`FIRST YEAR CALCULUSW W L CHENcW W L Chen, 1982, 2008.This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners.Chapter 19SEQUENCES19.1. Introduction A sequence (of numbers) is a set of numbers occurring in order. In simple cases, a sequence is defined by an explicit formula giving the n-th term xn in terms of n. We shall simply refer to the sequence xn . For example, xn = 1/n represents the sequence1 1 1, 2 , 3 , 1 , . . . . 4We shall only be concerned with the case when all the terms of a sequence are real, so that throughout this chapter, xn represents a real sequence. It is not necessary to start the sequence with x1 . However, the set of all natural numbers is a convenient tool to indicate the order with which the numbers occur. Remark. Formally, a real sequence is a function of the form f : N  R, where for every n  N, we write f (n) = xn . Let us now investigate how a sequence may behave. We begin by looking at three examples. Example 19.1.1. Consider the real sequence xn = 1/n. We are interested in the behaviour of xn as n gets large. It is easy to see that as n gets larger, then xn gets smaller. In fact, as n gets very large, then xn gets very close to 0. In this case, we say that xn  0 as n  . Example 19.1.2. Consider the real sequence xn = n2 . It is easy to see that as n gets larger, then xn also gets larger. In fact, xn can get arbitrarily large, as long as n is large enough. In this case, we say that xn   as n  .Chapter 19 : Sequences page 1 of 11First Year CalculuscW W L Chen, 1982, 2008Example 19.1.3. Consider the real sequence xn = (-1)n . It is easy to see that as n gets larger, then xn alternates between the values ±1, and does not get close to any real number or become infinite. Definition. We say that a real sequence xn converges to a finite limit x  R, denoted by xn  x as n   or byxlim xn = x, whenever n &gt; N .if, given any&gt; 0, there exists N  R such that |xn - x| &lt;Note that the quantity |xn - x| measures the difference between xn and its intended limit x. The definition thus says that this difference can be made as small as we like, provided that n is large enough. Note here that the choice of the real number N may well depend on the choice of the number . Definition. We say that a real sequence xn is convergent if it converges to some finite limit x as n  . Otherwise, we say that xn is divergent. Example 19.1.4. Consider the sequence xn = 1/n. Then xn  0 as n  , since |xn - 0| = 1/n &lt; whenever n &gt; N = 1 .Example 19.1.5. Consider the sequence xn = 1/n2 . Then xn  0 as n  , since |xn - 0| = 1/n2 &lt; whenever n &gt; N = 1 .Example 19.1.6. Consider the sequence xn = (n + 2)/n. Then xn  1 as n  , since 2 n+2 -1 = &lt; n n whenever n &gt; N = 2 .Example 19.1.7. Consider the sequence xn = n+1 -1 = nn+1 n(n + 1)/n. Then xn  1 as n  , since &lt; 1 &lt; 2n whenever n &gt; N = 1 . 2-1 +1n+1 nExample 19.1.8. Consider the sequence xn = (2n + 3)/(3n + 4). Then xn  2/3 as n  , since 2n + 3 2 1 1 - &lt; &lt; = 3n + 4 3 3(3n + 4) 9n whenever n &gt; N = 1 . 