Mathematical Reflections 2013, Issue 1 - Problem U253

Problem:
Evaluate $$\sum_{n>1} \dfrac{3n^2+1}{(n^3-n)^3}.$$

Proposed by Titu Andreescu.

Solution:
We observe that $$\dfrac{3n^2+1}{(n^3-n)^3}=\dfrac{1}{2}\left(\dfrac{1}{n^3(n-1)^3}-\dfrac{1}{(n+1)^3n^3}\right).$$
Hence, $$\begin{array}{lcl}\displaystyle \sum_{n=2}^\infty \dfrac{3n^2+1}{(n^3-n)^3}&=& \displaystyle \dfrac{1}{2} \sum_{n=2}^\infty \left(\dfrac{1}{n^3(n-1)^3}-\dfrac{1}{(n+1)^3n^3}\right)\\ &=& \displaystyle \dfrac{1}{2} \lim_{n \to \infty} \left(\dfrac{1}{8}-\dfrac{1}{(n+1)^3n^3}\right)\\&=& \dfrac{1}{16}. \end{array}$$