Find the Fourier series for $f(x) := |\sin(x)|$ and the sum of $\sum_{n=1}^{\infty} \frac {(-1)^{n+1}} {4n^2-1}$. 
Find the Fourier series for $f(x) := |\sin(x)|$ and the sum of $\sum_{n=1}^{\infty} \frac {(-1)^{n+1}} {4n^2-1}$.

I have computed $$c_n = \frac 1 {2\pi} \int^{\pi}_{-\pi} |\sin(x)|e^{-inx} dx =  \frac 1 {2\pi} \int^{\pi}_{0} |\sin(x)|e^{-inx} + |\sin(-x)|e^{-in(-x)} dx = \frac {1} {\pi} \frac {1} {n^2-1} ((-1)^{n+1} -1)$$
However, this doesn't correspond that well with the sum of the series I must find. What have I done wrong ?
 A: The exponential Fourier series for this particular $f(x)$ is
\begin{align}
|\sin(ax)| = \sum_{n= -\infty}^{\infty} c_{n} e^{-i nx},
\end{align}
where 
\begin{align}
c_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} |\sin(ax)| \ e^{-i nx} dx.
\end{align}
This integral can be calculated as follows.
\begin{align}
c_{n} &= \frac{1}{2\pi} \int_{-\pi}^{\pi} |\sin(ax)| \ e^{-i nx} dx \\
&= \frac{1}{2\pi} \int_{-\pi}^{0} |\sin(ax)| \ e^{-i nx} dx + \frac{1}{2\pi} \int_{0}^{\pi}
|\sin(ax)| \ e^{-i nx} dx \\
&= \frac{1}{2\pi} \int_{0}^{\pi} |\sin(-ax)| \ e^{i nx} dx + \frac{1}{2\pi} \int_{0}^{\pi}
|\sin(ax)| \ e^{-i nx} dx \\
&= \frac{1}{\pi} \int_{0}^{\pi} \sin(ax) \cos(nx) dx \\
&= \frac{1}{2\pi} \int_{0}^{\pi} \left[ \sin(a+n)x + \sin(a-n)x \right] dx\\
&= \frac{1}{2\pi} \left[ \frac{2a}{a^{2} - n^{2}} - \frac{\cos(a\pi + n \pi)}{a+n} 
- \frac{\cos(a\pi - n \pi)}{a-n} \right] \\
c_{n} &= \frac{a}{\pi (a^{2} - n^{2})} \left[ 1 - (-1)^{n} \cos(a\pi) \right].
\end{align}
From this the Fourier series is
\begin{align}
|\sin(ax)| = \frac{a}{\pi} \sum_{-\infty}^{\infty} \frac{(1 - (-1)^{n} \cos(a\pi))}
{\pi (a^{2} - n^{2})} \ e^{-i nx}.
\end{align}
The summation can be changed and is seen in the following.
\begin{align}
\sum_{-\infty}^{\infty} b_{n} e^{-i nx} &= \sum_{-\infty}^{-1} b_{n} e^{-i nx} + b_{0}
+ \sum_{1}^{\infty} b_{n} e^{-i nx} \\
&= b_{0} + \sum_{n=0}^{\infty} \left( b_{-n} e^{i nx} + b_{n} e^{-i nx} \right).
\end{align}
From this it is seen that
\begin{align}
|\sin(ax)| = \frac{1 - \cos(a\pi)}{a \pi} - \frac{2a}{\pi} \sum_{n=0}^{\infty}
\frac{(1-(-1)^{n} \cos(a\pi))}{n^{2}-a^{2}} \ \cos(nx) 
\end{align}
or
\begin{align}
|\sin(ax)| = \frac{\sin^{2}\left(\frac{a\pi}{2}\right)}{2 a \pi} - \frac{2a}{\pi} \sum_{n=0}^{\infty}
\frac{(1-(-1)^{n} \cos(a\pi))}{n^{2}-a^{2}} \ \cos(nx). 
\end{align}
Now, when $a=1$ this reduces to
\begin{align}
|\sin(x)| &= \frac{2}{\pi} - \frac{2}{\pi} \sum_{n=0}^{\infty} \frac{1+(-1)^{n}}{n^{2}-1} \ \cos(nx) \\
&= \frac{2}{\pi} - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\cos(2nx)}{4n^{2}-1}.
\end{align}
In order to evaluate the series in question let $x = \pi/2$ for which the Fourier series 
is reduced to
\begin{align}
|\sin(\pi/2)| &= \frac{2}{\pi} - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\cos(n \pi)}{4n^{2}-1} \\
&= \frac{2}{\pi} + \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{4n^{2}-1}.
\end{align}
which leads to 
\begin{align}
\sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{4n^{2}-1} = \frac{\pi - 2}{4}.
\end{align}