9Remark. Note that the inequality |xn - x| &lt; is equivalent to the inequalities x - &lt; xn &lt; x + . Note also that the convergence of a sequence is not affected by the initial terms. A simple and immediate consequence of our definition of convergence is the following result which we shall prove in Section 19.4. PROPOSITION 19A. The limit of a convergent real sequence is unique.Chapter 19 : Sequences page 2 of 11First Year CalculuscW W L Chen, 1982, 2008Definition. A real sequence xn is said to be bounded if there exists a number M  R such that |xn |  M for every n  N. Example 19.1.9. The sequence xn = 1/n is bounded. Clearly |xn |  1 for every n  N. Example 19.1.10. The sequence xn = 1/n2 is bounded. Clearly |xn |  1 for every n  N. Example 19.1.11. The sequence xn = (n + 2)/n is bounded. Clearly |xn | = n+2 2 =1+ 3 n n for every n  N.Example 19.1.12. The sequence xn = |xn | =(n + 1)/n is bounded. Clearly 1+  1  2 n for every n  N.n+1 = nExample 19.1.13. The sequence xn = (2n + 3)/(3n + 4) is bounded. Clearly |xn | = 2n + 3 2n + 3 2 1 5  = +  3n + 4 3n 3 n 3 for every n  N.Note that the bounded sequences in Examples 19.1.9­19.1.13 are also the convergent sequences in Examples 19.1.4­19.1.8 respectively. These are examples which illustrate the fact that convergence implies boundedness. More precisely, we have the following result which we shall prove in Section 19.4. PROPOSITION 19B. A convergent real sequence is bounded. The next example shows that a bounded real sequence is not necessarily convergent. Example 19.1.14. The sequence xn = (-1)n is bounded. Clearly |xn |  1 for every n  N. We now show that this sequence is not convergent. Let x be any given real number. We shall show that the sequence xn does not converge to x. Note first of all that for every n  N, we have |xn+1 - xn | = 2. We next use the triangle inequality, that for any ,   R, we have | + |  || + ||. By taking  = xn+1 - x and  = x - xn , we have 2 = |xn+1 - xn | = |xn+1 - x + x - xn |  |xn+1 - x| + |x - xn | = |xn+1 - x| + |xn - x|. It follows that for every n  N, at least one of the two inequalities |xn+1 - x|  1 and |x - xn |  1 must hold. This clearly shows that the condition for convergence cannot be satisfied with = 1. The next result shows that we can do arithmetic on limits. See Section 19.4 for proofs. PROPOSITION 19C. Suppose that xn  x and yn  y as n  . Then (a) xn + yn  x + y as n  ; (b) xn yn  xy as n  ; and (c) if y = 0, then xn /yn  x/y as n  . Remark. Let yn = 1/n and zn = (-1)n . Then yn  0 as n  , but zn does not converge as n  . On the other hand, it is easy to check that xn = yn zn  0 as n  . Note now that zn = xn /yn , but since yn  0 as n  , we cannot use Proposition 19C(c). Definition. We say that xn   as n   if, for every E &gt; 0, there exists N  R such that |xn | &gt; E whenever n &gt; N . In this case, we say that the sequence xn diverges to  as n  .Chapter 19 : Sequences page 3 of 11First Year CalculuscW W L Chen, 1982, 2008Remarks. (1) It can be shown that xn   as n   if and only if 1/xn  0 as n  . (2) Note that Proposition 19C does not apply in the case when a sequence diverges to . Example 19.1.15. The sequences xn = n, xn = n2 and xn = (-1)n n all satisfy xn   as n  . Example 19.1.16. Suppose that xn is a sequence of positive terms such that xn  0 as n  . For every fixed m  N, we have xm  0 as n  , in view of Proposition 19C(b). For every negative n integer m, we have xm   as n  , noting that xn &gt; 0 for every n  N. How about m = 0? n19.2. Special Results for Real Sequences Note that our discussion up to this point can be extended to sequences of complex numbers. However, real sequences are particularly interesting since the real numbers are ordered (unlike the complex numbers). This enables us to establish special results for convergence which apply only to real sequences. Detailed proofs will be given in Section 19.4. We begin with a simple example. Imagine that you have a ham sandwich, and you do the most disgusting thing of squeezing the two slices of bread together. Where does the ham go? PROPOSITION 19D. (SQUEEZING PRINCIPLE) Suppose that xn  x and yn  x as n  . Suppose further that xn  an  yn for every n  N. Then an  x as n  . Example 19.2.1. Consider the sequence an = Then 4n 4n + 3 4n + 3 + n-1 1 1 = 2 &lt; 2 &lt; = . 