# hw95 - 9.5 Pages 478-479 # 1-29 odd, 33 1. ( 1) n For n =1...

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9.5 Pages 478-479 # 1-29 odd, 33 1. For () 1 2 1 31 n n n = + , 2 lim 0 n n →∞ = + , so the alternating series converges. The error made by using 9 S to approximate the sum is not more than 10 22 0.065 310 1 31 a == + . 3. For 1 1 1 ln 1 n n n = + , 1 lim 0 ln 1 n n = + , so the alternating series converges. The error made by using 9 S to approximate the sum is not more than 10 11 0.417 ln 10 1 ln11 a + . 5. For 1 1 ln 1 n n n n + = , ln 1/ lim lim 0 1 nn n , so the alternating series converges. The error made by using 9 S to approximate the sum is not more than 10 ln10 0.230 10 a =≈ . 11. For 1 1 1 1 1 n n + = + we will first look at the absolute value of the series, 1 1 1 n = + , using the Limit Comparison Test with 1 1 n a = + , 2 1 n b n = . We know 2 1 1 n n = converges (p-series with 2 p = ). 2 1 lim lim lim 1 1 n n n a bn n n →∞ →∞ →∞  = ⋅= =  ++  Since 01 < <∞ , 1 n n a = and 1 n n b = “do the same thing.” So, 1 1 1 n = + converges; Limit Comparison Test. This means that the original series, 1 1 1 1 1 n n + = + , converges absolutely. 7. For 1 3 4 n n = we will use the Absolute Ratio Test. 1 1 3/4 33 lim lim lim 1 44 n n n n n a a + + →∞ = < So the series converges absolutely. 13. For 1 1 1 1 5 n n n + = we will first look at the alternating series. We have 1 lim 0 5 n n →∞ = , so 1 1 1 1 5 n n n + = converges. Now we will look at the absolute value of the series, 111 55 ∞∞ = . This diverges because it is a multiple of the harmonic series.

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## This note was uploaded on 02/25/2010 for the course MATH 201 taught by Professor Kim during the Spring '07 term at SIU Carbondale.

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hw95 - 9.5 Pages 478-479 # 1-29 odd, 33 1. ( 1) n For n =1...

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