) are both strictly decreasing. sequence (1) is both increasing and decreasing. sequence «(- 1)") is not monotone. It is neither increasing nor. Theorem. (i) If (xn) is increasing and bounded above, then it converges to its smallest upper bound. (ii) If (x n> is decreasing and bounded below, then it converges to its largest lower bound.. Proof If (xn) is decreasing and bounded below, then (- xn) is increasing and bounded above. Hence it is only necessary to prove (i). Suppose that (xn) increases and is bounded above with smallest upper bound B. We must show that Xn -+ B as n -+ Let E > 0 be given. Since B - E is not an upper bound, there exists at least one term XN of the sequence such that 00.. xN>B-E. But (xn) increases. Hence, for any n. Xn ~xN>B-E.. > N, x., ~. XN.. Thus, for any n. > N,.

(43) 35. Convergent sequences Also B is an upper bound for the sequence. Hence Thus, for any n > N,. i.e.. B-E<Xn. ~B.. B-E<Xn. <B+e. IXn - BI. x., ~. B (n. = 1,2,. ... ).. < E.. Given any E > 0, we have found an N such that, for any n Hence Xn ~ Bas n ~ 00.. > N,. IXn - BI. < E.. Bri----------------------II--------~~~ '-1-----------I I I. I I I I I I. I I. I I. I. N. 4.18 Examples (i) The sequence «(n - 1)/ n>increases and we know from exercise 3.6(1) that it is bounded above with smallest upper bound 1. By theorem 4.17 it follows that n-l --. -+. i as n. -+. 00.. n. (ii) If 0 < x < I, the sequence (xn> decreases and is bounded below with largest lower bound 0 (exercise 3.6(2)). Thus. 4.19 Example sequence. The examples above are nothing new. But consider the. «(1 + n-l)n>. We first show that this sequence increases. In the inequality for the geometric and arithmetic means, take al = a2 = ... = an-l = 1+ (n _1)-1 and an = I. Then.

(44) 36. Convergent sequences 1 )<n-lJ/n (1 +-n-l. Hence. (1 +. n~r-'. (n-l)(1+(n-1tl)+l. ~. =. n. (n=. <; ~ +~". 2,3, ... ). and so the sequence increases. To be able to deduce the convergence of the sequence from theorem 4.17, we also need to prove that the sequence is bounded above. By the binomial theorem,. 1. ~l+l+-+-+. 1. 2! 3!. ~ 1 + 1 +2+1. =. 1+. 1. -. 1. n!. 212+ ... + 2n1-t (b ecause. (it. 1-. ... +-. 21. = 1. + 2(1. -. 2n -I,,;:: f) "",n.. (!t) < 3.. 2. Hence the sequence is bounded above by 3. From theorem 4.17 it follows that «(1 + n-I)n) converges to a limit Band B ~ 3. (In fact, B = e = 2·718 ... ). 4.20. Exercise (1) Prove that n3+Sn2+2 2n3 + 9. ~ ~ as n ~. 00.. (2) Decide for what values of x the limit "1i.m~(~ :. ~:l. exists. Draw a graph which plots the value of the limit against the value ofx. (3) Use the sandwich theorem to prove that v(n+. l)-Vn~Oasn~oo..