2n 8n 4n + 3n + 1 4n2 + 3n + 1 n Writing xn = 1 2n and yn = 1 , n 4n + 3 . 4n2 + 3n + 1we have that xn  0 and yn  0 as n  . Hence an  0 as n  . Example 19.2.2. Consider the sequence an = n-1 cos n. Writing xn = -1/n and yn = 1/n, we have xn  an  yn for every n  N. Since xn  0 and yn  0 as n  , we have an  0 as n  . Example 19.2.3. It is important that xn and yn converge to the same limit. For example, if xn = -1 and yn = 1 for every n  N, then both xn and yn converge as n  . Let an = (-1)n . Then xn  an  yn for every n  N. Note from Example 19.1.14 that an does not converge as n  . In this case, the hypotheses of Proposition 19D are not satisfied. Note that xn and yn converge to different limits, so no &quot;squeezing&quot; occurs. Example 19.2.4. · If a = 1, then · If a = 0, then · If a &gt; 1, then Consider the sequence xn = an , where a  R. There are various cases: xn = 1 for every n  N, so that xn  1 as n  . xn = 0 for every n  N, so that xn  0 as n  . a = 1 + k, where k &gt; 0. Then xn   as n  , since |an | = (1 + k)n  1 + kn &gt; EChapter 19 : Sequencesfor every n &gt;E-1 . kpage 4 of 11First Year CalculuscW W L Chen, 1982, 2008· If 0 &lt; a &lt; 1, then a = 1/b, where b &gt; 1. Hence 1/xn   as n  . It follows that xn  0 as n  . · If -1 &lt; a &lt; 0, then a = -b, where 0 &lt; b &lt; 1. We then have bn  0 as n  . Also, -bn  xn  bn for every n  N. It follows from the Squeezing principle that xn  0 as n  . · If a = -1, then xn = (-1)n does not converge as n  . · If a &lt; -1, then a = 1/b where -1 &lt; b &lt; 0. Hence 1/xn  0 as n  . It follows that xn   as n  . Our next task is to study monotonic sequences. Definition. Let xn be a real sequence. (1) We say that xn is increasing if xn+1  xn for every n  N. (2) We say that xn is decreasing if xn+1  xn for every n  N. (3) We say that xn is bounded above if there exists B  R such that xn  B for every n  N. (4) We say that xn is bounded below if there exists b  R such that xn  b for every n  N. Remark. Note that a real sequence is bounded if and only if it is bounded above and below. PROPOSITION 19E. Suppose that xn is an increasing real sequence. (a) If xn is bounded above, then xn converges as n  . (b) If xn is not bounded above, then xn   as n  . PROPOSITION 19F. Suppose that xn is a decreasing real sequence. (a) If xn is bounded below, then xn converges as n  . (b) If xn is not bounded below, then xn   as n  . Example 19.2.5. The sequence xn = 3 - 1/n is increasing and bounded above. It is not too difficult that the smallest real number B  R such that xn  B for every n  N is 3. It is easy to show that xn  3 as n  . Example 19.2.6. Consider the sequence xn = 1 + 1 1 1 + + ... + . 1! 