Prove the Identity $\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $

Question

By considering the fact that $f(\pi/2)=1$, prove the identity

$\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $

This question was is a subsection in a chapter on Fourier series, can I use my understanding of Fourier series to prove the identity, or is there an easier way of making the proof?

EDIT Another part of this question asked to find the Fourier series of the periodic function $f(x)$ defined by $f(x)= |sin(x)|$, this is the function they are probably referring to in the proof.

Answer 1

Fourier series are not really needed to prove the identity: $$\begin{align*}\sum_{m\geq 1}\frac{(-1)^m}{4m^2-1}&=\frac{1}{2}\sum_{m\geq 1}(-1)^m\left(\frac{1}{2m-1}-\frac{1}{2m+1}\right)\\&=\frac{1}{2}\sum_{m\geq 1}(-1)^m\int_{0}^{1}\left(x^{2m-2}-x^{2m}\right)\,dx\\&=\frac{1}{2}\int_{0}^{1}\frac{x^2-1}{x^2+1}\,dx\\&=\frac{1}{2}\left(1-2\arctan 1\right)\\&=\color{red}{\frac{2-\pi}{4}}.\end{align*}$$

Answer 2

I'll play around and see what happens.

We want $\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $

$\begin{array}\\ S &=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}}\\ &=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2}\frac1{1-\frac{1}{4m^2}}\\ &=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2}\sum_{j=0}^{\infty}\frac{1}{(4m^{2})^j}\\ &=4\sum_{m=1}^{\infty}(-1)^m\sum_{j=0}^{\infty}\frac{1}{(4m^{2})^{j+1}}\\ &=4\sum_{m=1}^{\infty}(-1)^m\sum_{j=1}^{\infty}\frac{1}{(4m^{2})^{j}}\\ &=4\sum_{j=1}^{\infty}\sum_{m=1}^{\infty}(-1)^m\frac{1}{(4m^{2})^{j}}\\ &=4\sum_{j=1}^{\infty}\frac1{4^j}\sum_{m=1}^{\infty}(-1)^m\frac{1}{(m^{2})^{j}}\\ &=4\sum_{j=1}^{\infty}\frac1{4^j}\sum_{m=1}^{\infty}(-1)^m\frac{1}{m^{2j}}\\ &=-4\sum_{j=1}^{\infty}\frac1{4^j}\zeta(2j)(1-2^{1-2j})\\ &=-4\sum_{j=1}^{\infty}\frac1{4^j}\zeta(2j)+8\sum_{j=1}^{\infty}\frac1{4^{2j}}\zeta(2j))\\ \end{array} $

According to https://en.wikipedia.org/wiki/Riemann_zeta_function, $\sum_{j=1}^{\infty}\frac{\zeta(2j)-1}{4^j} =\frac16 $ and $\sum_{j=1}^{\infty}\frac{\zeta(2j)-1}{16^j} =\frac{13}{30}-\frac{\pi}{8} $. Therefore

$\begin{array}\\ \sum_{j=1}^{\infty}\frac{\zeta(2j)}{4^j} &=\frac16+\sum_{j=1}^{\infty}\frac{1}{4^j}\\ &=\frac16+\frac{1/4}{1-1/4}\\ &=\frac16+\frac13\\ &=\frac12\\ \end{array} $

and

$\begin{array}\\ \sum_{j=1}^{\infty}\frac{\zeta(2j)}{16^j} &=\frac{13}{30}-\frac{\pi}{8}+\sum_{j=1}^{\infty}\frac{1}{16^j}\\ &=\frac{13}{30}-\frac{\pi}{8}+\frac{1/16}{1-1/16}\\ &=\frac{13}{30}-\frac{\pi}{8}+\frac{1}{15}\\ &=\frac12-\frac{\pi}{8}\\ \end{array} $.

Putting this together,

$\begin{array}\\ S &=-4\sum_{j=1}^{\infty}\frac1{4^j}\zeta(2j)+8\sum_{j=1}^{\infty}\frac1{4^{2j}}\zeta(2j)\\ &=-4(\frac12)+8(\frac12-\frac{\pi}{8})\\ &=-2+4-\pi\\ &=2-\pi\\ \end{array} $.

Whew!