How can I evaluate the sum of this series?

Question

I need to calculate the sum of this series.

$\sum_{n=2}^{\infty}\frac{1}{n(n^2-1)} $

By using the partial fraction, I obtained this sum: $\sum_{n=2}^{\infty}-\frac{1}{n}+\frac{1}{2(n+1)}+\frac{1}{2(n-1)} $

Proceeding in the replacement of incremental values of n I noticed that some element is simplified, but I can not get a useful form to calculate the sum. any help?

Answer 1

Hint: for $n\geq 2$, $$\frac1{n(n^2-1)}=\frac1{(n-1)n(n+1)}=\frac12\left(\frac1{(n-1)n}-\frac1{n(n+1)}\right)$$