Sequences Limit Review Sequences Convergence and Divergence 1

In Section 2.1, we consider (infinite) sequences, limits of sequences, and bounded and monotonic sequences of real numbers. In addition to certain basic properties of convergent sequences, we also study divergent sequences and in particular, sequences that tend to positive or negative infinity. We present a number of methods to discuss convergent sequences together with techniques for calculating their limits. Also, we prove the bounded monotone convergence theorem (BMCT), which asserts that every bounded monotone sequence is convergent. In Section 2.2, we define the limit superior and the limit inferior. We continue the discussion with Cauchy sequences and give ex-amples of sequences of rational numbers converging to irrational numbers. As applications, a number of examples and exercises are presented. 2.1 Sequences and Their Limits An infinite (real) sequence (more briefly, a sequence) is a nonterminating collection of (real) numbers consisting of a first number, a second number, a third number, and so on: a1 , a2 , a3 , . . . . Specifically, if n is a positive integer, then an is called the nth term of the sequence, and the sequence is denoted by {a1 , a2 , . . . , an , . . .} or, more simply, {an } . For example, the expression {2n} denotes the sequence 2, 4, 6, . . .. Thus, a sequence of real numbers is a special kind of function, one whose domain is the set of all positive integers or possibly a set of the form {n : n ≥ k} for some fixed k ∈ Z, and the range is a subset of R. Let us now make this point precise. Definition 2.1. A real sequence {an } is a real-valued function f defined on a set {k, k + 1, k + 2, . . .}. The functional values 23 24 f (k), f (k + 1), f (k + 2), . . . are called the terms of the sequence. It is customary to write f (n) = an for n ≥ k, so that we can denote the sequence by listing its terms in order; thus we write a sequence as {an }n≥k or ∞ {an+k−1 }∞ n=1 or {an }n=k or {ak , ak+1 , . . .}. The number an is called the general term of the sequence {an } (nth term, especially for k = 1). The set {an : n ≥ k} is called the range of the sequence {an }n≥k . Sequences most often begin with n = 0 or n = 1, in which case the sequence is a function whose domain is the set of nonnegative integers (respectively positive integers). Simple examples of sequences are the sequences of positive integers, i.e., the sequence {an } for which an = n for n ≥ 1, {1/n}, {(−1)n }, {(−1)n + 1/n}, and the constant sequences for which an = c for all n. The Fibonacci sequence is given by a0 , a1 = 1, a2 = 2, an = an−1 + an−2 for n ≥ 3. The terms of this Fibonacci sequence are called Fibonacci numbers, and the first few terms are 1, 1, 2, 3, 5, 8, 13, 21. 2.1.1 Limits of Sequences of Real Numbers A fundamental question about a sequence {an } concerns the behavior of its nth term an as n gets larger and larger. For example, consider the sequence whose general term is 1 n+1 =1+ . n n It appears that the terms of this sequence are getting closer and closer to the number 1. In general, if the terms of a sequence can be made as close as we please to a number a for n sufficiently large, then we say that the sequence converges to a. Here is a precise definition that describes the behavior of a sequence. an = Definition 2.2 (Limit of a sequence). Let {an } be a sequence of real numbers. We say that the sequence {an } converges to the real number a, or tends to a, and we write a = lim an n→∞ or simply a = lim an , if for every  > 0, there is an integer N such that |an − a| <  whenever n ≥ N. 2.1 Sequences and Their Limits 25 In this case, we call the number a a limit of the sequence {an }. We say that the sequence {an } converges (or is convergent or has limit) if it converges to some number a. A sequence diverges (or is divergent) if it does not converge to any number. For instance, in our example above we would expect n+1 = 1. lim n→∞ n The notions of convergence and limit of a sequence play a fundamental role in analysis. If a ∈ R, other notations for the convergence of {an } to a are lim (an − a) = 0 n→∞ and an → a as n → ∞. The notation a = lim an means that eventually the terms of the sequence {an } can be made as close to a as may be desired by taking n sufficiently large. Note also that |an − a| <  for n ≥ N ⇐⇒ an ∈ (a − , a + ) for n ≥ N . That is, a sequence {an } converges to a if and only if every neighborhood of a contains all but a finite number of terms of the sequence. Since N depends on , sometimes it is important to emphasize this and write N () instead of N . Note also that the definition requires some N , but not necessarily the smallest N that works. In fact, if convergence works for some N then any N1 > N also works. To motivate the definition, we again consider an = (n + 1)/n. Given  > 0, we notice that    1 n + 1 1    n − 1 = n <  whenever n >  . Thus, N should be some natural number larger than 1/. For example, if  = 1/99, then we may choose N to be any positive integer bigger than 99, and we conclude that   n + 1    <  = 1 whenever n ≥ N = 100. − 1  n  99 Similarly, if  = 2/999, then 1/ = 499.5, so that   n + 1  2    n − 1 <  = 999 whenever n ≥ N = 500. Thus, N clearly depends on . The definition of limit makes it clear that changing a finite number of terms of a given sequence affects neither the convergence nor the divergence of the sequence. Also, we remark that the number  provides a quantitative measure of “closeness,” and the number N a quantitative measure of “largeness.” We now continue our discussion with a fundamental question: Is it possible for a sequence to converge to more than one limit? 26 2 Sequences: Convergence and Divergence Theorem 2.3 (Uniqueness of limits). The limit of a convergent sequence is unique. Proof. Suppose that a = lim an and a = lim an . Let  > 0. Then there exist two numbers N1 and N2 such that |an − a| <  for n ≥ N1 and |an − a | <  for n ≥ N2 . In particular, these two inequalities must hold for n ≥ N = max{N1 , N2 }. We conclude that |a − a | = |a − an − (a − an )| ≤ |an − a| + |an − a | < 2 for n ≥ N . Since this inequality holds for every  > 0, and |a − a | is independent of , we must have |a − a | = 0, i.e., a = a . Also, as a direct consequence of the definition we obtain the following: If an → a, then an+k → a for any fixed integer k. Indeed, if an → a as n → ∞, then for a given  > 0 there exists an N ∈ N such that |an − a| <  for all n ≥ N . That is, |an+k − a| <  for all n + k ≥ N + k = N1 or |am − a| <  for m ≥ N1 , which is same as saying that am → a as m → ∞. Definition 2.4. A sequence {an } that converges to zero is called a null sequence. Examples 2.5. (i) The sequence {n} diverges because no matter what a and  we choose, the inequality a −  < n < a + , i.e., |n − a| < , can hold only for finitely many n. Similarly, the sequence {2n } diverges. (ii) The sequence defined by {(−1)n } is {−1, 1, −1, 1, . . .