Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 43

$\frac{3}{2}$

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