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# hw96 - 9.6 Pages 483 1-25 odd 28 1 For the series = lim n...

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9.6 Pages 483 # 1-25 odd, 28 1. For the series 1 ( 1)! n n x n = we have 1 ( 1)! lim lim 0 ! n n n n n x x n n x ρ + →∞ →∞ = = = . So, the series converges for all x values. 3. For the series 2 1 n n x n = , we have 2 1 2 2 2 lim lim ( 1) ( 1) n n n n n x n x x n x n ρ + →∞ →∞ = = = + + . We want 1 x < , so we have convergence on 1 1 x − < < . For 1 x = , we have the series 2 1 1 n n = , which is a p-series with p=2 and the series converges. For 1 x = − , we have the series ( ) 2 1 1 n n n = , which is an alternating p-series with p=2 and the series converges. Therefore, the series 2 1 n n x n = converges on [ 1,1] . 5. This is the alternating series for problem 3; thus it converges on [ 1,1] by the Absolute Convergence Test. 7. 9. For the series ( ) ( ) 1 2 1 n n n x n = , we have ( ) ( ) ( ) 1 2 2 lim lim 2 1 ( 1) 2 n n n n n x n x x n n x ρ + →∞ →∞ = = = + + . We want 2 1 x < , so we have convergence on 1 2 1 1 3 x x − < < < < . For 1 x = , we have the series ( ) ( ) 1 1 1 1 1 n n n n n n = = = , which is the harmonic series, which diverges. For 3 x = , we have the series ( ) 1 1 n n n = , which is an alternating p-series with p=1 and the series converges. Therefore, the series ( ) ( ) 1 2 1 n n n x n =

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