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Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 3 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 2.2.1, 2.2.2, 2.2.8, 2.3.2, 2.3.3, 2.3.8, 2.3.11 Solution 2.2.1. (a) Given > 0, let N be any integer larger than s 1 1 −1 . 6 In this case, for any integer n ≥ N , we have 1 1 2 −1 n > 6 so that 1 6n2 + 1 > . Equivalently, 1 1 = − 0 6n2 + 1 6n2 + 1 < . This implies that lim 1/(6n2 + 1) = 0, as desired. (b) Given > 0, let N be any integer larger than 13 − 10 . 4 In this case, for any integer n ≥ N , we have 4n > 13 − 10, in which case > 13/(4n + 10). For such n, we have 3 3n + 1 3(n + 25 ) 3n + 1 − 2 2n + 5 = 2n + 5 − 2n + 5 13 2 13 < , = = 2n + 5 4n + 10 3n+1 so that we may conclude that lim 2n+5 = 3/2. 1 2 (c) Given > 0, let N be any integer larger than for any integer n ≥ N , we have 4 n + 3 > 2, √ so that n + 3 > 2/. For such n, we have 2 = √ 2 √ − 0 < , n+3 n+3 4 2 − 3. In this case, 2 so that we may conclude lim √n+3 = 0. Solution 2.2.2. Given a sequence of real numbers (an ) and a real number x, we claim that (an ) verconges to x if and only if (an ) is bounded: Suppose that (an ) is bounded, so that there exists an M with |an | < M for all n. In this case, note that |an − x| ≤ |an | + |x| < M + |x| for all n. Let = M + |x|. Then it is true that for all N , if n ≥ N (in fact, for any n) we have |an −x| < M +|x| = . Thus (an ) verconges to x. On the other hand, if (an ) verconges to x, then there is an so that (taking N = 1), for every n ∈ N we have |an − x| < . On the other hand, we observe that |an | = |an − x + x| ≤ |an − x| + |x|, so that |an | < + |x| for all n ∈ N, and (an ) is bounded. Finally, any divergent bounded sequence of real numbers will suffice to see a vercongent divergent sequence. For example, an = (−1)n diverges, but since (an ) is bounded, it verconges to any real number x. Solution 2.2.8. (a) The sequence ((−1)n ) is frequently in the set {1}: For every N ∈ N, there is an even number n ≥ N so that (−1)n = 1 ∈ {1}. On the other hand, ((−1)n ) is not eventually in {1}: For every N ∈ N there is an odd number n ≥ N so that (−1)n = −1 ∈ / {1}. (b) If (an ) is a sequence that is eventually in A, then (an ) is frequently in A: By assumption, there is an N0 ∈ N so that for every n ≥ N0 , we have an ∈ A. This means that, given any N ∈ N, we may choose 3 n = max{N, N0 }. This n satisfies n ≥ N , and we have an ∈ A, since n ≥ N0 . On the other hand (an ) may be frequently in A, but not eventually in A: The example in part (a) is one such example. (c) Using the definition of ‘eventually’, we rephrase Definition 2.2.3B: Definition. A sequence (an ) converges to a if, given any -neighborhood V (a) of a, the sequence (an ) is eventually in V (a). (d) If (xn ) is a sequence of real numbers so that infinitely many of them are equal to 2, then (xn ) is frequently in the set (1.9, 2.1): For every N ∈ N, there must exist n ≥ N so that xn = 2, otherwise the number of terms xn equal to 2 would be finite. On the other hand, it is not necessary that (xn ) is eventually in (1.9, 2.1). For example, if xn = 1 + (−1)n , then the sequence (xn ) is frequently in the set {−2}, which is disjoint from (1.9, 2.1). It is thus not eventually in (1.9, 2.1), as in part (a). Solution 2.3.2. (a) Given > 0, since (xn ) → 0 there ispan N ∈ N so √ that for every n ≥ N we have |xn | < 2 . Thus | xn | = |xn | < for every n ≥ N , and we conclude that (xn ) → 0. (b) By Theorem 2.3.4 part (i), xn ≥ 0 implies that x ≥ 0. We may also assume that x 6= 0, since we’ve already proven the case x = 0 in part (a). Thus we proceed with the assumption that