## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 11 - Section 11.5 - Alternating Series - 11.5 Exercises - Page 736: 23

#### Answer

We need the first five terms. Thus, $\Sigma_{n=1}^{5}\frac{(-1)^{(n+1)}}{n^{6}}\approx 0.9856$

#### Work Step by Step

Alternating series test: Suppose that we have a series $\Sigma a_n$, such that $a_{n}=(-1)^{n}b_n$ or $a_{n}=(-1)^{n+1}b_n$, where $b_n\geq 0$ for all $n$. Then if the following two conditions are satisfied, the series is convergent. 1. $\lim\limits_{n \to \infty}b_{n}=0$ 2. $b_{n}$ is a decreasing sequence. In the given problem, $b_{n}=\frac{1}{n^{6}}$ which satisfies both conditions of the Alternating Series Test as follows: 1. $b_{n}=\frac{1}{n^{6}}$, is decreasing because the denominator is increasing. 2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{1}{n^{6}}=0$ Hence, the given series is convergent by the Alternating Series Test. Now compute the terms $a_{n}=\frac{(-1)^{n+1}}{n^{6}}$ until we get to one where $|a_{n}|\lt 0.00005$ $a_{1}=1$ $a_{2}=-\frac{1}{64}=-0.015625$ $a_{3}=\frac{1}{729}\approx 0.00137174211$ $a_{4}=-\frac{1}{4096}\approx -0.00024414062$ $a_{5}=\frac{1}{15625}\approx 0.000064$ $a_{6}=-\frac{1}{46656}=-0.000021$ $|a_{6}|\lt 0.00005$ We need the first five terms to get the right accuracy. Thus, $\Sigma_{n=1}^{5}\frac{(-1)^{(n+1)}}{n^{6}}\approx 0.9856$

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