Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises: 43

Answer

$\frac{3}{2}$

Work Step by Step

$\Sigma \frac{2}{n^2-1}=\Sigma (\frac{1}{n-1}-\frac{1}{n+1})$ $=(\frac{1}{1}-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...=\frac{1}{1}+\frac{1}{2}=\frac{3}{2}$
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