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 - 11.2 Exercises - Page 735: 43


The series is convergent and the sum is $\frac{3}{2}$

Work Step by Step

$\Sigma^{\infty}_{n=2} \frac{2}{n^{2}-1}$ $=\Sigma^{\infty}_{n=2} \frac{2}{(n-1)(n+1)}$ $s_{n}=\Sigma^{n}_{i=2}\frac{2}{(i-1)(i+1)}$ $=\frac{2}{1 \times 3}+\frac{2}{2 \times 4}+ \frac{2}{3 \times 5}+...+ \frac{2}{(n-1)(n+1)}$ Thus $s_{n}=\Sigma^{n}_{i=2}\frac{2}{(i-1)(i+1)}$ $s_{n}=\Sigma^{n}_{i=2} (\frac{1}{i-1} - \frac{1}{i+1})$ Use partial fraction decomposition $=(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+...+(\frac{1}{n-1}-\frac{1}{n+1})$ $=1+\frac{1}{2}$ $=\frac{3}{2}$ And so $\lim\limits_{n \to \infty}s_{n} = \lim\limits_{n \to \infty} \frac{3}{2}$ Therefore the series is convergent and the sum is $\frac{3}{2}$
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