2! n!Clearly xn is an increasing sequence. On the other hand, xn = 1 + 1 + 1 1 1 + + ... + 1·2 2·3 (n - 1)n 1 1 1 =1+1+ 1- + - + ... + 2 2 31 1 - n-1 n=3-1 &lt; 3, nso that xn is bounded above. Unfortunately, it is very hard to find the smallest real number B  R such that xn  B for every n  N. While Proposition 19E tells us that the sequence xn converges, it does not tell us the precise value of the limit. In fact, the limit in this case is the number e. Example 19.2.7. Consider the sequence xn = 1 + a + a2 + . . . + an . Then xn = n + 1 if a = 1 and xn = 1 - an+1 1-a if a = 1.Suppose that a &gt; 0. Then xn is increasing. If 0 &lt; a &lt; 1, then xn &lt; 1/(1 - a) for all n  N, and so xn converges as n  . If a  1, then xn is not bounded above, so that xn   as n  . In fact, if a = 1, then the convergence or divergence of xn depends on the convergence and divergence of an+1 , which we have considered before in Example 19.2.4.Chapter 19 : Sequences page 5 of 11First Year CalculuscW W L Chen, 1982, 200819.3. Recurrence Relations In practice, it may not always be convenient to define a sequence explicitly. Sequences may often be defined by a relation connecting two or more successive terms. Here we shall not make a thorough study of such relations, but confine our discussion to two examples. Example 19.3.1. Suppose that x1 = 3 and xn+1 = 4xn + 2 xn + 3for every n  N. Note first of all that 0 &lt; x2 &lt; x1 . Suppose that n &gt; 1 and 0 &lt; xn &lt; xn-1 . Then clearly xn+1 &gt; 0. Furthermore, xn+1 - xn = 4xn + 2 4xn-1 + 2 10(xn - xn-1 ) - = &lt; 0. xn + 3 xn-1 + 3 (xn + 3)(xn-1 + 3)It follows from the Principle of induction that xn is a decreasing sequence and bounded below by 0, so that xn converges as n  . Suppose that xn  x as n  . Then x = lim xn+1 = limnn4x + 2 4xn + 2 = . xn + 3 x+3Hence x = 2. Note that the other solution x = -1 has to be discounted, since xn &gt; 0 for every n  N. Example 19.3.2. Let s &gt; 0. Suppose that x1 &gt; 0 and that for n &gt; 1, we have xn = 1 2 xn-1 + s xn-1 .It is not difficult to show that xn &gt; 0 for every n  N. On the other hand, for n &gt; 1, we have x2 = n so that x2 - s = n and so xn+1 - xn = 1 2 xn + s xn - xn = 1 2 s - xn xn = s - x2 n  0. 2xn 1 4 x2 + n-1 s2 x2 n-1 - 2s = 1 4 xn-1 - s xn-121 4x2 + n-1s2 x2 n-1+ 2s , 0,It follows that, with the possible exception that x2  x1 may not hold, the sequence xn is decreasing and bounded below, so that xn converges as n  . Suppose that xn  x as n  . Then x = lim xn = limnn1 2xn-1 +s xn-1=1 s x+ , 2 xso that x2 = s. This gives a proof that s has a square root.Chapter 19 : Sequences page 6 of 11First Year CalculuscW W L Chen, 1982, 200819.4. Further Discussion In this section, we first give formal proofs of the various results stated in the earlier sections. Proof of Proposition 19A. Suppose that xn  x and xn  x as n  . Then given any there exist N , N  R such that |xn - x | &lt; and |xn - x | &lt; whenever n &gt; N . whenever n &gt; N , &gt; 0,Let N = max{N , N }  R. It follows that whenever n &gt; N , we have |x - x |  |xn - x | + |xn - x | &lt; 2 . Now |x - x | is a non-negative constant less than any 2 &gt; 0, so we must have |x - x | = 0, whence x =x . Proof of Proposition 19B. Suppose that xn  x as n  . Then there exists N  N such that |xn - x| &lt; 1 for every n &gt; N . Hence |xn | &lt; |x| + 1 whenever n &gt; N .Let M = max{|x1 |, . . . , |xN |, |x| + 1}. Then clearly |xn |  M for every n  N. Proof of Proposition 19C. (a) We shall use the inequality |(xn + yn ) - (x + y)|  |xn - x| + |yn - y|. Given any &gt; 0, there exist N1 , N2  R such that |xn - x| &lt; /2 and |yn - y| &lt; /2 whenever n &gt; N2 . whenever n &gt; N1 ,Let N = max{N1 , N2 }  R. It follows that whenever n &gt; N , we have |(xn + yn ) - (x + y)|  |xn - x| + |yn - y| &lt; . (b) We shall use the inequality |xn yn - xy| = |xn yn - xn y + xn y - xy| = |xn (yn - y) + (xn - x)y|  |xn ||yn - y| + |y||xn - x|. Since xn  x as n  , there exists N1  R such that |xn - x| &lt; 1 so that |xn | &lt; |x| + 1Chapter 19 : Sequenceswhenever n &gt; N1 ,whenever n &gt; N1 .page 7 of 11First Year CalculuscW W L Chen, 1982, 2008On the other hand, given any&gt; 0, there exist N2 , N3  R such that |xn - x| &lt; 2(|y| + 1) whenever n &gt; N2 ,and |yn - y| &lt; 2(|x| + 1) whenever n &gt; N3 .Let N = max{N1 , N2 , N3 }  R. It follows that whenever n &gt; N , we have |xn yn - xy|  |xn ||yn - y| + |y||xn - x| &lt; .(c) We shall first show that 1/yn  1/y as n  . To do this, we shall use the identity |yn - y| 1 1 = - . yn y |yn ||y| Since y = 0 and yn  y as n  , there exists N1  R such that |yn - y| &lt; |y|/2 so that |yn | &gt; |y|/2 On the other hand, given any whenever n &gt; N1 . whenever n &gt; N1 ,&gt; 0, there exists N2  R such that |yn - y| &lt; y 2 /2 whenever n &gt; N2 .Let N = max{N1 , N2 }  R. It follows that whenever n &gt; N , we have 1 1 |yn - y| 2|yn - y| - =  &lt; . yn y |yn ||y| |y|2 We now apply part (b) to xn and 1/yn to get the desired result. Proof of Proposition 19D. By Proposition 19C, yn - xn  0 as n  . It follows that given any &gt; 0, there exist N , N  R such that |yn - xn | &lt; /2 and |xn - x| &lt; /2 whenever n &gt; N . whenever n &gt; N ,Let N = max{N , N }  R. It follows that whenever n &gt; N , we have |an - x|  |an - xn | + |xn - x|  |yn - xn | + |xn - x| &lt; . Hence an  x as n  .Chapter 19 : Sequences page 8 of 11First Year CalculuscW W L Chen, 1982, 2008Proof of Proposition 19E. (a) Suppose that the sequence xn is bounded above. Then the set S = {xn : n  N} is a non-empty set of real numbers which is bounded above. Let x = sup S. We shall show that xn  x as n  . Given any &gt; 0, there exists N  N such that xN &gt; x - . Since the sequence xn is increasing and bounded above by x, it follows that whenever n &gt; N , we have x  xn  xN &gt; x - , so that |xn - x| &lt; . (b) Suppose that the sequence xn is not bounded above. Then for every E &gt; 0, there exists N  N such that xN &gt; E. Since the sequence xn is increasing, it follows that |xn | = xn  xN &gt; E for every n &gt; N . Hence xn   as n  . We conclude this chapter by discussing subsequences. Heuristically, a subsequence is obtained from a sequence by possibly omitting some of the terms, and keeping the remainder in the original order. We can make this more formal in the following way. Definition. Suppose that x1 , x2 , x3 , . . . , xn , . . . is a real sequence. Suppose further that n1 &lt; n2 &lt; n3 &lt; . . . &lt; np &lt; . . . is an infinite sequence of natural numbers. Then the sequence xn1 , xn2 , xn3 , . . . , xnp , . . . is called a subsequence of the original sequence. Example 19.4.1. The sequence 2, 4, 6, 8, . . . of even natural numbers is a subsequence of the sequence 1, 2, 3, 4, . . . of natural numbers. Example 19.4.2. The sequence 2, 3, 5, 7, . . . of primes is not a subsequence of the sequence 1, 3, 5, 7, . . . of odd natural numbers. Example 19.4.3. The sequence 1, 2, 3, 4, . . . of natural numbers is a subsequence of the sequence     1, 2, 3, 4, . . . . We shall establish the following important result