Evaluating sums question $\sum_{k=2}^n \frac{1}{k^2-1}$ [duplicate]

Timbuc gave you the right way to approach the problem. If you take a deeper look to the results obtained for low values of $n$, you should easily find that the general term is just given by $$S_n=\sum_{k=2}^n\frac{1}{k^2-1}=\frac12\sum_{k=2}^n\left(\frac1{k-1}-\frac1{k+1}\right)=\frac{(n-1) (3 n+2)}{4 n (n+1)}$$

$$\frac12\sum_{k=2}^N\left(\frac1{k-1}-\frac1{k+1}\right)=$$

$$=\frac12\left(1-\color{red}{\frac13}+{\frac12}-\color{green}{\frac14}+\color{red}{\frac13}-\color{blue}{\frac15}+\color{green}{\frac14}-\frac16+\ldots+\color{purple}{\frac1{N-1}}-\frac1{N+1}\right)=\ldots$$