}, and this sequence diverges by oscillation because the nth term is always either 1 or −1. Thus an cannot approach any one specific number a as n grows large. Also, we note that if a is any real number, we can always choose a positive number  such that at least one of the inequalities a −  < −1 < a +  or a −  < 1 < a +  is false. For example, the choice  = |1 − a|/2 if a = 1, and  = |1 + a|/2 if a = −1, will do. If a = 1 or −1, choose  to be any positive real number less than 1. Thus the inequality |(−1)n − a| <  will be false for infinitely many n. Hence {(−1)n } diverges. (iii) The sequence {sin(nπ/2)}n≥1 diverges because the sequence is {1, 0, −1, 0, 1, 0, . . .}, and hence it does not converge to any number, by the same reasoning as above. (iv) The sequence {(−1)n /n} converges to zero, and so it is a null sequence. • 2.1 Sequences and Their Limits 27 Definition 2.6. A sequence {an } is bounded if there exists an R > 0 such that |an | ≤ R for all n. A sequence is unbounded if it is not bounded. Since a convergent sequence eventually clusters about its limit, it is fairly evident that a sequence that is not bounded cannot converge, and hence the next theorem is not too surprising; it will be used in the proof of Theorem 2.8. Theorem 2.7. Every convergent sequence is bounded. The converse is not true. Proof. Let {an }n≥1 converge to a. Then there exists an N ∈ N such that |an − a| < 1 =  for n ≥ N . It follows that |an | < 1 + |a| for n ≥ N . Define M = max{1 + |a|, |a1 |, |a2 |, . . . |aN −1 |}. Then |an | < M for every n ∈ N. To see that the converse is not true, it suffices to consider the sequence {(−1)n }n≥1 , which is bounded but not convergent, although the odd terms and even terms both form convergent sequences with different limits. 2.1.2 Operations on Convergent Sequences The sum of sequences {an } and {bn } is defined to be the sequence {an + bn }. We have the following useful consequences of the definition of convergence that show how limits team up with the basic algebraic operations. Theorem 2.8 (Algebra of limits for convergent sequences). Suppose that limn→∞ an = a and limn→∞ bn = b, where a, b ∈ R. Then • • • • limn→∞ (ran + sbn ) = ra + sb, r, s ∈ R. [Linearity rule for limn→∞ (an bn ) = ab. [Product rule for provided b = 0. [Quotient rule for limn→∞ an /bn = a/b, √ √ √ limn→∞ m an = m a, provided m an is defined for all n and sequences] sequences] sequences] √ m a exists. Proof. The linearity rule for sequences is easy to prove. The quotient rule for sequences is easy if we prove the product rule for sequences (see also Questions 2.44(33) and 2.44(34)). We provide a direct proof. We write an bn − ab = (an − a)bn + (bn − b)a. Since every convergent sequence must be bounded, there exists an M > 0 such that |bn | ≤ M (say), for all n. Let  > 0 be given. Again, since bn → b as n → ∞, there exists an N2 such that |bn − b| <  2(|a| + 1) for n ≥ N2 . (We remark that we could not use /2|a| instead of /[2(|a| + 1)] because a could be zero.) 28 2 Sequences: Convergence and Divergence Also by the hypothesis that an → a as n → ∞, there exists an N3 such that  for n ≥ N3 . |an − a| < 2M Finally, for n ≥ max{N2 , N3 } = N , we have |an bn − ab| ≤ |an − a| |bn | + |bn − b| |a|     M+ |a| < + = . < 2M 2(|a| + 1) 2 2 The product rule clearly follows. The proof of third part follows from Lemma 2.9. The proof of the final part is left as a simple exercise (see Questions 2.44(16)). Lemma 2.9 (Reciprocal rule). If limn→∞ bn = b and b = 0, then the reciprocal rule holds: 1 1 = . n→∞ bn b Proof. The proof is easy, and so we leave it as a simple exercise. lim Note that if an = (−1)n and bn = (−1)n−1 , then {a2n } and {an + bn } both converge, although individual sequences {an } and {bn } diverge. Example 2.10. Find the limit of each of these convergent sequences:  2   6    n − 2n + 3 n + 3n4 − 