asst9soln

# asst9soln - MATH 138 Calculus 2, Solutions to Assignment 9...

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MATH 138 Calculus 2, Solutions to Assignment 9 Problem 1: For each series below determine whether it converges absolutely, conditionally, or not at all. (a) X n =1 ( - 1) n n 10 2 n Solution: Let a n = n 10 2 n . Then ± ± ± ± a n +1 a n ± ± ± ± = ( n + 1) 10 2 n +1 · 2 n n 10 = ² n + 1 n ³ 10 · 1 2 = 1 2 ( 1 + 1 n ) 10 1 2 as n → ∞ . Thus | a n | converges by the Ratio Test. In other words, a n converges absolutely. (b) X n =1 ( - 1) n +1 n n + 1 Solution: Let a n = ( - 1) n +1 n n + 1 . Let b n = n n = 1 n . Then | a n | b n = n n + 1 ÷ n n = 1 1 + 1 n 1 as n → ∞ , and b n diverges ( its a p -series with p = 1 2 ) , so | a n | diverges by the Limit Comparison Test. Thus a n is not absolutely convergent, but we still need to determine whether it is conditionally convergent. Let f ( x ) = x x + 1 so that | a n | = f ( n ). Then f 0 ( x ) = 1 2 x ( x + 1) - x ( x + 1) 2 = 1 - x 2 x ( x + 1) 0 for x 1 , so f ( x ) is decreasing for x 1, and hence {| a n |} is decreasing for n 1. Also, lim n →∞ | a n | = lim n →∞ n n + 1 = lim n →∞ 1 n 1 + 1 n 0 as n → ∞ . Thus a n converges by the Alternating Series Test, so we have shown that a n converges conditionally. 1

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Problem 2: Find the radius as well as the interval of convergence of the following power series, and give reasons for your answers. (a) X n =0 2 n n ! x n Solution: Let a n = 2 n n ! x n . Then for all x , we have ± ± ± ± a n +1 a n ± ± ± ± = ± ± ± ± 2 n +1 x n +1 ( n + 1)! · n ! 2 n x n ± ± ± ± = 2 | x | n + 1 0 as n → ∞ , so | a n | converges for all x by the Ratio Test. Thus the radius of convergence is R = , and the interval of convergence is I = ( -∞ , ). (b) X n =0 1 3 n ( x + 1) n Solution: This series is geometric with ratio r = x +1 3 , so it converges if and only if ± ± x +1 3 ± ± < 1, that is when | x + 1 | < 3. Thus the radius of convergence is R = 3 and the interval of convergence is I = ( - 4 , 2). (c) X n =2 1 (ln n ) 2 x n Solution: Let a n = 1 (ln n ) 2 x n . Then for all x , we have ± ± ± ± a n +1 a n ± ± ± ± = ± ± ± ± x n +1 (ln( n + 1)) 2 · (ln n ) 2 x n ± ± ± ± = ² ln n ln( n + 1) ³ 2 | x | -→ | x | as n → ∞ since replacing n by the continuous variable t then using l’Hˆopital’s Rule gives lim n →∞ ln n ln( n + 1) = lim t →∞ ln t ln( t + 1) = lim t →∞ 1 /t 1 / ( t + 1) = lim t →∞ t + 1 t = lim t →∞ ( 1 + 1 t ) = 1 . By the Ratio Test,
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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.

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asst9soln - MATH 138 Calculus 2, Solutions to Assignment 9...

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