Evaluate the limit $\lim\limits_{n\rightarrow \infty}(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\dots+\frac{1}{n^{3}-n})$ $$\displaystyle\lim_{n\rightarrow \infty} \left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\dots+\frac{1}{n^{3}-n}\right)$$
I am not able to find any technique to proceed. It might be simple but I am not able to understand. If the hint and the concept used is explained it will be very helpful. 
Thanks in advance

To complement Ákos' answer:
$\begin{align}&\sum_{n=2}^\infty\frac 1 {n^3-n}=\sum_{n=2}^\infty \frac{1}{2}\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)\\
&=\frac 1 2 \left[ \left(\frac 1 2 +\frac 1 6 + \frac 1 {12} + \cdots \right)-\left(\frac 1 6 +\frac 1 {12} + \cdots \right)\right]\\
&=\frac 1 2 \times \frac 1 2=\frac 1 4
\end{align}$
