Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.3 Exercises - Page 745: 34

Answer

(a) $\frac{\pi^{2}-6}{6}$ (b) $\frac{6\pi^{2}-49}{36}$ (c)$ \frac{\pi^{2}}{24}$

Work Step by Step

(a) $\Sigma_{n=2}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}-\frac{1}{1^{2}}=\frac{\pi^{2}-6}{6}$ (b) $\Sigma_{n=3}^{\infty}\frac{1}{(n+1)^{2}}=\Sigma_{n=4}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}-1-\frac{1}{4}-\frac{1}{9}=\frac{6\pi^{2}-49}{36}$ (c) $\Sigma_{n=1}^{\infty}\frac{1}{(2n)^{2}}=\Sigma_{n=1}^{\infty}\frac{1}{4n^{2}}=\frac{1}{4} \cdot \frac{\pi^{2}}{6}=\frac{\pi^{2}}{24}$
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