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

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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 ) ! lim lim 0 ! n n nn nx x n ρ + →∞ →∞ == = . So, the series converges for all x values. 3. For the series 2 1 n n x n = , we have 21 2 22 lim lim ) ) n n x n + →∞ →∞ = ++ . We want 1 x < , so we have convergence on 11 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 ] by the Absolute Convergence Test. 7. 9. For the series 1 2 1 n n n x n = , we have 1 lim lim 2 1 ) 2 n n x n + →∞ →∞ −− = + +− . We want x < , so we have convergence on 12 1 13 xx −< − < → < < . For 1 x = , we have the series 1 1 1 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 = converges on ( ] 1, 3 .
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hw96 - 9.6 Pages 483 # 1-25 odd, 28 1. For the series = lim...

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