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Unformatted text preview: SOLUTIONS TO HOMEWORK # 10 15.1 (a) ( 1) n n satisfies the conditions of the alternating series test and thus converges. (b) As we proved in the class, using the Ratio Test, the series 2 n n ! converges and hence lim 2 n n ! = 0. Therefore ( 1) n n ! 2 n does not converge to 0, and ( 1) n n ! 2 n diverges by Corollary 14.5. A 1) The partial sums s n = 1 if n is even and s n = 0 if n is odd. Thus the series diverges. Another approach: ( 1) n does not converge to 0 so the series cannot converge. 2) This is an alternating series. But the main condition of the Alternating Series Theorem (AST), namely a n 0, is not fulfilled here so there is no contradiction to this Theorem. B 1) The series n =2 1 n (log n ) p converges for every real p . Proof a) lim 1 n (log n ) p = 0. Case 1. If p 0 this follows from the fact that logn + and hence lim n (log n ) p = + . Therefore our statement holds. Case 2. If p < 0 we have 1 n (log n ) p = (log n )  p  n ; consider the functions f ( x ) := log x x 1  p  ,g ( x ) := (log x )  p  x . By the LHospital Rule, since  p  > 0, lim x f ( x ) = lim x 1 x 1  p  x 1  p  1 =  p  lim x 1 x 1  p  = 0 ....
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This note was uploaded on 04/28/2008 for the course MATH 312 taught by Professor Olsavskygregorm during the Spring '08 term at Pennsylvania State University, University Park.
 Spring '08
 OLSAVSKYGREGORM

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