in analysis. PROPOSITION 19G. Every bounded sequence of real numbers has a convergent subsequence. Proof. We say that n  N is a &quot;peak&quot; point if xn &gt; xm for every m &gt; n. There are two possibilities: (i) Suppose that there are infinitely many peak points n1 &lt; n2 &lt; n3 &lt; . . . &lt; np &lt; . . . . Then xn1 &gt; xn2 &gt; xn3 &gt; . . . &gt; xnp &gt; . . . is a decreasing subsequence, clearly bounded below, and is therefore convergent by Proposition 19F. (ii) Suppose that there are no or only finitely many peak points. Let n1 = 1 if there are no peak points, and let n1 = N + 1 if N represents the largest peak point. Then n1 is not a peak point, and so there exists n2 &gt; n1 such that xn1  xn2 . On the other hand, n2 is not a peak point, and so there exists n3 &gt; n2 such that xn2  xn3 . Continuing inductively, we conclude that there exists an infinite sequence n1 &lt; n2 &lt; n3 &lt; . . . &lt; np &lt; . . . of natural numbers such that xn1  xn2  xn3  . . .  xnp  . . . is an increasing subsequence, clearly bounded above, and is therefore convergent by Proposition 19E.Chapter 19 : Sequences page 9 of 11First Year CalculuscW W L Chen, 1982, 2008Problems for Chapter 19 1. Use the -N definition to prove each of the following convergence as n  : 3n + 7 3 3 a) xn =  b) xn = 2  0 2n + 9 2 n 2. Use the arithmetic of limits to find the limit of each of the following sequences: n2 + 1 3n2 + 4n + 5 a) xn = 2 b) xn = 2 n +5 2n - 3n + 7 3. Use the Squeezing principle to find the limit of each of the following sequences: 1 n n a) xn = sin cos b) xn = 1/n if n is prime n 3 4 0 otherwise 1/n if n is odd c) xn = -1/n2 if n is even 4. Find the limit of each of the following sequences, and try to justify your assertions: 1 + 2 + ... + n n b) xn = a) xn = n 2 n2 n (-1) 2n + 1 1 3n + 4 c) xn = + d) xn = n + 4n + 3 3n + 2 2 2n + 9 n (-1) 2n + 3 n 2 f) xn = 2 cos e) xn = + n n2 3n + 4 6 2 n if n  10 g) xn = 1/n if n &gt; 10 5. For what values of a, b  R does the sequence xn = a + b(-1)n converge? 6. Find a real sequence xn that satisfies the following conditions simultaneously: a) 0 &lt; xn &lt; 1 for every n  N; b) xn = 1/2 for every n  N; and c) xn  1/2 as n  . 7. Suppose that x is a real number. Discuss the convergence of the sequence xn = to distinguish the four cases |x| &gt; 1, |x| &lt; 1, x = 1 and x = -1. x + xn , taking care 1 + xnHarder Problems for Chapter 19  8. A sequence xn is defined inductively by x1 = 1 and xn+1 = xn + 6 for every n  N. a) Prove by induction that xn is increasing, and xn &lt; 3 for every n  N. b) Deduce that xn converges as n   and find its limit. 9. Suppose that x1 &lt; x2 and xn+2 = 1 (xn+1 + xn ) for every n  N. Show that 2 a) xn+2 &gt; xn for every odd n  N; b) xn+2 &lt; xn for every even n  N; and c) xn  1 (x1 + 2x2 ) as n  . 3 1 10. Suppose that an  L as n  , and that sn = (a1 + . . . + an ) for every n  N. Show that sn  L n as n  . [Hint: Consider first the case L = 0.]Chapter 19 : Sequences page 10 of 11First Year CalculuscW W L Chen, 1982, 200811. Show that the sequence xn =1 is increasing and bounded above. n [Remark: Hence it converges. The limit is e.] 1+n12. For each of the following sequences xn , find monotonic subsequences: 1 if n even a) xn = an + b b) xn = 0 if n odd 3n c) xn = cos d) xn = 1/n if n prime 4 0 otherwiseChapter 19 : Sequencespage 11 of 11`

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