2 1 (p > 0). (b) . (c) (a) . np 5n3 n6 + 2n + 3 Solution. (a) As n grows arbitrarily large, 1/n (and hence 1/np ) gets smaller and smaller for p > 0. Thus, limn→∞ 1/np = 0. Also, we note that if  > 0, then |(1/np ) − 0| <  or n > 1/(1/p ). Thus, if N is any integer greater than 1/(1/p ), then |(1/np ) − 0| <  for all n ≥ N . Thus, for each p > 0, n−p → 0 as n → ∞. That is, {1/np } is a null sequence for each p > 0. (b) We cannot use the quotient rule of Theorem 2.8 because neither the limit for the numerator nor that for the denominator exists. On the other hand, we can divide the numerator and denominator by n3 and then use the linearity rule and the product rule. We then have   2 n2 − 2n + 3 1 1 3 − = + 3 → 0 as n → ∞. 5n3 5 n n2 n (c) Divide the numerator and denominator by n6 , the highest power of n that occurs in the expression, to obtain 1+ n6 + 3n4 − 2 = lim 6 n→∞ n + 2n + 3 n→∞ 1 + lim 3 n2 2 n5 − + 2 n6 3 n6 = 1. 2.1 Sequences and Their Limits 29 In fact, if we set an = 1 + 3 2 − 6 n2 n and bn = 1 + 2 3 + 6, n5 n then the linearity rule gives that an → 1 and bn → 1 as n → ∞. Finally, the quotient rule gives the desired limit, namely, an = 1. lim • n→∞ bn Suppose that {an } is a sequence of real numbers such that an > 0 for all but a finite number of n. Then there exists an N such that an > 0 for all n ≥ N . If the new sequence {1/an+N }n≥0 converges to zero, then we say that {an } diverges to ∞ and write lim an = ∞. Equivalently, if lim an does not exist because the numbers an > 0 become arbitrarily large as n → ∞, we write limn→∞ an = ∞. We summarize the discussion as follows: Definition 2.11 (Divergent sequence). For given sequences {an } and {bn }, we have (a) limn→∞ an = ∞ if and only if for each R > 0 there exists an N ∈ N such that an > R for all n ≥ N . (b) limn→∞ bn = −∞ if and only if for each R < 0 there exists an N ∈ N such that bn < R for all n ≥ N . We do not regard {an } as a convergent sequence unless lim an exists as a finite number, as required by the definition. For instance, √ lim n3 = ∞, lim (−n) = −∞, lim 3n = ∞, lim ( n + 5) = ∞. n→∞ n→∞ n→∞ n→∞ We do not say that the sequence {n2 } “converges to ∞” but rather that it “diverges to ∞” or “tends to ∞.” To emphasize the distinction, we say that {an } diverges to ∞ (respectively −∞) if lim an = ∞ (respectively −∞). We note that lim(−1)n n is unbounded but it diverges neither to ∞ nor to −∞. Definition 2.12 (Oscillatory sequence). A sequence that neither converges to a finite number nor diverges to either ∞ or −∞ is said to oscillate or diverge by oscillation. An oscillating sequence with finite amplitude is called a finitely oscillating sequence. An oscillating sequence with infinite amplitude is called an infinitely oscillating sequence. For instance, {(−1)n }, {1 + (−1)n }, {(−1)n (1 + 1/n)} oscillate finitely. We remark that an unbounded sequence that does not diverge to ∞ or −∞ oscillates infinitely. For example, the sequences {(−1)n n}, {(−1)n n2 }, are all unbounded and oscillate infinitely. {(−n)n } 30 2 Sequences: Convergence and Divergence Example 2.13. Consider an = (n2 + 2)/(n + 1). Then  an = n 1 + n22 1 + n1  . From the algebra of limits we observe that 1 + n22 = 1. n→∞ 1 + 1 n lim On other hand, limn→∞ an does not exist. Indeed, we can show that an → ∞ as n → ∞. According to the definition, we must show that for a given R > 0, there exists an N such that an > R for all n ≥ N . Now we observe that an > R ⇐⇒ n + 1 + 3 > R + 2, n+1 which helps to show that an > R if n ≥ R + 2. So we can choose any positive integer N such that N ≥ R + 2. We then conclude that an → ∞ as n → ∞. Similarly, we easily have the following: (1) As in Example 2.10(c), we write n7 + 2n3 − 1 = lim n→∞ n6 + n2 + 3n + 1 n→∞ lim 1 n 1 + n24 − n17 + n15 + n36 + 1 n7 . The numerator tends to 1 as n → ∞, whereas the denominator approaches 0. Hence the quotient increases without bound, and the sequence must diverge. We may rewrite in the present notation, n7 + 2n3 − 1 = ∞. n→∞ n6 + n2 + 3n + 1 lim (2) {n/3 + 1/n}, {n3 − n}, {(n2 + 1)/(n + 1)}, and {(n3 + 1)/(n + 1)} all diverge to∞.  (3) (−1)n n2 diverges but neither to −∞ nor to ∞. (4) an → ∞ =⇒ a2n → ∞. (5) If an > 0 for all large values of n, then an → 0 =⇒ 1/an → ∞. Is the converse true? • √ Finally, we let an = n2 + 5n − n and consider the problem of finding lim an . It would not be correct to apply the linearity property for sequences (because neither lim n2 + 5n nor lim n exists as a real number). At this place it important to remember√that the linearity rule in Theorem 2.8 cannot be applied to {an }, since lim n2 + 5n = ∞ and lim n = ∞. It is also not correct to use this as a reason to say that the limit does not exist. The supporting argument is as follows. Rewriting an algebraically as 2.1 Sequences and Their Limits an = √ 5n n2 + 5n + n 2 = √ = n + 5n − n √ 2 2 n + 5n + n n + 5n + n we obtain limn→∞ 5 1+ 5 n 31 , +1 √ n2 + 5n − n = 5/2. Remark 2.14. We emphasize once again that Theorem 2.8 cannot be applied to sequences that diverge to ∞ or −∞. For instance, if an = n + 1, bn = n, and cn = n2 for n ≥ 1, then it is clear that the sequences {an }, {bn }, and {cn } diverge to ∞, showing that the limits do not exist as real numbers. Also, it is tempting to say that an − b n → ∞ − ∞ = 0 and cn − bn → ∞ − ∞ = 0 as n → ∞. Note that ∞ is not a real number, and so it cannot be treated like a usual real number. In our example, we actually have an − bn = 1 for all n ≥ 1, and cn − bn = n(n − 1) → ∞ as n → ∞. • 2.1.3 The Squeeze/Sandwich Rule In the following squeeze rule, the sequence {bn } is “sandwiched” between the two sequences {an } and {cn }. Theorem 2.15 (Squeeze/Sandwich rule for sequences). Let {an }, {bn }, and {cn } be three sequences such that an ≤ bn ≤ cn for all n ≥ N and for some N ∈ N. If lim an = lim cn = L, n→∞ n→∞ then limn→∞ bn = L. If bn → ∞, then cn → ∞. Also, if cn → −∞, then bn → −∞. Proof. Let  > 0 be given. By the definition of convergence, there exist two numbers N1 and N2 such that |an − L| <  for n ≥ N1 and |cn − L| <  for n ≥ N2 . In particular, since an ≤ bn ≤ cn for all n ≥ N , we have L −  < an ≤ b n ≤ cn < L +  for n ≥ N3 = max{N, N1 , N2 }, showing that |bn − L| <  for n ≥ N3 , as required. We leave the rest as a simple exercise. Corollary 2.16. If {cn } is a null sequence of nonnegative real numbers, and |bn | ≤ cn for all n ≥ N , then {bn } is a null sequence. 32 2 Sequences: Convergence and Divergence √ √ √ For instance, since {1/ n} is null and 1/(1 + n) < 1/ n for all n ≥ 1, √ {1/(1 + n)} is also a null sequence. Similarly, comparing 1/3n with 1/n, it follows easily that {1/3n} is a null sequence. Corollary 2.17. If limn→∞ an = 0 and |bn − L| ≤ an for all n ≥ N , then limn→∞ bn = L. Proof. By the last corollary, it follows that {bn − L} is a null sequence, and so the desired conclusion follows. Alternatively, it suffices to observe that |bn − L| ≤ an ⇐⇒ L − an ≤ bn ≤ L + an and apply the squeeze rule. For instance, using the squeeze rule, we easily have the following: (a) limn→∞ cos n2 /n = 0, because −(1/n) ≤ cos n2 /n ≤ 1/n. With the same reasoning, one has sin(nπ/2) = 0. lim n→∞ n √  √ √ √ √ (b) limn→∞ n + 1 − n = 0 and limn→∞ n( n + 1 − n) = 1/2. Moreover, √ √ 1 1 n= √ √ < √ . 