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Unformatted text preview: the Alternating Series Test. We know that the error of the n th partial sum of a series is bounded by  a n +1  , so the remaining question boils down to, “for what value of n is 1 n 5 n < . 0001?” 1 n 5 n < . 0001 n 5 n > 10000 n ≥ 5 , since 4 · 5 4 = 2500 and 5 · 5 5 = 15 , 625. Thus, we only need to add the ﬁrst four terms of the series to approximate the sum within the allotted error. Exercise 32 : For what values of p is the following series convergent? ∞ X n =1 (1) n1 n p Solution : We apply the Alternating Series Test: for all p > 0, b n = 1 n p → as n → ∞ . Similarly, for all p > 0 we have n p < ( n + 1) p ⇒ 1 ( n + 1) p < 1 n p . Thus, for all p > 0 the series converges by the Alternating Series Test, and for all p ≤ 0 the series diverges by the same....
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This note was uploaded on 07/26/2011 for the course MATH 153 taught by Professor Rempe during the Spring '08 term at Ohio State.
 Spring '08
 REMPE
 Math

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