How to find sum and convergence of series $\sum_{n = 1}^{\infty}\frac{1}{(2n-1)^{2}\cdot(2n+1)^{2}}$ I laid this shot that: 
$$\frac{1}{(2n-1)^2(2n+1)^2}=1/4\, \left( 2\,n+1 \right) ^{-2}-1/4\, \left( 2\,n-1 \right) ^{-1}+1/
4\, \left( 2\,n-1 \right) ^{-2}+1/4\, \left( 2\,n+1 \right) ^{-1}
$$
But I don't know what to do next. Help me please
 A: Note that:
$$\sum_{n=1}^\infty\frac1{(2n-1)^2}=1+\sum_{n=1}^\infty\frac1{(2n+1)^2}$$
$$\sum_{n=1}^\infty\frac1{2n-1}=1+\sum_{n=1}^\infty\frac1{2n+1}$$
Thus, your series reduces as follows:
$$\sum_{n=1}^\infty\frac1{(2n-1)^2(2n+1)^2}=-\frac12+\frac12\sum_{n=1}^\infty\frac1{(2n-1)^2}$$
Since everything cancels except that.  Now let $S$ be the following sum:
$$S=\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$$
It can be seen that
$$S=\sum_{n=1}^\infty\left[\frac1{(2n-1)^2}+\frac1{(2n)^2}\right]=\frac14S+\sum_{n=1}^\infty\frac1{(2n-1)^2}$$
Thus,
$$\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac34S=\frac{\pi^2}8$$
and finally,
$$\sum_{n=1}^\infty\frac1{(2n-1)^2(2n+1)^2}=-\frac12+\frac12\left(\frac{\pi^2}8\right)=\frac{\pi^2-8}{16}$$
A: By partial fraction decomposition
$$ \frac{1}{(2x-1)^2 (2x+1)^2} = \frac{\frac{1}{16}}{\left(x-\frac{1}{2}\right)^2}+\frac{\frac{1}{8}}{x-\frac{1}{2}}-\frac{\frac{1}{8}}{x+\frac{1}{2}}+\frac{\frac{1}{16}}{\left(x+\frac{1}{2}\right)^2} \tag{1}$$
hence it follows that:
$$ \sum_{n\geq 1}\frac{1}{(2n-1)^2 (2n+1)^2}=\frac{1}{4}+\frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n-1)^2}+\frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n+1)^2} = \color{red}{\frac{\pi^2-8}{16}}\tag{2}$$
