Help summing the telescoping series $\sum_{n=2}^{\infty}\frac{1}{n^3-n}$. I know a priori that the series $$\sum_{n=2}^{\infty}\frac{1}{n^3-n}$$ converges. However, I am tasked with summing the series by treating it as a telescoping series. 
By partial fraction decomposition, the series can be written as: $$\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$$
$s_n=\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)=\left(\frac{1}{6}+\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{8}+\frac{1}{4}-\frac{1}{3}\right)+...+\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$
I then grouped the terms by their position in each partial sum, 
First terms: $\frac{1}{6},\frac{1}{8},\frac{1}{10},\frac{1}{12}...$
Second terms: $\require\cancel{\cancel{\frac{1}{2}}},\cancel{\frac{1}{4}},\cancel{\frac{1}{6}},\cancel{\frac{1}{8}}...$
Third terms: $\cancel{-\frac{1}{2}},-\frac{1}{3},\cancel{-\frac{1}{4}},-\frac{1}{5}...$
Cancelling leaves the series: $$\sum_{n=3}^{\infty}\frac{1}{2n}-\sum_{n=1}^{\infty}\frac{1}{2n+1}$$
However I'm stuck here since I see a divergent harmonic series summed with another harmonic series but I know the original series in question is convergent to $\frac{1}{4}$. What can I do? I suspect my error can be fixed somehow by adjusting the bounds of the sums...? Thanks in advance. 
 A: Your partial fraction decomposition is fine, but you need not to break up into separate positive and negative sums.  Instead, you need to cancel terms within the summation.
$$\sum_{n=2}^{\infty}\bigg(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\bigg)=\\\frac 12\sum_{n=2}^{\infty}\bigg(\frac{1}{(n+1)}+\frac{1}{(n-1)}-\frac{2}{n}\bigg)=\\\frac 12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right)-\left(\frac{1}{n}-\frac{1}{n+1}\right)$$
Now notice that the negative part of the $n$ term of the sum cancels with the positive part of the $n+1$ term of the sum, so all the terms disappear except the positive part of the $n=2$ term.  Our sum then equals
$$\frac 12\left(\frac 11-\frac 12\right)=\frac 14$$
A: $$\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$$
$$=\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{2n}-\frac{1}{2n}\right)$$
$$=1/2\sum_{n=2}^{\infty}\left(\frac{1}{(n+1)}-\frac{1}{n}+\frac{1}{(n-1)}-\frac{1}{n}\right)$$
Do  you  see two telescoping series?
A: Alt. hint:   it is not necessary to do a full partial fractions decomposition, it's enough to telescope:
$$
\frac{1}{n^3-n} = \frac{1}{2}\frac{(n+1) - (n-1)}{(n-1)n(n+1)} = \frac{1}{2}\left(\frac{1}{(n-1)n} - \frac{1}{n(n+1)}\right)
$$