(45) Convergent sequences. 37. * (4) Let x be a positive number, and let N be the smallest natural number satisfying N > x. Prove that. ::~:-:)! (~rN". (n "N).. Deduce that x'[n: -+ 0 as n -+ 00. * (5) Let ~ be any positive rational number and let lx] < 1. Show that there exists a natural number N such that (1 + 1/N)Ct+ 1 lx] ~ 1. Deduce that InCt+1xnl ~ INCt+1XNI. (n ~N).. Hence show that nCtxn-+ 0 as n -+ 00. * (6) Prove that the sequence tn 1In) decreases for n ~ 3. [Hint: example 4.19.J Hence prove that the sequence converges. (The convergence of (n 1In) can also be established by a method like that of example 4.12. If n = (1 + hn)", then it folluws from the binomial theorem that n(n - 1) 2. 2. hn. Therefore h ;. < n. -+. 0 as n. -+. 00.). 4.21. Some simple properties of convergent sequences. 4.22. Theorem A sequence can have at most one limit. Proof Suppose that x., -+ I as n -+ 00 and Xn -+ m as n. -+. 00.. Let. E. > 0 be. given. Then I/-ml. =. I/-xn+xn-ml~l/-xnl+lxn-ml<E+E. =. 2E. (triangle inequality) provided that n is sufficiently large. But, from example 1.7 it follows that; if o ~ W - m I < E for every E > 0, then II- m I = 0, i.e. I = m.. 4.23. Theorem Suppose that (i) Ifxn~a(n=I,2, (ii) If Xn ~ b (n. =. 1,2,. Xn -+. I as n. -+. 00.. ),then/~a. ), then I ~ b.. Proof If x., ~ b (n = 1,2, ... ), then - x., ~ - b (n = 1,2, ... ). Hence we need only prove (i). Let E > O. Then there exists an N such that, for any n > N,. <E E < x.; N, a ~ Xn < 1 + e. Hence, given any e > 0, a < 1 + €. From example 1.7 it follows that a < I.. Example Suppose that x.; -+ I as n -+ If x.; > a (n = 1,2, ... ), it is tempting to conclude that I > a. But this may not be true. For example, Iln -+ as n -+ and 11n > (n = 1,2, ... ). But we certainly cannot conclude that 0>0.. 4.24. 00.. °. 00. 4.25. °. Theorem Any convergent sequence is bounded. Proof Let Xn. I as n -+ We have to find a K such that IXnl ~ K (n = 1,2, ... ). (See proposition 2.3.) It is true that, for any € > 0, there exists an N such that, for any n > N, Ixn -II < e. In particular, this is true when e = 1, i.e. there exists an N I such that, for any n > NI, IXn. -II. -+. 00.. < 1.. From theorem 1.18it follows that, for any n Ixnl-III. ~. IXn. [x»] < III + 1. i.e.. > NI,. <1 (n > NI).. -II. The result now follows if we take. 4.26. Divergent sequences. A divergent sequence is a sequence which does not converge. Any unbounded sequence is therefore divergent (theorem 4.25). However bounded divergent sequences exist as well.. 4.27. Example The sequence «(- 1)") diverges. Proof The sequence is obviously bounded. Suppose that (- l )". -+ I as Then, given any € > 0, we can find an N such that, for any /I > N, I(- 1)" - II < €. But there are both even and odd values of n greater than Nand so 11-/1< € and 1- 1 -II < €. These inequalities must be true for any e > and so we obtain the contradiction I = 1 and I = - 1.. n. -+. 00.. °. We say that a sequence (xn) diverges to + and write Xn for any H> 0, we can find an N such that, for any n > N, 00. -+. +. 00. as n. -+. 00. if,.

