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Unformatted text preview: Math 3110 Homework 7 Solutions Exercise 8.1.1 In each case, let a n denote term n of the series. (a) We compute a n +1 a n =  x  n +1 2 n +1 n + 1 2 n n  x  n =  x  2 n n + 1 . Since n n + 1 1 as n , we want  x  2 < 1, so the radius of convergence is 2. (b) We compute a n +1 a n = ( n + 1)!( n + 1)!  x  n +1 (2 n + 2)! (2 n )! ( n )!( n )!  x  n = ( n + 1) 2 (2 n + 2)(2 n + 1)  x  = n + 1 4 n + 2  x  . Since n + 1 4 n + 2 1 4 as n , we want 1 4  x  < 1, so the radius of convergence is 4. (c) We use the root test to get that (  a n  ) 1 n =  x  ( n ) 1 n 2 . Since n 1 n 2 1 as n , we want  x  < 1, so the radius of convergence is 1. (d) We compute a n +1 a n =   1  n +1  x  2( n +1) 4 n +1 ( n + 1)!( n + 1)! 4 n ( n )!( n )!   1  n  x  2 n = x 2 4( n + 1) 2 . For any constant x , this goes to 0 as n , so the radius of convergence is infinite....
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This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).
 '08
 RAMAKRISHNA
 Math

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