Mathematical Reflections 2011, Issue 4 - Problem U201

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

Proposed by Titu Andreescu.

Solution:
At first, we observe that $$\dfrac{3n^2-1}{(n^3-n)^2} = \dfrac{1}{2} \left(\dfrac{2n-1}{n^2(n-1)^2} - \dfrac{2(n+1)-1}{(n+1)^2n^2}\right).$$ So,
$$\sum_{n=2}^{+\infty} \dfrac{3n^2-1}{(n^3-n)^2} = \dfrac{1}{2} \sum_{n=2}^{+\infty} \left(\dfrac{2n-1}{n^2(n-1)^2} - \dfrac{2(n+1)-1}{(n+1)^2n^2}\right).$$ This is a telescopic sum, so we obtain
$$\dfrac{1}{2} \sum_{n=2}^{+\infty} \left(\dfrac{2n-1}{n^2(n-1)^2} - \dfrac{2(n+1)-1}{(n+1)^2n^2}\right) = \lim_{n \to +\infty} \dfrac{1}{2} \left(\dfrac{3}{4} - \dfrac{2n+1}{(n+1)^2n^2}\right) = \dfrac{3}{8}.$$