Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.5 - Alternating Series - 11.5 Exercises - Page 736: 28

Answer

$0.9856$

Work Step by Step

The given series is $\Sigma_{n=1}^\infty \dfrac{(-1)^{n+1}}{n^6}$ The sums of the given series are as follows: $S_1=\Sigma_{n=1}^3 \dfrac{(-1)^{n+1}}{n^6} \approx 9.85747$ $S_2=\Sigma_{n=1}^4 \dfrac{(-1)^{n+1}}{n^6} \approx 9.85502$ $S_3=\Sigma_{n=1}^5 \dfrac{(-1)^{n+1}}{n^6} \approx 9.85567$ $S_4=\Sigma_{n=1}^6 \dfrac{(-1)^{n+1}}{n^6} \approx 9.85545$ Hence, the sum of the series is $0.9856$ when approximated up to four decimals.
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