Find the sum: $\sum_{n=2}^\infty \frac{1}{n^2-1}$ 
Evaluate :  $$\sum_{n=2}^\infty \frac{1}{n^2-1}$$ 

I've tried to rewrite the questions as $$\sum _{n=2}^{\infty \:\:}\left(-\frac{1}{2\left(n+1\right)}+\frac{1}{2\left(n-1\right)}\right)$$   but still couldnt't get any answer as when I substitute numbers $(n)$ into the $Sn \left(-\frac{1}{6}+\frac{1}{2}-\frac{1}{8}+\frac{1}{4}-\frac{1}{10}\right)$ , $Sn$ just keeps going on and on.  So how do I solve this problem and what does this series converge to? 

$$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2(n-1)}-\frac{1}{2(n+1)}$$
Taking $\frac{1}{2}$  common ,
$$\sum_{n=2}^{\infty}\frac{1}{n^2-1}=\frac12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$
Write the first few terms of the series as :
$\frac{1}{2}( 1 - \frac{1}{3} +\frac{1}{2} - \frac{1}{4}  + \frac{1}{3}  - \frac{1}{5} + \frac{1}{4} - \cdot \cdot \cdot$
You can see that except $1$ and $\frac{1}{2}$ every terms get cancelled out and the $n$-th term tends to zero. So the result inside the bracket is $\frac{3}{2}$ . But there is a $\frac{1}{2}$ outside the bracket which needs to be multiplied. So the answer is $\frac{3}{4}$.
