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Unformatted text preview: Math 3110 Homework 6 Solutions Exercise 7.2.1 Let a n = 1 n 2 . Define e ( n ) by e ( n ) = 1 if n is even and e ( n ) = 0 if n is odd. Because a n ≥ 0, we get 0 ≤ e ( n ) a n ≤ a n and 0 ≤ (1 e ( n )) a n ≤ a n for all n . Then by Theorem 7.2D, because ∞ X n =1 a n converges, so do ∞ X n =1 e ( n ) a n and ∞ X n =1 (1 e ( n )) a n . By Theorem 7.2C, ∞ X n =1 e ( n ) a n + ∞ X n =1 (1 e ( n )) a n = ∞ X n =1 e ( n ) a n + (1 e ( n )) a n = ∞ X n =1 a n = π 2 6 . We can relabel the terms to get ∞ X n =1 e ( n ) a n = ∞ X n =1 1 (2 n ) 2 ∞ X n =1 (1 e ( n )) a n = ∞ X n =0 1 (2 n + 1) 2 Furthermore, by Corollary 7.2, ∞ X n =1 1 (2 n ) 2 = ∞ X n =1 1 4 1 n 2 = 1 4 ∞ X n =1 1 n 2 = π 2 24 . Now we plug in the known values for two sums to get π 2 6 = ∞ X n =0 1 (2 n + 1) 2 + π 2 24 , from which ∞ X n =0 1 (2 n + 1) 2 = π 2 8 . 1 Exercise 7.3.3 (a) Define q ( n ) by q ( i ) = 1 if there is some j such that i = n j and q ( i ) = 0 otherwise. We have ≤ q ( n ) ≤...
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This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell.
 '08
 RAMAKRISHNA
 Math

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