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Unformatted text preview: 1. Thus, by the Alternating Series Test, our series converges conditionally . 2. [6 points] X n =1 (3) n n 2 n ! Let us apply the Ratio Test: lim n (3) n +1 ( n +1) 2 ( n +1)! (3) n n 2 n ! = lim n 3 n +1 ( n + 1) 2 ( n + 1)! n ! 3 n n 2 = lim n 3( n + 1) 2 ( n + 1) n 2 = lim n 3( n + 1) n 2 = 0 As this limit is less than one, we can say that our series converges absolutely . 3. [6 points] X n =1 (2 n ) n 5 n Let us apply the Root Test: lim n n s (2 n ) n 5 n = lim n 2 n 5 = Since this limit is not less than one, we can say that our series diverges ....
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This note was uploaded on 07/26/2011 for the course MATH 153 taught by Professor Rempe during the Spring '08 term at Ohio State.
 Spring '08
 REMPE
 Math

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