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# hw7 - Math 3110 Homework 7 Solutions Exercise 8.1.1 In each...

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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. (e) We use the root test to get that ( | a n | ) 1 n = 1 + 2 n | x | . As n → ∞ , 1 + 2 n 1, so we want | x | < 1, and the radius of convergence is 1. 1

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(f) We use the root test to get ( | a n | ) 1 n = | x | ln( n ) . For any constant x , this goes to 0 as n → ∞ , so the radius of convergence is infinite.
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hw7 - Math 3110 Homework 7 Solutions Exercise 8.1.1 In each...

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