(47) Convergent sequences. 39. Xn >H. Similarly, a sequence (xn) diverges to - 00 and x n -+ H> 0, we can find an N such that, for any n > N,. -. 00. as n. -+. 00. if, for any. xn<-H. o. I. N. I I I I I. I. H ---------~------. I. -H. o XII. xn. ~oo. N as n --+-. I I. --------~-----. 00. Xl/~-OO. as n. -+. 00. A divergent sequence <xn> which does not satisfy xn -+ + 00 as n -+ 00 or 00 as n -+ 00 is said to oscillate. An example is the sequence «(- l)").. -+ -. 4.28 n -+ 00.. Example Let. Q. Proof Let H. >. be a positive rational number. Prove that n". °. -+. +. 00. as. be given. We have to find an N such that, for any. n>N, nCt.>H i.e.. n. > HI/Ct... We simply choose N = H"Ct..Then, for any n. > N, n > HI/a. and hence nCt.> H.. A common error is to suppose that proposition 4.8 (about combining con00. For example, it is tempting to suppose that, if x.; -+ 00 as n -+ 00 and j/., -+ 00 as n -+ 00, then xn - Yn -+ 00 - 00 = as n -+ 00. But 00 is not a real number and cannot be treated as such. Indeed, one should immediately be warned that something is wrong by the appearance of the totally meaningless expression 00 - 00. To press the point home, we consider the example x., = n2 (n = 1,2, ... ) and y., = n (n = 1,2, ... ). Then x., -+ 00 as n -+ 00 and Yn -+ 00 as n -+ 00, cut Xn - Yn -+ 00 as n -+00. Or, again, take x., = (n + 1) (n = 1,2, ... ) and y., =n (n = 1,2, ... ). This time xn - Yn -+ 1 as n -+ 00.. vergent sequences) applies to sequences which diverge to + 00 or -. °. f'} J. I! ,I.

(48) 40. Convergent sequences. 4.29. Exercise (1) Suppose that Xn ~ I as n ~ (i) IXn -II ~ 0 as n ~ 00 (ii) IXnl ~ III as n ~ 00. (2) Suppose that xn ~ I as n ~ that, for any n > N, Xn. 00.. Prove that. 00.. If I> 0, prove that there exists an N such. > !/.. (3) Prove that (ii) 00 as n ~ 00 (i) 2n ~ + 00 as n ~ 00 (iii) «(-It/!) oscillates. (4) Let xn > 0 (n = I, 2, ... ). Prove that x., ~ 0 as n ~ 00 if and only if l/xn ~ 00 a~~n ~ 00. (5) Suppose that (xn) increases and is unbounded above. Prove that Xn ~ + 00 as n ~ 00. If (xn) decreases and is unbounded below, prove that Xn ~ - 00 as n ~ 00. * (6) Let S be a non-empty set of real numbers and suppose that d(~, S) = 0 (see exercise 2.l3(4)). Show that, for each n EN, we can find an x; E S such that I~- xnl < l/n. Deduce that Xn ~ ~ as n ~ 00. If S is bounded above, show that a sequence of points of S can be found which converges to its supremum. If S is unbounded above, show that a sequence of points of S can be found which diverges to + 00.. vn ~ -.

(49) 5. SUBSEQUENCES. 5.1. Subsequences. Suppose that (xn) is a sequence and that ( n; ) is a strictly increasing sequence of natural numbers. Then the sequence (xn ) r. is called a subsequence of (xn>. The first few terms of some subsequences of. X2, X4, X6,. (X3r+s>. = = =. (X2r>. =. X2,X4,XS,X16,···. (Xr+1> (X2r>. (xn. > are listed below.. X2,X3,X4,··· .••. xS,XU,X14,X17,'". Roughly speaking, if we think of a sequence as a list of its terms, then we obtain a subsequence by crossing out some of the terms. For example, (X2n> may be obtained from (x n > by crossing out terms as below.. J. The insistence that t.n; > is strictly increasing is of some importance. It means, for example, that the sequence whose first few terms are. is not a subsequence of (xn>. The fact that ( n; > is strictly increasing has the consequence that n; ~ r. (r = 1,2, ... ).. This is easily proved by induction. 5.2 of (xn. Theorem Suppose that. Xn -+. I as n -+. 00. and that (x,) is a subsequence. >. Then Xn. r. -+ I. as r. -+. 00.. Proof Let E > 0 be given. Since x., -+ I as n that, for any n > N,. -+. 00,. there exists an N such. 41.

