Sum of infinite series: $\frac{1}{3^2+1} + \frac{1}{4^2+2} + \frac{1}{5^2+3} + \cdots \text{ad inf.}$ How to find the sum of the following infinite series:
$$ S = \frac{1}{3^2+1} + \frac{1}{4^2+2} + \frac{1}{5^2+3} + \cdots \text{ad inf.} $$
I only need to prove that $\left[\frac{1}{S}\right] = 2$.

Hint: Use $\dfrac{1}{n^2+n-2} = \dfrac{1}{(n-1)(n+2)} = \dfrac{1}{3}\Big(\dfrac{1}{n-1}-\dfrac{1}{n+2} \Big).$
