Question: I'm supposed to determine whether this sum diverges or converges and if it converges then find its value: 
$$
\sum\_{n=2}^\infty \frac{1}{n^2-1}.
$$

Using the comparison test I eventually showed that this converges. But I can't figure out how to show what this sum converges to. The only sums we actually found values for in my notes are geometric series which this clearly isn't.

I saw that I could use partial fraction decomposition to represent the terms as 
$$\frac{1/2}{n-1}- \frac{1/2}{n+1} $$
but that just gets me $\infty - \infty$, so this isn't the way to do it.

I'm not sure how to find the value of this sum. I don't need the full solution but a hint would be appreciated. Thanks. :)

Answer: 
Use partial fraction decomposition to represent the sum and notice that it is a telescoping series.
**hint**: $\dfrac{1}{n^2-1} = \dfrac{1}{2}\cdot \left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right) = \dfrac{1}{2}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right) + \dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$. From this you see that there are $2$ sums you calculate, and using telescoping the first sum is $\dfrac{1}{2}$, and the second is $\dfrac{1}{4}$, thus the answer is $\dfrac{3}{4}$ as claimed.