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Unformatted text preview: MATH 138 Winter 2011 Assignment 9 Solutions 1. Determine whether the series is absolutely convergent, conditionally con vergent, or divergent. (a) X n =1 ( 1) n 1 2 n 4 We consider the series of absolute values, X n =1 2 n 4 = 2 X n =1 1 n 4 , which clearly converges ( pseries with p = 4 > 1 ). Hence the series X n =1 ( 1) n 1 2 n 4 converges absolutely . (b) X n =1 n ! 100 n We use the ratio test with a n = n ! 100 n . Then lim n a n +1 a n = lim n ( n + 1)! 100 n +1 100 n n ! = lim n n + 1 100 = . Therefore, by the ratio test, the series diverges . (c) X n =1 ( 1) n n n 3 + 2 To check for absolute convergence, if we focus on the series of abso lute values, we have X n =1 n n 3 + 2 , which we can show diverges using the limit comparison test, comparing it with X n =1 1 n 1 2 (this is a divergent series  a pseries with p = 1 / 2 < 1 ). That is, lim n n n 3 +2 1 n = lim n n n n 3 +2 = lim n n 3 n 3 +2 = lim n 1 1+2 /n 3 = 1 . Since this limit exists, is not zero, and is not infinite, we conclude the two 1 series have the same convergence behaviour. Namely, we can con clude that X n =1 n n 3 + 2 diverges, so the original alternating series X n =1 ( 1) n n n 3 + 2 does not absolutely converge. It remains to check if it converges conditionally. Let b n = n n 3 +2 . If this is a postivie decreasing sequence which ap proaches zero, then the alternating series will converge (by the alter nating series test). Lets check: Clearly b n is positive, and clearly lim n b n = 0 (since the degree of the denominator is greater than that of the numerator). To check if b n is decreasing, we consider the function f ( x ) = x x 3 +2 . Then f ( x ) = x 3 +2 x 1 2 x 3 +2 (3 x 2 ) x 3 +2 = 2( x 3 +2) 3 x 3 2( x 3 +2) 3 / 2 = 4 x 3 2( x 3 +2) 3 / 2 . Although this is positive for x = 1 , we can see that for x = 2 and any larger integer values (or noninteger values, for that matter), this derivative is negative, meaning that the function is decreasing. This is what we need. We can now conclude that b n is a decreasing sequence for all n N , n 2 . Thus we have satisfied all of the conditions in the alternating series test, and conclude that this alternating series X n =1 ( 1) n n n 3 + 2 converges. We already showed above that it does not converge absolutely, so we say that the series X n =1 ( 1) n n n 3 + 2 converges conditionally . (d) X n =1 ( 1) n +1 n 2 2 n n !...
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 Winter '07
 Anoymous
 Math

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