Fourier Series for $|\cos(x)|$ I'm having trouble figuring out the Fourier series of $|\cos(x)|$ from $-\pi$ to $\pi$.
I understand its an even function, so all the $b_n$s are $0$
$$a_0 = \frac 2 \pi \int_0^\pi |\cos(x)|\,dx = 0$$
$$a_n = \frac 2 \pi \int _0^\pi  |\cos(x)| \cos(nx) \, dx = \frac 2 \pi \int_0^\pi \cos^2(x)\,dx.$$
since for all $j,k$ not equal the integral is zero.
so only $a_1$ remains. is this correct? 
How would I evaluate $\sum_{n=1}^\infty (-1)^{n-1} /(4n^2 - 1)\  {}$?

Although $ \int_0^\pi \cos(x)\,dx = 0$, $a_0\ne 0$ because $$\int_0^{\pi/2} |\cos(x)|\,dx=\int_{\pi/2}^{\pi} |\cos(x)|\,dx. $$
We can evaluate it as follows, as can be seen in the plot below
$$a_0 = \frac 1 \pi \int_{-\pi}^\pi |\cos(x)|\,dx=\frac 2 \pi \int_0^\pi |\cos(x)|\,dx=\frac 4 \pi \int_0^{\pi/2} |\cos(x)|\,dx = \frac 4 \pi \int_0^{\pi/2} \cos(x)\,dx=\frac 4 \pi.$$
$$\tag{1}$$
Plot of $\cos x$ (doted line) and $|\cos x|$ (solid line) in the interval $[-\pi,\pi]$. 

The coefficients $b_n=0$ as you concluded. As for the $a_n$ coefficients only the odd ones are equal to $0$ (see below). The functions $\cos(x)$ and $\cos(nx)$ are orthogonal in the interval $[-\pi,\pi]$, but $|\cos(x)|$ and $\cos(nx)$ are not. Since
\begin{equation*}
\left\vert \cos (x)\right\vert =\left\{ 
\begin{array}{c}
\cos (x) \\ 
-\cos (x)
\end{array}
\begin{array}{c}
\text{if} \\ 
\text{if}
\end{array}
\begin{array}{c}
0\leq x\leq \pi /2 \\ 
\pi /2\leq x\leq \pi, 
\end{array}
\right. \tag{2}
\end{equation*}
we have that
\begin{eqnarray*}
a_{n} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }\left\vert \cos (x)\right\vert\cos (nx)\,dx=\frac{2}{\pi }\int_{0}^{\pi }\left\vert \cos (x)\right\vert
\cos (nx)\,dx \\
&=&\frac{2}{\pi }\int_{0}^{\pi /2}\left\vert \cos (x)\right\vert \cos
(nx)\,dx+\frac{2}{\pi }\int_{\pi /2}^{\pi }\left\vert \cos (x)\right\vert
\cos (nx)\,dx \\
&=&\frac{2}{\pi }\int_{0}^{\pi /2}\cos (x)\cos (nx)\,dx-\frac{2}{\pi }
\int_{\pi /2}^{\pi }\cos (x)\cos (nx)\,dx. \\
a_{1} &=&\frac{2}{\pi }\int_{0}^{\pi /2}\cos ^{2}(x)\,dx-\frac{2}{\pi }\int_{\pi /2}^{\pi }\cos ^{2}(x)\,dx=0.
\end{eqnarray*}
Using the following trigonometric identity, with  $a=x,b=nx$, 
\begin{equation*}
\cos (a)\cos (b)=\frac{\cos (a+b)+\cos (a-b)}{2},\tag{3}
\end{equation*}
we find
\begin{eqnarray*}
a_{2m} &=&\frac{4}{\pi \left( 1-4m^{2}\right) }\cos (\frac{2m\pi }{2})=\frac{
4}{\pi \left( 1-4m^{2}\right) }(-1)^{m} \\
a_{2m+1} &=&\frac{4}{\pi ( 1-4(2m+1)^{2}) }\cos (\frac{(2m+1)\pi 
}{2})=0,\qquad  m=1,2,3,\ldots.\tag{4}
\end{eqnarray*}
The expansion of $\left\vert \cos (x)\right\vert $ into a trigonometric
Fourier series in the interval $[-\pi ,\pi ]$ is thus
\begin{equation*}
\left\vert \cos x\right\vert =\frac{a_{0}}{2}+\sum_{n=1}^{\infty }\left(
a_{n}\cos (nx)+b_{n}\sin (nx)\right) =\frac{2}{\pi }+\frac{4}{\pi }
\sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}\cos (2mx)\tag{5}
\end{equation*}

$$|\sin(x)|\ \text{(blue) and the partial sum  }\frac{2}{\pi }+\frac{4}{\pi }
\sum_{m=1}^{5 }\frac{(-1)^{m}}{1-4m^{2}}\cos (2mx) \ \text{(red) in }[-\pi,\pi]$$
Setting $x=0$ in $(5)$, we obtain
\begin{equation*}
1=\frac{2}{\pi }+\frac{4}{\pi }\sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}=\frac{2}{\pi }-\frac{4}{\pi }\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{1-4n^{2}}.\tag{6}
\end{equation*}
Hence
\begin{equation*}
\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{1-4n^{2}}=\frac{1}{2}-\frac{\pi }{4}.\tag{7}
\end{equation*}
