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asst8soln

# asst8soln - MATH 138 ASSIGNMENT 8 SOLUTIONS WINTER 2010 1 1...

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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 0 . 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 0 = 1. 1

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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 p -series. (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 2 ln 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 . By the Comparison Test, X n = K 1 (ln n ) ln n converges since X n = K 1 n 2 converges.
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