$$\displaystyle\lim_{n\rightarrow \infty} \left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+\dots+\frac{1}{n^{3}-n}\right)$$
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
Best Answer
Since: $$ \frac{1}{2}\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)=\frac{1}{n^3-n} $$ We have: $$ \sum_{n=2}^M\frac{1}{n^3-n}=\sum_{n=2}^M\frac{1}{2}\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)=\frac{1}{4}-\frac{1}{M(M+1)} $$ So the limit is $\frac{1}{4}$.