Calculating the infinite sum of one over the odd numbers squared: $\sum_{i \ge 0} \frac{1}{(2i+1)^2}$ Can someone tell me how to calculate the following infinite sum?
$$
(1/1^2)+(1/3^2)+(1/5^2)+(1/7^2)+(1/9^2)+(1/11^2)+ \cdots
$$
Don't give me the answer. Can you tell me if this is a geometric series? 
 A: By absolute convergence, you may write
$$
\sum_{n=1}^{\infty}\frac{1}{{n}^2}=\sum_{p=1}^{\infty}\frac{1}{{(2p)}^2}+\sum_{p=1}^{\infty}\frac{1}{{(2p-1)}^2}=\frac14\sum_{p=1}^{\infty}\frac{1}{p^2}+\sum_{p=1}^{\infty}\frac{1}{{(2p-1)}^2}.
$$ Can you take it from here?
A: To test if it's geometric, compare the ratio of adjacent terms. For example, $\frac{1/3^2}{1/1^2}$ and $\frac{1/5^2}{1/3^2}$. Are these two quantities equal?
Hint:
$$
\left( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \cdots\right) \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \frac{1}{16^2} + \cdots \right) \\
= \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \cdots
$$
A: METHODOLGY $1$

Here, we present a way forward that does not require prior knowledge of the value of the series $\sum_{n=1}\frac{1}{n^2}=\frac{\pi^2}{6}$, the Riemann-Zeta Function, or dilogarithm function.  Rather, we apply straightforward analysis that includes application of the residue theorem.

To that end, note that we can write the series of interest as
$$\begin{align}
\sum_{n=0}^\infty \frac{1}{(2n+1)^2}&=\sum_{n=1}^N \int_0^1 x^{2n}\,dx\int_0^1y^{2n}\,dy\\\\
&=\int_0^1\int_0^1 \sum_{n=0}^\infty(x^2y^2)^n\,dx\,dy\\\\
&=\int_0^1\int_0^1 \frac{1}{1-x^2y^2}\,dx\,dy\\\\
&=\frac12\int_0^1\frac{\log(1+x)-\log(1-x)}{x}\,dx\tag 1
\end{align}$$
Then, we enforce the substitution $x\to \frac{x-1}{x+1}$ in $(1)$ to obtain
$$\frac12\int_0^1\frac{\log(1+x)-\log(1-x)}{x}\,dx=\int_1^\infty \frac{\log(x)}{x^2-1}\,dx\tag 2$$
Next, enforcing the substitution $x\to 1/x$ in $(2)$ reveals
$$\int_1^\infty \frac{\log(x)}{x^2-1}\,dx=\int_0^1 \frac{\log(x)}{x^2-1}\,dx \tag 3$$
Adding $(2)$ and $(3)$ and dividing by $(2)$ yields
$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx$$

Moving to the complex plane, we evaluate the integral $J$ defined by 
$$J=\oint_C \frac{\log^2(z)}{z^2-1}\,dz$$
where $C$ is the classical keyhole contour with (i) the branch cut along the non-negative real axis and (ii) with deformations around $z=1$.  Applying the residue theorem, it is easy to see that $J=i\pi^3$.  Therefore, we find that 
$$\begin{align}
J&=i\pi^3\\\\
&=\int_{0}^{\infty}\frac{\log^2(x)}{x^2-1}\,dx-\text{PV}\int_0^\infty \frac{\left(\log(x)+i2\pi\right)^2}{x^2-1}\,dx\\\\
&=-i4\pi\int_0^\infty \frac{\log(x)}{x^2-1}\,dx\\\\
&+\color{blue}{(4\pi^2)\text{PV}\left(\int_0^\infty \frac{1}{x^2-1}\,dx\right)}\\\\
&+\color{red}{(4\pi^2)\lim_{\epsilon \to 0^+}\int_{\pi}^{2\pi} \frac{1}{(1+\epsilon e^{i\phi})^2-1}\,(i\epsilon e^{i\phi})\,d\phi}\\\\
&=-i4\pi\int_0^\infty \frac{\log(x)}{x^2-1}\,dx+\color{blue}{0}+\color{red}{i2\pi^3}\tag 4
\end{align}$$
Finally, solving $(4)$ for the integral of interest yields
$$\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$ 
and hence we find that the series of interest is 
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}}$$

METHODOLGY $2$

I thought it might be instructive to include another way forward that relies strictly on a well-known method of contour integration. The true potential usefulness of this posted solution is that it presents a methodology that can be applied to a wide class of problems for which some of the simpler ways forward to evaluate the specific series of interest fail.  It is to that end that we proceed.


Observe that the function $\pi \cot(\pi z)$ has simple poles at $z=n\in \mathbb{Z}$.  Let $\displaystyle f(z)=\frac{\pi \cot(\pi z)}{(2z+1)^2}$, which has a second order pole at $z=-1/2$ and simple poles at $z=n$. 


Residues of $f(z)$:

The residues of $f(z)$ at $z=n$ are 
$$\begin{align}
\text{Res}(f(z),z=n)&=\lim_{z\to n}\frac{(z-n)\pi \cot(\pi z)}{(2z+1)^2}\\\\
&=\frac{1}{(2n+1)^2}
\end{align}$$ 
while the residue of $f(z)$ at $z=-1/2$ is 
$$\begin{align}
\text{Res}(f(z),z=-1/2)&=\lim_{z\to -1/2}\frac{d}{dz}\left(\frac{(z+1/2)^2\pi \cot(\pi z)}{(2z+1)^2}\right)\\\\
&=-\frac{\pi^2}{4}
\end{align}$$ 


Contour Integration of $f(z)$:

We choose a contour $C$ to be the circle of radius $N+1/2$, centered at the origin.  The Residue Theorem guarantees that 
$$\oint_C f(z)\,dz=2\pi i \left(\sum_{n=-N}^N \frac{1}{(2n+1)^2}-\frac{\pi^2}{4}\right) \tag 1$$ 
In addition, as $N\to \infty$, the contour integral on the left-hand side of $(1)$ approaches zero.  
Putting it all together, we see that
$$\begin{align}
\sum_{-\infty}^\infty\frac{1}{(2n+1)^2}&=2\sum_{n=0}^\infty\frac{1}{(2n+1)^2}\\\\
&=\frac{\pi^2}{4}
\end{align}$$
whereby we obtain the coveted result

$$\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$

And we are done!
A: Not a geometric series.
Hint: consider Basel Problem. Consider sum over even terms. Subtract it.