(50) 42. Subsequences. Ix., -II < €. Let R = N. Then, for any r. > R,. Ixn-ll<f. r. We have shown that, given any e that, for any r > R,. nr;;:' r > R = N and so n;. > 0, we can. > N.. Thus. find an R (namely R. = N). such. Ixnr-ll<€.. Hence xn. -jo. r. I as r. -jo. 00.. 5.3 Example We have seen that the bounded sequence < (- l)")diverges. A simpler proof can be based on theorem 5.2. Obviously the subsequence of odd terms tends to - 1 and the subsequence of even terms to + 1. But, if < (- l )" ) converged, all its subsequences would tend to the same limit. Example In example 4.4 we showed that, if x > 0, then xlln 1 as We prove this again by a different method. As in example 4.4 we need only consider the case x ;;:.1 . If x;;:. 1, then xlln ;;:. 1 (n = 1, 2, ... ) and hence the sequence < x'!") is bounded below by 1. Moreover. 5.4 n. -jo. -jo. 00.. xlln/xl/(n+l). = x(n+l-n)/n(n+l). = x1ln(n+1);;:'1. (n = 1,2, ... ). and therefore (Xlln) decreases. From theorem 4.17 it follows that t.x'!") converges to a limit 1 and 1 ;;:. 1. What is the value of /? By theorem 5.2, we know that all subsequences of (Xl In) must also tend to /. Hence Xl/2l1-jo/asn-jooo. Using proposition 4.8, it follows that xlln. =. x'l?": x'l?" -jo/.1. =. /2 as n. -jo. 00.. But a sequence can only have one limit. Thus. / = 12. °. and so / = or 1 = 1. But I;;:' 1 and hence 1 = 1. We have shown that xlln-jolasn-jooo.. Note: A warning is appropriate here. Do not use this method of working out the value of a limit until you have shown that the sequence converges. For example, if you thoughtlessly tried to evaluate the limit of the sequence «(-1)n) by looking at its subsequences, you would obtain / = + 1 = - 1..

(51) Subsequences 5.5 Example inductively by. Let a. Xn+! = !exn. Obviously Xn. 43. > 0 and let Xl>. +. ... ). But further. +a. -_2 xn. 2 xn+!xn. (n = 1,2, ... ).. ax;;!). > 0 (n = 1,2,. O. Define the rest of the sequence (xn >. We know in advance that this quadratic equation in x.; has a real solution. Thus 'b2 - 4ac;> 0', i.e.. > 0, it follows ;> va (n =. that. Since xn+! xn+l. We wish to prove that Xn - Xn+l. (xn). 1,2, ... ). decreases and so we consider. = Xn - HXn. + ax;;l). ;> 0 (n. 2,3, ... ).. =. Hence the sequence (xn) decreases and is bounded below by we omit its first term). From theorem 4.1 7 it follows that X n -+. 1 as n. -+. Va (provided. 00. va.. where I;> What is the value of I? By theorem 5.2 we also have Xn+l -+ 1 as n xn+!. = HXn. Now 1:::/=0 (because I;> xn+l. = !exn. +. -+. But. 00.. ax;;!).. va). Hence, by proposition + ax;;!). -+. ~(l. + ai-I). as n. -+. 4.8, 00.. Since a sequence can have at most one limit, I = t(l i.e.. + arl). 12 = a.. va. Thus 1 = or 1 = therefore shown that Xn. -+. va as n. va. But we know that I;> va. Hence 1 = va. We have -+. 00..

(52) Subsequences. 44. 5.6 Example We attack the previous example by a different method. Again let a> 0 but this time let Xl be unspecified for the moment. Define the rest of the sequence inductively by Now. xn+l = HXn +ax;;l). (n = 1,2, ... ).. Xn+l - va = !exn + ax;;l) - va. 1. 1. 2Xn (2Xn_1. =. 1 2 (2 ) Xn-2 )4 (Xn-2. - va)8. 1 1 1 . / 2n -2x.; (2X _ )2' .. (2XI )2n-1 (XI - va) . n 1. If we assume that XI ~ -Ja, then it follows, as in example 5.5, that Xn ~ va = 1, 2, ... ). Hence. (n. IXn+1 - va I <;. =. 2va (XI 2va. val2n. We know that, if I y I < 1, then y n --+ 0 as n --+ that y2n --+ 0 as n --+ Hence our argument shows that Xn --+ va as n. 00.. By theorem 5.2 it follows. 00.. XI -. va. /. --+. 00. provided that. <1.. 2va We have already assumed that XI ~ va and so we have shown that Xn --+ va as n --+ provided that va <; X I < 3 va. Of course we already know from example 5.5 that this was true under the weaker hypothesis that XI> O. However, the 00.

