<s>[INST] 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? 
 [/INST] 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
$.
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