Answer
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 $\frac{3}{2}$
