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Unformatted text preview: MATH 138 ASSIGNMENT 8 SOLUTIONS WINTER 2010 1. The Integral Test can be applied since f ( x ) = 1 (2 x +1) 7 is continuous, positive and decreasing on [1 , ). By the remainder estimate for the Integral Test, we want to find n such that Z n 1 (2 x + 1) 7 dx < 10 5 . We compute that Z n 1 (2 x + 1) 7 dx = lim t  1 12(2 x + 1) 6 t n = 1 12(2 n + 1) 6 . We need to solve 1 12(2 n + 1) 6 < 1 10 5 , i.e., 12(2 n + 1) 6 > 10 5 . The solution is n > 1 2 10 5 12 1 / 6 1 1 . 75. Since n is an integer, we can take n = 2. The corresponding partial sum estimate is 2 X n =1 1 (2 n + 1) 7 = 1 3 7 + 1 5 7 = 80312 170859375 . 00047005 . 2. (a) We apply the Limit Comparison Test to a n = n 2 +5 n +3 n 2 +2 n +7 and b n = 1 n . a n b n = n 2 + 5 n + 3 n 2 + 2 n + 7 n 1 = n 2 q 1 + 5 n + 3 n 2 n 2 ( 1 + 2 n + 7 n 2 ) = q 1 + 5 n + 3 n 2 1 + 2 n + 7 n 2 Thus lim n a n b n = 1. By the Limit Comparison Test, X n =1 n 2 + 5 n + 3 n 2 + 2 n + 7 di verges since X n =1 1 n diverges. (b) We apply the Limit Comparison Test to a n = 2 1 /n n 2 and b n = 1 n 2 . a n b n = 2 1 /n n 2 n 2 1 = 2 1 /n Thus lim n a n b n = 2 = 1. 1 2 MATH 138 ASSIGNMENT 8 SOLUTIONS By the Limit Comparison Test, X n =1 2 1 /n n 2 converges since X n =1 1 n 2 is a conver gent pseries. (c) If n 2, then n ! n n = n n n 1 n 3 n 2 n 1 n 1 1 1 2 n 1 n = 2 n 2 . If n = 1, then n ! n n 2 n 2 also holds. Thus X n =1 n ! n n X n =1 2 n 2 . By the Comparison Test, X n =1 n ! n n converges since X n =1 2 n 2 = 2 X n =1 1 n 2 con verges. (d) Let K be any integer that is greater than e e 2 . Then n K = n e e 2 = ln n e 2 = (ln n ) ln n e 2ln n = n 2 = 1 (ln n ) ln n 1 n 2 . Thus X n = K 1 (ln n ) ln n X n = K 1 n 2 ....
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This note was uploaded on 03/16/2011 for the course MATH 138 taught by Professor Anoymous during the Winter '07 term at Waterloo.
 Winter '07
 Anoymous
 Calculus, Remainder

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