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.
 A: 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*}
A: by using 
$$\frac{\pi^2}{6}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...$$
$$\frac{\pi^2}{6}=1+\frac{1}{3^2}+\frac{1}{5^2}+..\frac{1}{2^2}(1+\frac{1}{2^2}+\frac{1}{3^2}+...)$$
$$\frac{\pi^2}{6}=1+\frac{1}{3^2}+\frac{1}{5^2}+..\frac{1}{2^2}(\frac{\pi^2}{6})$$
$$\frac{\pi^2}{6}-\frac{\pi^2}{24}=1+\frac{1}{3^2}+\frac{1}{5^2}+..$$
$$\frac{\pi^2}{8}=1+\frac{1}{3^2}+\frac{1}{5^2}+..$$
A: 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 $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}=\frac{3}4 \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{3}4 \times\frac{\pi ^2}{6}=\frac{\pi ^2}{8}$$
A: We can use the residue theorem:
$$\sum_{n=1}^{\infty} \frac1{(2 n-1)^2} = \frac12 \sum_{n=-\infty}^{\infty} \frac1{(2 n-1)^2} = -\frac{\pi}{2} \operatorname*{Res}_{z=1/2} \frac{\cot{\pi z}}{(2 z-1)^2} = \frac{\pi^2}{8 \sin^2{(\pi/2)}} = \frac{\pi^2}{8}$$
A: To use $\zeta(2)=\frac{\pi^2}{6}$ (have a look at this historical question) is a fast way to go, but plenty of approaches are possible. An Euler-style proof follows.
Step 1. The function $f(x)=\cos(\pi x)$ is an entire function having simple roots at $\mathbb{Z}+\frac{1}{2}$.
Step 2. The Mittag-Leffler theorem then gives the Weierstrass product:
$$ \cos(\pi z) = \prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2}\right) \tag{2a}$$
where the LHS can be written as a Taylor series:
$$ \cos(\pi z) = 1-\frac{\pi^2 z^2}{2}+\frac{\pi^4 z^4}{24}-\ldots\tag{2b}$$
Step 3. By comparing the coefficient of $z^2$ in the RHS of $(2a)$ and in the RHS of $(2b)$ we get:
$$ -\sum_{n\geq 0}\frac{4}{(2n+1)^2} = -\frac{\pi^2}{2}\tag{3} $$
and your claim immediately follows.
