Infinite sums of reciprocal power: $\sum\frac1{n^{2}}$ over odd integers [duplicate]

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.

Consider $$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\sum_{n=1}^{\infty}\frac{1}{(2n)^{2}}+\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}=\frac{1}4\sum_{n=1}^{\infty}\frac{1}{n^2}+\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}$$ So $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}=\frac{3}4 \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{3}4 \times\frac{\pi ^2}{6}=\frac{\pi ^2}{8}$$

We have $$ \sum_{n=1}^\infty \frac 1{n^2} = \frac{\pi^2}6 \tag+$$ Hence the even numbers sum to $$ \sum_{n=1}^\infty \frac 1{(2n)^2} = \frac 14 \sum_{n=1}^\infty \frac 1{n^2} = \frac{\pi^2}{24} $$ So, the difference is $$\sum_{n=1}^\infty \frac 1{(2n-1)^2} = \frac{\pi^2}6 - \frac{\pi^2}{24} = \frac{\pi^2}8 $$ The fact that $(+)$ holds, is "well-known", hence perhaps something you could refer to (looking it up in a table), or you compute $$ \int_0^1 \int_0^1 \frac 1{1- xy}\, dy\,dx $$ in two ways, expanding $\frac 1{1-xy} = \sum_{n=0}^\infty (xy)^n$ gives $$ \int_{[0,1]^2} \frac{1}{1-xy} \, d(x,y) = \sum_{n=1}^\infty \frac 1{n^2}$$ On the other hand, let $u = \frac 12(x+y)$, $v= \frac 12(y-x)$, then \begin{align*} \int_{[0,1]^2} \frac 1{1-xy}\, d(x,y) &= 4 \int_0^{1/2}\int_0^u \frac 1{1 - u^2+ v^2} \,dv \, du + 4 \int_{1/2}^1 \int_0^{1-u} \frac1{1 - u^2 + v^2}\, dv\, du\\ &= \frac{\pi^2}6 \end{align*}

by using $$\frac{\pi^2}{6}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...$$

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