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Unformatted text preview: MAA 4200: Homework 3 Solutions June 17, 2009 [1.34] (5 points) If a n = ( 1) n (1 1 /n ), then the subsequence a 2 k = (1 1 / 2 k ) → 1 as k → ∞ by the Algebra of Limits Theorem and the fact that 1 /k → 0 as k → ∞ . [1.35] (5 points) Let E = { a n : n ∈ N } with accumulation point x . Applying Theorem 1.17 on page 52 to the set E gives that there is a sequence of elements of E which converges to x . Any sequence of elements of E must CONTAIN a subsequence of { a n } ∞ n =1 . For example, Theorem 1.17 may give you a sequence like a 3 ,a 4 ,a 2 ,a 6 ,a 7 ,... this is NOT a subsequence of { a n } ∞ n =1 because the order is changed; note that a 3 came before a 2 . This cannot happen in a subsequence. To give a nice proof of this problem it is best to prove it from scratch. We will follow the proof of Theorem 1.17 with one slight change. Proof: Let k = 1. Since x is an accumulation point of E , there exists an element y ∈ E such that  y x  < 1. Since y ∈ E there must be a subscript n 1 so that y = a n 1 ....
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 Spring '07
 thieme
 Logic, Algebra, Limits, accumulation point, 2k, ek, Preliminary Work

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