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Hwk5_Solns

# Hwk5_Solns - Math 3110 Spring 09 Homework 5 Selected...

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Math 3110 Spring ’09 Homework 5: Selected Solutions [7.2-1]: Claim : X 0 1 (2 n + 1) 2 = π 2 8 Proof We know that X 0 1 n 2 = π 2 6 , and X 0 1 (2 n + 1) 2 = X 0 1 n 2 - X 0 1 (2 n ) 2 . So X 0 1 (2 n + 1) 2 = π 2 6 - 1 4 X 0 1 n 2 = π 2 6 - 1 4 π 2 6 = π 2 8 [7.2-1]: Claim : If X 0 a n is absolutely convergent and { b n } is bounded, then X 0 a n b n is convergent. Proof Since { b n } is bounded, there is a B such that | b n | ≤ B for all n . So X 0 | a n b n | ≤ B X 0 | a n | , which is ﬁnite since X 0 a n is absolutely convergent. Thus X 0 a n b n also converges. [7.2-1]: Claim : Let { a n } be a sequence and { a n i } any subsequence. (a) If X 0 a n is absolutely convergent, then X 0 a n i is absolutely convergent. (b) It is not the case that X 0 a n convergent implies that X 0 a n i is convergent. Proof (a) Since { a n i } is a subsequence and X 0 a n is absolutely convergent, for some L < we clearly have k X i =0 | a n i | ≤ n k X n =0 | a n | ≤ X 0 | a n | ≤ L for all k . So X 0 a n i is absolutely convergent as claimed. (b) On the other hand, the series

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Hwk5_Solns - Math 3110 Spring 09 Homework 5 Selected...

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