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?

Wolfram Alpha gives me the partial fraction expansion:
$$\frac{n^2}{(4 - n^2)^2} = \frac{1}{8}\left(\frac 1{n-2}-\frac{1}{n + 2}\right) + \frac{1}{4}\left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}\right)$$
So the first part telescopes, and the second part will be some modified version of $\zeta(2)$.
In more detail: $$\begin{align}\sum_{n\text{ odd}} \left(\frac 1{n-2}-\frac{1}{n + 2}\right)&=\frac{1}{-1}-\frac{1}{3}+\frac{1}{1}-\frac{1}{5}+\frac{1}{3}-\frac{1}{7}+\cdots\\
&=-1+1=0\end{align}$$
since all the other terms cancel out.
And $$\begin{align}\sum_{n\text{ odd}}\left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}\right)&=\frac{1}{(-1)^2}+\frac{1}{3^2}+\frac{1}{1^2}+\frac{1}{5^2}+\frac{1}{3^2}+\frac{1}{7^2}+\cdots\\
&=2\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots\right)
\end{align}$$
It's a famous result that $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots=\frac{\pi^2}{8}$. You can can prove it if you know:
$$\frac{\pi^2}{6}=\zeta(2) = \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$$
and thus $$\zeta(2)-\frac1{2^2}\zeta(2) = \frac{1}{1^2}+\frac{1}{3^3}+\frac{1}{5^2}+\cdots$$
So $$\sum \frac{n^2}{(4-n^2)^2}=0 + \frac{1}{4}\cdot 2\cdot \frac{\pi^2}{8}=\frac{\pi^2}{16}$$