(53) 45. Subsequences. argument given here is interesting because it allows us to estimate how 'good' an approximation x; is to ../a. For example, suppose we want to estimate the value of ../2. It is obvious that 1 < ../2 < 2 (because 12 = 1 < 2 and 22 = 4> 2). For a better estimate tab a = 2 and Xl = 2 in the argument above. Then 8. [x. ../21 ~ 2../2 (2 - ../2J. -. =. 2~. 4. -1)) 2~. 8. 2../2 (../2(../2. 2../2 =28(../2-1)8. Using the crude estimate ../2 < 2 we obtain IX4 -: ../21 ~ 2-6 = i4 < J·016. Hence X4 = = 1-414 ... differs from ../2 by at most 0·016. (A better estimate for ../2 may be obtained by evaluating Xs or by starting with Xl = ~and using the estimate ../2 <. m. n. 5.7. Exercise (I). Given that. t n'!"). converges (see exercise 4.20(6)), show that. by considering the subsequence «(2n )1/2n); (2) A sequence (xn) is defined by Xl = hand. =. Xn+1. x~ + k. where 0 < k < iand h lies between the roots a and b of the equation X2_X +k =0. Prove that a < Xn+1 ~ Xn 0 and x.> 0 and (xn) is defined inductively by x. n+1. ---. k 1+x ' n. show that one of the sequences (X2n) and (X2n-1) is increasing and the other decreasing. Prove that both sequences converge to the limit I which is the posi. tive root of the equation X2 + x = k. What conclusion may be drawn concerning (xn)? * (4) Two sequences ( x n) and (y n) are defined inductively by Xl = !and Y1 = 1 and Xn. = "/(xn-1Yn-1). (n = 2,3, ... ).

(54) 46. Subsequences. (n. =. 2,3, ... ).. <r,. Prove that xn_1 <xn <Yn-I (n = 2, 3, ... ) and deduce that both sequences converge to the same limit I, where t < I < 1. (Actually I =. rr/4.). t (5) Let Y be any real number. Prove that the sequence «(1 + yn-I)n) . is increasing for those values of n which satisfy n ~ 1 - y. [Hint: example 4.19.] Show also that the sequence is bounded above and hence converges, [Hint: From exercise 4.20(4), (2y)n/n! -+ 0 as n -+ and therefore we can find an Nsuch that, for any n ~N,yn/n! ~(DIl.] Show that the product of the limits of the two sequences «(1 + xn -I) n) and «(1 -xn-I)n) is equal to 1. ' 00. 5.8. Bolzano- Weierstrass theorem. 5.9. Theorem. Every sequence has a monotone subsequence.. Proof Let (xn ) be any sequence of real numbers. We must construct a . subsequence (xn r ) which is either increasing or decreasing. We distinguish two . cases. (i) Every set {xn: n > N} has a maximum. In this case we call find a sequence ( nr) of natural numbers such that = maxxn n>1. and so on. Obviously n 1 < n2 < n3 < .~ .. and so, at each stage, we are taking the maximum of a smaller set than at the previous stage. Hence (xnr) is a decreasing subsequence of (xn). (ii) Suppose that it is not true that all of the sets {xn: n > N} have a maximum. Then, for some N1, the set {xn: n > N1} has no maximum. It follows that, given any xm with m > N1, we can find an xn following Xm such that x n > X m. (Otherwise the biggest of x N 1 + I, ... , X m would be a maximum for {Xn:n >N.})..

