Determining limit of a sequence If  $$\lim_{n\to \infty}\left(\frac{1}{1^2}  + \frac{1}{2^2} +\cdots+ \frac{1}{n^2} \right)=\frac{\pi^2}{6}$$
If $$\lim_{n\to \infty}\left(\frac{1}{1^3\cdot 2^3}  + \frac{1}{3^3\cdot 4^3} +\cdots+ \frac{1}{n^3\cdot(n+1)^3} \right)=S$$
How to determine $S$?
I tried making a general pattern
~     $ \frac{n+1-n}{n^3\cdot(n+1)^3}   $ = $ \frac{1}{n^3\cdot(n+1)^2}  - \frac{1}{n^2\cdot(n+1)^3}   $   this is not giving me any cancellation of terms neither any involvement of above series

HINT: $$\dfrac{1}{n^3(n+1)^3}=6\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)-3\left(\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}\right)+\left(\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right).$$ Now, telescope.
