Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 726: 35

Answer

(a) $\frac{9\pi^{4}}{10}$ (b) $\frac{\pi^{4}}{90}-\frac{17}{16}$

Work Step by Step

(a) $\Sigma_{n=1}^{\infty}(\frac{3}{n})^{4}=\Sigma_{n=1}^{\infty}\frac{3^{4}}{n^{4}}=\Sigma_{n=1}^{\infty}\frac{81}{n^{4}}$ $=81\Sigma_{n=1}^{\infty}\frac{1}{n^{4}}$ $=81\times\frac{\pi^{4}}{90}$ $=\frac{9\pi^{4}}{10}$ (b) $\Sigma_{k=5}^{\infty}\frac{1}{(k-2)^{4}}=\Sigma_{k=3}^{\infty}\frac{1}{k^{4}}$ $\Sigma_{k=3}^{\infty}\frac{1}{k^{4}}=\frac{\pi^{4}}{90}-\frac{1}{1^{4}}-\frac{1}{2^{4}}$ $=\frac{\pi^{4}}{90}-\frac{17}{16}$
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