(55) Subsequences. 47. We define Xn 1 = XN 1.+1 and then let Xn be the first term following x., 1 for which Xn 2 > x., 1 . Now let Xn 3 be the first term following Xn • for which Xn 3 >xn 2 . And so on. We obtain an increasing subsequence <xn r )of<xn)·. 5.10. Theorem. (Balzano-Weierstrass. theorem). Every bounded sequence of. real numbers has a convergent subsequence.. Proof Let <xn) be a bounded sequence. By theorem 5.9, <xn) has a monotone subsequence <xn r ). Since <xn) is bounded, so is <xn r ). Hence, by theorem 4.17, < xnr) converges.. 5.11. Examples. (i) The sequence < (- l )") is bounded. Two convergent subsequences are the sequence - 1, - 1, -, 1, ... of odd terms and the sequence 1, 1, 1, 1, ... of even terms. (ii) A more sophisticated example is the sequence ( r n) whose first few terms are. Every term of the sequence is a rational number between 0 and l. Hence the sequence is bounded. It has many convergent subsequences. For example 1 2 :1 4 2,4,6, g, .... ,. !,t!,!, .. tttt S.12t. .. Urn sup and lim inf. We shall not have a great deal of use for the subject matter of this section. However, it is material which has to be learned eventually and it is natural to deal with it here. Suppose that (xn > is a bounded sequence and let L denote the set of all real numbers which 'are the limit of some subsequence of ( x n ). We know from the Balzano-Weierstrass theorem that L is not empty (i.e. L =1=0). Of course, if (xn) converges, L will just consist of a single point (theorem 5.2).. 5.13t. Proposition Let < xn) be a bounded sequence and let L be the set of all real numbers which are the limit of some subsequence of < xn). Then L has a maximum and a minimum. That L is a bounded set follows from theorem 4.23. Hence L certainly has a supremum and infimum. Proposition 5.13 asserts that these actually belong to the set L. The proof is relegated to the appendix. Readers who prefer to omit.

(56) 48. Subsequences. the proof altogether will encounter it again at a later stage since it is a special case of a general theorem concerning 'closed sets'. _ If (xn) is a bounded sequence, let us denote the maximum of L by I and the minimum by 1. Then 7 is the largest number to which a subsequence of (xn) converges and { is the smallest number to which a subsequence of (xn) converges. We calli the limit superior of (xn ) and write. I =. lim sup. Xn•. n-+oo. Similarly, {is the limit inferior of (xn). and we write. { = lim inf Xn. n-+oo. (Occasionally you will see such expressions as lim sup n-+oo. means that. (xn). x., = +. is unbounded above. Similarly lim inf. xn = -. 00.. 00. This simply means that. n-+OO. (xn). 5.14 t. is unbounded below.). Examples. (i) The sequence (1In) converges with limit O. By theorem 5.2, all its subsequences converge to O. Hence L contains the single point 0 and 1 1 1 lim sup - = lim inf - = lim - = O. n-+oo n n -+ n n->oo n 00. (ii) The sequence «( - l )") is bounded. The set L consists of two points and -1 (i.e. L = {+ 1, -1}). We have limsup(-I)n. = l;liminf(-I)n. +1. =-1.. n-+oo. (iii) Let ( r n) be the sequence whose first few terms are t l, i,t t 1, t, ~,t 0 < r n < 1 (n = 1,2, ... ). From theorem 4.23 it follows that L is a subset of [0, 1] (i.e. L C [0, 1D. But the subsequence !, l. con~erges to 0. Thus L = 0. Also the subsequence t i, t t ...converges to 1. Thus I = 1. We have shown that. t t ...Obviously. lim sup r n = 1;. t ~,.... lim inf rn =. n->oo. 5.15 t. °. Exercise. t (1) Calculate (i) lim sup {(-l)n(l. +. n-1)}. (ii) lim inf {(-l)n(l n-+oo. + n-1)}.

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