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?
 A: 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 $(1/2)\pi^2/8=\pi^2/16$ 
A: 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 $$\sum \frac{n^2}{(4-n^2)^2}=0 + \frac{1}{4}\cdot 2\cdot \frac{\pi^2}{8}=\frac{\pi^2}{16}$$
A: HINT
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(n^2-4)^2}=\sum_{n=1}^{\infty} \frac{(2n-1)^2}{((2n-1)^2-4)^2}$$
Using partial fraction expansion, note $$\frac{(2n-1)^2}{((2n-1)^2-4)^2}=\left(\frac{1}{4(2n+1)^2}+\frac{1}{4(2n-3)^2}\right)-\left(\frac{1}{8(2n+1)}-\frac{1}{8(2n-3)}\right)$$
Note that the second part has cancelling terms. 
