Prove that $\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$ I am asked to prove that $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}.$$
However, I am asked to prove it using the fact that
$$\frac{\pi}{2}\tan\left(\frac{\pi}{2}z\right)=\sum_{m \text{ odd}}\left(\frac{1}{m-z}-\frac{1}{m+z}\right),$$
where $z\in \mathbb{C}$, which is something I proved in a previous exercise.
My first thought was using the fact that
$$\frac{1}{m-z}-\frac{1}{m-z}=\frac{2z}{m^2-z^2}$$
and therefore
$$\sum_{m \text{ odd}}\left(\frac{1}{m-z}-\frac{1}{m+z}\right)=\sum_{m \text{ odd}}\frac{2z}{m^2-z^2}=\sum_{n=0}^\infty \frac{2z}{(2n+1)^2-z^2}.$$
This last series is similar to the one I am aiming at, but I don't know how to transform it into the one that I want. Can someone help me?
 A: Choose $z=\frac{a}{2}$ and take the limit as $a\rightarrow 0$.
We know that
$$
\lim_{x\rightarrow 0}\frac{\tan(x)}{x}=1,
$$
so
$$
\lim_{a\rightarrow 0}\frac{\frac{\pi}{2}\tan(\frac{\pi a}{4})}{a}=\frac{\pi^2}{8},
$$
which could be verified by L'Hospital's rule. We can replace $\tan(\frac{\pi a}{4})$ with the above summation, i.e.,
$$
\frac{\pi}{2}\tan\left(\frac{\pi a}{4}\right)=\sum_{n=0}\frac{a}{(2n+1)^2-\frac{a^2}{4}}.$$
Now, we can substitute and clean up.
$$
\begin{align}
\lim_{a\rightarrow 0}\frac{\sum_{n=0}\frac{a}{(2n+1)^2-\frac{a^2}{4}}}{a}&=\frac{\pi^2}{8}\\
\lim_{a\rightarrow 0}\sum_{n=0}\frac{1}{(2n+1)^2-\frac{a^2}{4}}&=\frac{\pi^2}{8}\\
\sum_{n=0}\lim_{a\rightarrow 0}\frac{1}{(2n+1)^2-\frac{a^2}{4}}&=\frac{\pi^2}{8}\\
\sum_{n=0}\frac{1}{(2n+1)^2}&=\frac{\pi^2}{8}
\end{align}
$$
Disclaimer: Thanks to Dan Velleman for his/her insight.
A: Following the comment of @RaymondManzoni, we obtain continuing OP's approach
\begin{align*}
\frac{\pi}{4z}\tan\left(\frac{\pi}{2}z\right)=\sum_{n=0}^\infty\frac{1}{(2n+1)^2-z^2}\tag{1}
\end{align*}

We can evaluate the right-hand side of (1) at $z=0$. Since $\tan\left(\frac{\pi}{2}z\right)$ has a series expansion for $|z|<1$ we obtain by taking the limit as $z\to 0$ and the tangent  series expansion:
\begin{align*}
\color{blue}{\sum_{n=0}^\infty\frac{1}{(2n+1)^2}}&
=\lim_{z\to 0}\frac{\pi}{4z}\tan\left(\frac{\pi}{2}z\right)\\
&=\lim_{z\to 0}\frac{\pi}{4z}\left(\frac{\pi}{2}z+O(z^3)\right)\\
&\,\,\color{blue}{=\frac{\pi^2}{8}}
\end{align*}

A: Definitely not as good as C.Koca's proof because this will just feel like it came out of nowhere (and it's also not the proof you asked for! Just adding it there because might as well, if someone's looking for it). But here's an alternative proof using Fourier series:
Start from the function $f : \mathbb{R} \to \mathbb{R}$ which is $x \mapsto 1 + x/\pi$ on $(-\pi, 0)$, $x \mapsto 1 - x/\pi$ on $(0, \pi)$, and $0$ everywhere else. It's not even continuous but it doesn't matter, you can still compute its Fourier decomposition in the usual way, and get:
$f(x) = \dfrac{1}{2} + \dfrac{4}{\pi^2}\displaystyle\sum_{n \geq 0} \dfrac{\cos\big((2n + 1)x\big)}{(2n+1)^2}$
You can then apply your function at an appropriate value of $x$ to make the right-hand side look like the sum you want it to. And the left-hand side if explicitly given by the original expression for $f$, so you get an equality from which you can deduce the value of your sum :).
