How to prove that $\sum_{n \, \text{odd}} \frac{n^2}{(4-n^2)^2} = \pi^2/16$? The series:
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(4-n^2)^2} = \pi^2/16$$
showed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the series was calculated. 
How do I prove this directly?

First, the partial fraction of the summand can be written
$$\begin{align}
\frac{n^2}{(4-n^2)^2}&=\frac14\left(\frac{1}{n-2}+\frac{1}{n+2}\right)^2\\\\
&=\frac14 \left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}+\frac{1/2}{n-2}-\frac{1/2}{n+2}\right)
\end{align}$$

Second, we note that 
$$\begin{align}
\sum_{n\,\,\text{odd}}\frac{1}{(n\pm 2)^2}&=\sum_{n=-\infty}^\infty \frac{1}{(2n-1)^2}\\\\
&=2\sum_{n=1}^\infty \frac{1}{(2n-1)^2}\\\\
&=2\left(\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty \frac{1}{(2n)^2}\right)\\\\
&=\frac32 \sum_{n=1}^\infty \frac{1}{n^2}\\\\
&=\frac{\pi^2}{4}
\end{align}$$

Third, it is easy to show that 
$$\sum_{n=-\infty}^\infty \left(\frac{1}{2n-3}-\frac{1}{2n+1}\right)=0$$

Putting it all together we have
$$\sum_{n,\,\,\text{odd}}\frac{n^2}{(4-n^2)^2}=\frac{\pi^2}{8}$$
If we sum over the positive odd only, then the answer is $(1/2)\pi^2/8=\pi^2/16$ 
