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Spring 2006 - Terras' Class - Exam 1 (Version A)

Spring 2006 - Terras' Class - Exam 1 (Version A) -...

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1 Solutions to Exam #1 - Version α 1) a) 2 n 2 n=1 n +1 (-1) 3n +1 diverges since the terms don’t approach 0 as n goes to infinity. The absolute value of a term is 1 3 1 2 2 + + n n which approaches 1/3 by l’Hôpital’s rule and thus the actual term has no limit. Then even terms go to 1/2 and the odd ones to -1/2. b) n=1 1 n(n +2) converges by the comparison test since 2 n 1 2) n(n 1 + and = 1 n 2 n 1 converges by the integral test. c) 1 (2 )! n n n n = converges by comparison with e = = 1 n n! 1 , since ! 1 ! 1 ) 1 ( ) 1 2 )( 2 ( n n n n n n n n + = " " (2n)! n n . e = = 1 n n! 1 converges by the ratio test as 1 1 1 ! 0 ( 1)! 1 n n a n a n n + = = + + i , as n goes to infinity. You could also use the ratio test on the original series, but that is a little harder. 2) f(x)= n n n n=1 x (-1) 3 is a geometric series. If we substitute u=-x/3, then f(x)= n n=1 1 u = u 1-u , converging for |u|<1. a) The radius of convergence of the series for f(x) is R=3, since we need |u|=|x/3|<1 or |x|<3. Or we can use the ratio test to find the radius of convergence.
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