Find $ \sum_{n=2}^{+\infty}$ $\frac{1}{n^{3}-n}$ Find:
$$ \sum_{n=2}^{+\infty}\frac{1}{n^{3}-n}$$
I tried to resolve into partial fractions to see if there are some cancellations ,but that did not helped me .How do i do this ?Thanks

Since $n^3 - n = n(n^2 - 1) = n(n-1)(n+1)$, $$\frac{1}{n^3 - n} = \frac{1}{n}\frac{1}{(n-1)(n+1)} = \frac{1}{2n}\left(\frac{1}{n-1} - \frac{1}{n+1}\right) = \frac{1}{2(n-1)n} - \frac{1}{2n(n+1)}.$$ Now $\sum_{n = 2}^\infty 1/(n^3 - n)$ telescopes to $1/4$.
