Summation of $\frac{n^2}{(n^2-4)^2}$ where $n$ runs over odds

Question

I want to evaluate the following sum over odds:

$$\sum_{n=1,3,5,\dots}\frac{n^2}{(n^2-4)^2}.$$

I tried to set $n=2k+1,k=0,1,2,\dots$ but that makes it complicated.

Answer 1

Since $\frac{2n}{n^2-4}=\frac{1}{n-2}+\frac{1}{n+2}$,$$\frac{8n^2}{(n^2-4)^2}=\frac{2}{(n-2)^2}+\frac{2}{(n+2)^2}+\frac{4}{n^2-4}=\frac{2}{(n-2)^2}+\frac{2}{(n+2)^2}+\frac{1}{n-2}-\frac{1}{n+2}.$$Summing this over odd positive $n$ gives$$2(\pi^2/8+1)+2(\pi^2/8-1).$$since the sum of $1/n^2$ for odd $n>0$ is $\pi^2/8$. (You can verify the linear terms vanish by telecoping.) Dividing out the factor of $8$, your sum is $\pi^2/16$.