Series Question: $\sum_{n=1}^{\infty}\frac{n^2}{(4n^2-1)^3}$ How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{n^2}{(4n^2-1)^3}$$
I tried to use partial fraction
$$\begin{align}\frac{n^2}{(4n^2-1)^3}&=\frac{1}{64(2n+1)}-\frac{1}{64(2n-1)}+\frac{1}{64(2n+1)^2}+\frac{1}{64(2n-1)^2}\\&-\frac{1}{32(2n+1)^3}+\frac{1}{32(2n-1)^3}\end{align}$$
I can compute
$$\sum_{n=1}^{\infty}\left[\frac{1}{64(2n+1)}-\frac{1}{64(2n-1)}\right]=-\frac{1}{64}$$
using telescoping series, but I cannot compute the rest. I believe there's a better way than this. Any help would be appreciated. Thanks in advance.

From where you left off...:
$\displaystyle \sum_{n=1}^\infty \dfrac{1}{32}\cdot \left(\dfrac{1}{(2n-1)^3} - \dfrac{1}{(2n+1)^3}\right) = \dfrac{1}{32}\cdot \left(1 - \dfrac{1}{3^3} + \dfrac{1}{3^3} - \dfrac{1}{5^3} + ... \right) = \dfrac{1}{32}$, and
$\displaystyle \sum_{n=1}^\infty \dfrac{1}{64}\cdot \left(\dfrac{1}{(2n-1)^2} + \dfrac{1}{(2n+1)^2}\right) = \dfrac{1}{32}\cdot \displaystyle \sum_{n=1}^\infty \dfrac{1}{(2n-1)^2} - \dfrac{1}{64} = \dfrac{1}{32}\cdot \dfrac{\pi^2}{8} - \dfrac{1}{64}$.
Thus adding these values and your partial answer to complete the calculation.
