$$\lim_{n\to\infty}\Bigg({\frac{1}{6} + \frac{1}{24} + \frac{1}{60} + \cdots + \frac{1}{n^3 - n}}\Bigg)$$
Actually, I was unable to think it in a proper way but it was a multiple choice question. So, I just added first few terms and checked that where does that converge and got the answer which was nearly $\displaystyle\frac{1}{4}$. What is a proper way to solve this.
Hint: telescope $\;\dfrac{1}{n^3-n}=\dfrac{1}{(n-1)\,n\,(n+1)}=\dfrac{1}{2}\left(\dfrac{1}{(n-1)\,n}-\dfrac{1}{n\,(n+1)}\right)\,$.
hint: $\dfrac{1}{n^3-n}= \dfrac{A}{n}+\dfrac{B}{n-1}+\dfrac{C}{n+1}$. You first have to find $A,B,C$ and rearrange these to telescope terms. After this, you should know the next move...
Guide:
Note that we have$$\frac{1}{k^3-k}=\frac{0.5}{k-1}-\frac{1}{k}+\frac{0.5}{k+1}$$
Now, use telescoping sum to evaluate the partial sum and then take limit to $\infty$.
