<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] 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 
$$\sum_{n=2}^{\infty}\frac{1}{n^2-1}=\frac12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)=\frac12\left(\sum_{n=1}^{\infty}\frac1n-\sum_{n=3}^{\infty}\frac1n\right)=\frac12\left(\frac11+\frac12\right)=\frac34$$
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