8 - MATH 138 Winter 2011 Assignment 8 Solutions 1. (a) We...

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MATH 138 Winter 2011 Assignment 8 Solutions 1. (a) We note that 0 < 2 n 3 n +1 < 2 n 3 n = ( 2 3 ) n for every positive integer n . By the Comparison Test, X n =1 2 n 3 n + 1 converges since X n =1 ± 2 3 ² n is a convergent geometric series (with a = r = 2 3 ). (b) We apply the Limit Comparison Test to a n = n +5 3 n 7 + n 2 and b n = 1 n 4 / 3 . a n b n = n + 5 3 n 7 + n 2 · n 4 / 3 1 = n 7 / 3 ( 1 + 5 n ) n 7 / 3 ( 1 + 1 n 5 ) 1 / 3 = ( 1 + 5 n ) ( 1 + 1 n 5 ) 1 / 3 . Thus lim n →∞ a n b n = 1 . By the Limit Comparison Test, X n =1 n + 5 3 n 7 + n 2 converges since X n =1 1 n 4 / 3 is a convergent p -series. (c) We apply the Limit Comparison Test to a n = 3 1 /n n 4 and b n = 1 n 4 . a n b n = 3 1 /n n 4 · n 4 1 = 3 1 /n Thus lim n →∞ a n b n = 3 0 = 1 . By the Limit Comparison Test, X n =1 3 1 /n n 4 converges since X n =1 1 n 4 is a convergent p -series. (d) Since
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This note was uploaded on 07/24/2011 for the course MATH 138 taught by Professor Anoymous during the Winter '07 term at Waterloo.

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8 - MATH 138 Winter 2011 Assignment 8 Solutions 1. (a) We...

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