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.
 A: Note that:
$$\sum _{n=1}^{\infty }{\frac {{n}^{2}}{ \left( 4\,{n}^{2}-1 \right) ^{3
}}}=\frac{1}{4}\,\sum _{n=1}^{\infty }\frac{1}{\left( 4\,{n}^{2}-1 \right)^3 }+
 \frac{1}{\left( 4\,{n}^{2}-1 \right)^2 }\tag{1}$$
then note that:
$$\sum _{n=1}^{\infty }  \frac{1}{\left( {n}^{2}-{x}^{2} \right)}=\frac{1}{2x^2}-\frac{1}{2}\,{\frac {\pi \,\cot \left( \pi \,x \right) }{x}}\tag{2}$$
insert $(2)$ into the differential equation:
$$ \left( {\frac {d^{2}}{d{x}^{2}}}f \left( x \right)  \right) a+
 \left( {\frac {d}{dx}}f \left( x \right)  \right) b \tag{3}$$
and evaluate $(3)$ at $x=1/2$ to get:
$$\frac{1}{4}\,\sum _{n=1}^{\infty }{\frac {a\,512}{ \left( 4\,{n}^{2}-1
 \right) ^{3}}}+{\frac {128\,a+64\,b}{ \left( 4\,{n}^{2}-1 \right) ^{2
}}}=a \left( 48-4\,{\pi }^{2} \right) +b \left( -8+{\pi }^{2} \right)\tag{4} $$
then let:
$$a=\frac{1}{512},\quad b=\frac{3}{256} \tag{5} $$ 
to get:

$$\sum _{n=1}^{\infty }{\frac {{n}^{2}}{ \left( 4\,{n}^{2}-1 \right) ^{3
}}}=\frac{1}{4}\,\sum _{n=1}^{\infty }\frac{1}{\left( 4\,{n}^{2}-1 \right)^3 }+
 \frac{1}{\left( 4\,{n}^{2}-1 \right)^2 }={\frac {1}{256}}\,{\pi }^{2} \tag{6} $$

A: HINT:
You have to know that:
$$\zeta(n)=\dfrac{2^n}{2^n-1}\sum\limits_{k=0}^{\infty}\dfrac{1}{(2k+1)^n}.$$
where $\zeta(\cdot)$ is the Riemann zeta function.
A: 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.
A: The polygamma functions are very smart to solve this kind of problem :

