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# homework_8 - Homework 8 14.2(a This diverges by comparing...

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Homework 8 14.2. (a) This diverges by comparing n - 1 n 2 1 / 2 n n 2 = 1 2 n for all n 2 because 1 2 n diverges. (b) This does not coverge by the corollary because lim( - 1) n = 0. It does not diverge to or -∞ because it is bounded by | ( - 1) n | ≤ 1. (c) This converges. In fact, this is 3 n n 3 = 3 1 n 2 = π 2 2 . (d) This converges absolutely by the ratio test since | a n +1 a n | = | ( n + 1) 3 3 n +1 3 n n 3 | = | 1 3 ( n + 1 n ) 3 | so lim | a n +1 a n | = 1 3 and hence lim sup | a n +1 a n | = 1 3 < 1. (e) This converges absolutely by the ratio test since | a n +1 a n | = | ( n + 1) 2 ( n + 1)! n ! n 2 | = | n + 1 n 2 | so lim | a n +1 a n | = 0 and hence lim sup | a n +1 a n | = 0 < 1. (f) This converges by comparing 1 n n 1 n 2 for all n 1 because 1 n 2 converges. (g) This converges absolutely by the root test since | a n | 1 /n = ( n 2 n ) 1 /n = 1 2 n 1 /n so lim | a n | 1 /n = 1 2 and hence lim sup | a n | 1 /n = 1 2 < 1. 14.6. (a) Suppose that | b n | ≤ M for all n N . We show that a n b n converges by the Cauchy criterion. Let > 0. Since | a n | converges, it satisfies the Cauchy criterion. Therefore there exists N N such that n m > N implies that n k = m | a k | < M . Then, n m > N also implies that | n k = m a k b k | ≤ n k = m | a k || b k | ≤ M n k = m | a k | < M M = 1

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2 using the triangle inequality and the boundedness of
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