How to find sum of the series Σ 1/ ((2n - 1)^2 (2n+1)^2) ?



How to find sum of series Σ 1/ ((2n - 1)^2 (2n+1)^2)
Mathematica gives the answer 
$$\sum \_{n=1}^{\infty } \frac{1}{(2 n-1)^2 (2 n+1)^2}=\frac{\pi^2-8}{16}.$$
An elementary way to prove this is to write 
$$\frac{4}{(2 n-1)^2 (2 n+1)^2}=
\frac{1}{2 n+1}-\frac{1}{2 n-1}+\frac{1}{(2 n-1)^2}+\frac{1}{(2 n+1)^2} 
$$
and then 
\begin{equation}
 \sum \_{n=1}^{\infty }\Big(\frac{1}{2 n+1}-\frac{1}{2 n-1}\Big)=-\frac{1}{2\times1-1}=-1
\end{equation}
by telescoping, 
$$\sum \_{n=1}^{\infty }\frac{1}{(2 n-1)^2}=\sum \_{m=1}^{\infty }\frac{1}{m^2}-\sum \_{n=1}^{\infty }\frac{1}{(2 n)^2}
=\frac34\,\sum \_{m=1}^{\infty }\frac{1}{m^2}=\frac{\pi^2}{8},
$$
$$\sum \_{n=1}^{\infty }\frac{1}{(2 n+1)^2}=-1+\sum \_{k=1}^{\infty }\frac{1}{(2 k-1)^2}=-1+\frac{\pi^2}{8}.
$$


