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Unformatted text preview: Math 23b Homework 8 Solutions 1. (a) Let a n = 1 n 2 . Given > 0, we have by the Archimedean Property that there is some n > √ so that n 2 > and  1 n 2  = 1 n 2 < . Hence, lim n →∞ a n = 0. Let b n = n . If b n → L then there exists N such that for all n ≥ N ,  n L  =  b n L  < 1 so that n < L + 1. But there exists an integer greater than L + 1 so we have a contradiction. Now consider a n b n = 1 n . By a similar argument to the one above, 1 n → 0. (b) Let a n = 1 n and b n = n . Then a n → 0, b n does not converge, and a n b n = 1. Clearly then a n b n → 1, since  a n b n 1  = 0 for all n ≥ 1. (c) Let a n = 1 n , b n = n 2 . Then a n → 0, b n does not converge, and a n b n = n does not converge. 2. (a) We will prove the contrapositive: all convergent sequences are bounded. Suppose x n → L . Then there exists N ∈ N such that, for all n ≥ N ,  x n L  < 1....
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This note was uploaded on 10/15/2011 for the course MATH 23b taught by Professor Bonglian during the Spring '09 term at Brandeis.
 Spring '09
 BongLian
 Math

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