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Unformatted text preview: Unofficial Math 31B Practice Final by TA Ning Khamsemanan Disclaimer: this practice exam is my attempt to help you study for the final. Things in here might not be in the real exam and vice versa. Don’t use this as your main study. There might be some typos and/or mistakes. 1. Find the radius of convergence and Interval of convergence of the fol lowing (a) ∑ ∞ n =1 3 n ( x 2) n ( n +1)! lim n →∞ a n +1 a n = lim n →∞ 3 n +1 ( x 2) n +1 ( n + 1 + 1)! · ( n + 1)! 3 n ( x 2) n = lim n →∞ 3( x 2) n + 2 =  x 2  lim n →∞ 3 n + 2 =  x 2 · 0 = 0 < 1 Thus R = ∞ and I = (∞ , ∞ ) (b) ∑ ∞ n =1 ( x 1) n n 2 n lim n →∞ a n +1 a n = lim n →∞ ( x 1) n +1 ( n + 1)2 n +1 · n 2 n ( x 1) n = lim n →∞ ( x 1) n 2( n + 1) =  x 1  2 · lim n →∞ n n + 1 =  x 1  2 < 1 Thus for the series is convergent if  x 1  2 < 1 ⇔  x 1  < 2 ⇔ 2 < x 1 < 2 ⇔  1 < x < 3. Therefore R = 2. Now, we need to check when x = 1 and x = 3. For x = 1, we get ∞ X n =1 ( 1 1) n n 2 n = ∞ X n =1 ( 2) n n 2 n = ∞ X n =1 2 2 n · 1 n = ∞ X n =1 ( 1) n n Let f ( x ) = 1 x , then f ( x ) = 1 x 2 < 0. Thus the b n is decreasing. Also lim n →∞ 1 n = 0 Hence, ∑ ∞ n =1 ( 1) n n is convergent by the Alternating Series Test. For x = 3, we get ∞ X n =1 (3 1) n n 2 n = ∞ X n =1 2 n n 2 n = ∞ X n =1 1 n 1 which is divergent by pseries....
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 Spring '09
 HOUDAYER
 Math, lim

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