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? 
 A: $$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2(n-1)}-\frac{1}{2(n+1)}$$
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 $\frac{3}{4}$.
A: As you correctly have: $$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2(n-1)}-\frac{1}{2(n+1)}$$ Now, observe that $$\sum_{n=2}^{\infty}\frac{1}{n-1}=\sum_{n=1}^{\infty}\frac{1}{n} \qquad \text{and}\qquad\sum_{n=2}^{\infty}\frac{1}{n+1}=\sum_{n=3}^{\infty}\frac{1}{n}$$ Hence 
$$\sum_{n=2}^{\infty}\frac{1}{n^2-1}=\frac12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)=\frac12\left(\sum_{n=1}^{\infty}\frac1n-\sum_{n=3}^{\infty}\frac1n\right)=\frac12\left(\frac11+\frac12\right)=\frac34$$
A: More generally,
if
$s_m(n)
=\sum_{k=m+1}^n \frac{1}{k^2-m^2}
$
then,
if $n > 3m$,
$\begin{array}\\
s_m(n)
&=\sum_{k=m+1}^n \frac{1}{k^2-m^2}\\
&=\sum_{k=m+1}^n \frac1{2m}(\frac{1}{k-m}-\frac1{k+m})\\
&=\frac1{2m}\sum_{k=m+1}^n \frac{1}{k-m}-\frac1{2m}\sum_{k=m+1}^n\frac1{k+m}\\
&=\frac1{2m}(\sum_{k=1}^{2m} \frac{1}{k}+\sum_{k=2m+1}^{n-m} \frac{1}{k})-(\frac1{2m}\sum_{k=2m+1}^{n-m}\frac1{k}+\frac1{2m}\sum_{k=n-m+1}^{n+m}\frac1{k})\\
&=\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}-\frac1{2m}\sum_{k=n-m+1}^{n+m}\frac1{k}\\
\end{array}
$
so
$\begin{array}\\
s_m(n)-\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}
&=-\frac1{2m}\sum_{k=n-m+1}^{n+m}\frac1{k}\\
\text{so that}\\
|s_m(n)-\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}|
&=\frac1{2m}|\sum_{k=n-m+1}^{n+m}\frac1{k}|\\
&\le\frac1{2m}|\sum_{k=n-m+1}^{n+m}\frac1{n-m+1}|\\
&=\frac1{2m}|\frac{2m}{n-m+1}|\\
&=\frac{1}{n-m+1}\\
&\to 0
\quad\text{ as } n \to \infty\\
\end{array}
$
Therefore
$\lim_{n \to \infty} s_m(n)
=\frac1{2m}\sum_{k=1}^{2m} \frac{1}{k}
$.
For $m=1$
the sum is
$\frac1{2}(\frac1{1}+\frac1{2})
=\frac34
$.
A: HINT:
$$\frac{1}{2[(n+2)-1]}-\frac{1}{2[n+1]}=0$$