x > 0. Note that we have for any non-negative numbers a, b ≥ 0, √ √ √ √ a − b = ( a + b)( a − b). √ √ Thus, as long as a + b 6= 0, we have √ √ a−b √ . a− b= √ a+ b √ √ √ Since xn + x ≥ x > 0, we conclude that xn − x √ √ |x − x| 1 √ = √ n √ ≤ √ |xn − x| | xn − x| = √ xn + x xn + x x √ for all n ∈ N. Given > 0, since x > 0 we have x > 0. As (xn ) → √x, there is an N ∈ N so that for every n ≥ N we have |xn − x| < x. 4 Thus, for any n ≥ N , we have √ √ √ 1 1 | xn − x| ≤ √ |xn − x| < √ · x = , x x √ √ and we conclude that xn → x. Solution 2.3.3. Since lim xn = lim zn = l, given > 0 there are integers N1 and N2 so that, for any n satisfying n ≥ N1 and n ≥ N2 we have |xn − l| < and |zn − l| < . Let N = max{N1 , N2 }, so that for any n ≥ N we have xn − x > − and zn − l < . The inequality xn ≤ yn ≤ zn implies that xn − l ≤ yn − l ≤ zn − l, so that − < xn − l ≤ yn − l ≤ zn − l < for n ≥ N . Equivalently, |yn − l| < for n ≥ N . We conclude that lim yn = l. Solution 2.3.8. (a) Let xn = (−1)n and yn = −(−1)n . In this case both sequences (xn ) and (yn ) diverge. On the other hand, the sum xn + yn = 0 for all n, so that (xn + yn ) converges to 0. (b) By Theorem 2.3.3 part (i), as (xn ) converges, we know that (−xn ) converges. By Theorem 2.3.3 part (ii), as (xn + yn ) and (−xn ) each converge, this implies that (xn + yn + (−xn )) = (yn ) converges, a contradiction. (c) Let bn = 1/n, so that bn 6= 0 for all n ∈ N, and (bn ) is a convergent sequence. On the other hand, 1/bn = n, so that (1/bn ) diverges. (d) By Theorem 2.3.2, the convergent sequence (bn ) is bounded, so that there exists a real number M1 so that |bn | < M1 for all n ∈ N. If (an − bn ) is bounded, there exists a real number M2 so that |an − bn | < M2 . Using the triangle inequality, this implies that |an | = |an − bn + bn | ≤ |an − bn | + |bn | < M1 + M2 so that the sequence (an ) is bounded, a contradiction. (e) Let (bn ) be any divergent sequence, and let an = 0 for all n ∈ N. In this case, both sequences (an ) and (an bn ) are the sequence all of whose terms are 0. Thus both are convergent, though (bn ) is divergent. 5 Solution 2.3.11. Since (xn ) is a convergent sequence, there is a real number x so that, given any > 0, there is an N1 ∈ N such that, for all n ≥ N1 we have |x − xn | < /2. Moreover, by Theorem 2.3.2 the sequence (xn ) is bounded, so that there exists a real number M such that |xn | < M for all n ∈ N. Note that by Theorem 2.3.4 part (iii) we know as well that |x| < M . For any n ≥ N1 we have: x1 + . . . + xn (x1 − x) + . . . + (xn − x) − x = n n (x1 − x) + . . . + (xN1 − x) (xN1 +1 − x) + . . . + (xn − x) = + n n x1 + . . . + xN1 − N1 x (xN1 +1 − x) + . . . + (xn − x) + ≤ n n |x1 | + . . . + |xN1 | + N1 |x| |xN1 +1 − x| + . . . + |xn − x| ≤ + n n M + . . . + M + N1 M /2 + . . . + /2 < + n n (n − N1 ) 2N1 M + = n n 2 2N1 M < + . n 2 Let N2 ∈ N be any integer larger than 4N1 M/, so that for n ≥ N2 we will have 2N1 M/n < /2. Let N = max{N1 , N2 }. By the above calculations, if n ≥ N , then n ≥ N1 and we have x1 + . . . + xn 2N1 M < − x + . n n 2 Moreover, n ≥ N2 , so that we have x1 + . . . + x n 2N1 M < − x + < + = . n n 2 2 2 We conclude that ((x1 + . . . + xn )/n) converges to x, as desired. Finally, consider the sequence (xn ) given by xn = (−1)n . This sequence is divergent (we showed in class that a very similar sequence doesn’t converge, and that same proof will work here. Alternatively, we will be developing a simple test soon to check divergence). On the other hand, x1 + . . . + xn −1 + 1 . . . + (−1)n = . n n These terms are given by yn = 0 for n even, and yn = 1/n for n odd. This sequence converges to 0. yn =