Infinite sums of reciprocal power: $\sum\frac1{n^{2}}$ over odd integers The infinite series I need to solve is 
$$\sum_{n=1,3,5...}^{\infty}\frac{1}{n^{2}}$$
and because the point of interest lies in the value of odd n, 
the infinite series can be expressed as
$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}$$
This came up in a quantum mechanics problem involving the expectation value of the Hamiltonian. 
Is there a good idea to verify the solution is indeed $$\frac{\pi^{2}}{8}$$
or is this something with which I must refer to a math table?
Any good ideas would be helpful.

by using 
$$\frac{\pi^2}{6}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...$$
$$\frac{\pi^2}{6}=1+\frac{1}{3^2}+\frac{1}{5^2}+..\frac{1}{2^2}(1+\frac{1}{2^2}+\frac{1}{3^2}+...)$$
$$\frac{\pi^2}{6}=1+\frac{1}{3^2}+\frac{1}{5^2}+..\frac{1}{2^2}(\frac{\pi^2}{6})$$
$$\frac{\pi^2}{6}-\frac{\pi^2}{24}=1+\frac{1}{3^2}+\frac{1}{5^2}+..$$
$$\frac{\pi^2}{8}=1+\frac{1}{3^2}+\frac{1}{5^2}+..$$