2 n n+1+ n √ Note that n. For√ n = 1, this √ the above inequality is useful in estimating √ gives 2 < 1.5, and for n = 2, 4, we have 3 < 1.875 and 5 < 2.25. Indeed, for n = 2, we have √ √ √ √ 5 2 5 × 1.5 7.5 2 = < = = 1.875. 3< 2+ 4 4 4 4 0< n+1− (c) limn→∞ n/2n = 0. Indeed, using induction we easily see that 2n ≥ n2 for n ≥ 4, so that n 1 0< n ≤ . 2 n √ √ √ 2 (d) limn→∞ bn = 1 if bn = 1/( n + 1) + 1/( n2 + 2) + · · · + 1/( n2 + n). We note that n n 1 1 √ . < bn < < bn < √ , i.e., n2 + n n2 + 1 1 + 1/n 1 + 1/n2 √ √ √ (e) limn→∞ cn = ∞ if cn = 1/( n + 1) + 1/( n + 2) + · · · + 1/( n + n). We note that √ n n cn > √ = √ = bn , n+n 2 where bn → ∞ as n → ∞. 2.1 Sequences and Their Limits 33 Using the squeeze rule, Theorem 2.8, and a few standard examples allows one to calculate limits of important sequences. Example 2.18. Show that (a) lim a1/n = 1 for a > 0. n→∞ (b) lim n1/n = 1. n→∞ (c) lim n→∞ n! = 0. nn Solution. (a) We consider the cases a > 1 and a < 1, since there is nothing to prove if a = 1. Suppose first that a > 1. Then a1/n ≥ 1, and so a1/n = 1 + xn for some sequence {xn } of positive real numbers. Then by the binomial theorem, a = (1 + xn )n ≥ 1 + nxn for all n ≥ 1, which is equivalent to 0 < a1/n − 1 ≤ a−1 n for all n ∈ N. Thus, a1/n → 1 as n → ∞ if a > 1. For 0 < a < 1, we have (1/a)1/n → 1 as n → ∞, and therefore, by the reciprocal rule, a1/n = 1 1 → =1 1/n 1 (1/a) as n → ∞. The sequence {a1/n } is referred to as the nth root sequence. (b) Clearly (1 + 1)n ≥ 1 + n > n, so that n1/n − 1 < 1 for n ≥ 1. Also, for n ≥ 1, we observe that n1/n ≥ 1, so that n1/n − 1 = xn with xn ≥ 0. In particular, using the binomial theorem, we deduce that n = (1 + xn )n ≥ 1 + nxn + n(n − 1) 2 n(n − 1) 2 xn ≥ 1 + xn , 2 2 which implies that 0 ≤ xn = n1/n − 1 ≤ 2 n for n ≥ 1. √ By the squeeze rule, xn → 0 as n → 0, since 1/ n → 0. We conclude that n1/n → 1 as n → ∞, as desired. (c) It follows that n! 1 0< n ≤ . n n The second inequality is true because n! = n(n − 1) · · · 2 · 1 < n · n · · · n · 1 = nn−1 . The squeeze rule (with an = 0, cn = 1/n) gives the desired conclusion. • 34 2 Sequences: Convergence and Divergence Remark 2.19. We observe that case (a) of Example 2.18 may be obtained as a special case of case (b). For instance, if a ≥ 1, then for n large enough we have 1 ≤ a < n. Taking roots on both sides, we obtain 1 ≤ a1/n < n1/n for large n. Again, by the squeeze rule, we see that limn→∞ a1/n = 1. As a consequence of (a) and (b) of Example 2.18 and the product rule for sequences, we can easily obtain that √ lim (2n)1/n = 1 and lim (3 n)1/2n = 1. • n→∞ n→∞ 2.1.4 Bounded Monotone Sequences Now we introduce some important terminology associated with sequences. A sequence {an } is said to be bounded above if there exists an M ∈ R such that an ≤ M for all n, bounded below if there exists an m ∈ R such that an ≥ m for all n, bounded if it is bounded both below and above, monotonically increasing (or simply increasing) if an ≤ an+1 for all n (see Figure 2.1), • monotonically decreasing (or simply decreasing) if an ≥ an+1 for all n (see Figure 2.2), • • • • y O y x Fig. 2.1. An increasing sequence. • • • • • O x Fig. 2.2. A decreasing sequence. strictly increasing if an < an+1 for all n, strictly decreasing if an > an+1 for all n, monotonic if it is either increasing or decreasing, strictly monotonic if it is either strictly increasing or strictly decreasing, alternating if an changes sign alternately. In other words, an is of the form an = (−1)n−1 bn or an = (−1)n bn (bn ≥ 0) for all n. That is, an an+1 < 0 for all n.