Fourier cosine series and sum help I have been having some problems with the following problem:
Find the Fourier cosine series of the function $\vert\sin x\vert$ in the interval $(-\pi, \pi)$. Use it to find the sums
$$ \sum_{n\: =\: 1}^{\infty}\:\ \frac{1}{4n^2-1}$$  and $$ \sum_{n\: =\: 1}^{\infty}\:\ \frac{(-1)^n}{4n^2-1}$$
Any help is appreciated, thank you.
edit:I have gotten as far as working out the Fourier cosine series using the equations for cosine series 
$$\phi (X) = 1/2 A_0 +  \sum_{n\: =\: 1}^{\infty}\:\ A_n \cos\left(\frac{n\pi x}{l}\right)$$ and 
$$A_m = \frac{2}{l} \int_{0}^{l} \phi (X) \cos\left(\frac{m\pi x}{l}\right) dx $$
I have found $$A_0 = \frac{4}{l}$$ but the rest of the question is a mess on my end and then I don't know how to relate the rest of it back to those sums.

There is an alternative way to calculating the Fourier series as follows. The Fourier series for a periodic function $f(x)$ of period $2\pi$ (written $f(x) \sim \sum a_ne^{inx}$) is the sum
$$\sum a_n e^{inx}$$
where the Fourier coefficients $a_n$ are defined by $a_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx} dx$. Now in your case, you can split the integral up into
$$\frac{1}{2\pi} \left(\int_{-\pi}^0 -\sin x e^{-inx} dx + \int_0^\pi \sin x e^{-inx} dx\right).$$
Now recall that $\sin x = \frac{e^{inx} - e^{-inx}}{2i}$ and chuck this into the integral to get the Fourier coefficient $a_n$ and hence you will obtain your Fourier series. You should get (I'm not spoiling anything because you can just chuck it into mathematica) that
$$f(x) \sim \sum_{n \in \Bbb{Z}} \frac{(1 + (-1)^n)e^{inx}}{\pi (1- n^2)} $$ Now to complete your problem in evaluating that sum, I suggest the following: The fourier series of $f(x)$ is absolutely convergent so we can say that 
$$f(x) = \sum_{n \in \Bbb{Z}} \frac{(1 + (-1)^n)e^{inx}}{\pi (1- n^2)}. $$
For $n$ odd, the sum is zero so actually you can write
$$f(x) = \sum_{n \in \Bbb{Z}}  \frac{2e^{2inx}}{\pi (1- (2n)^2)}.$$
Now put $x = \pi$ so that $f(\pi) =0.$ Then we get that
$$\begin{eqnarray*} 0 = \sum_{n \in \Bbb{Z}}  \frac{2e^{2in\pi}}{\pi (1- (2n)^2)} &=& \sum_{n\in\Bbb{Z}} \frac{2}{\pi (1- (2n)^2)} \\
&=&\frac{2}{\pi} + \sum_{n\in\Bbb{Z}, n \geq 1}\frac{4}{\pi (1- (2n)^2)} \\
\implies \sum_{n\in\Bbb{Z}, n \geq 1}\frac{1}{4n^2 - 1} &=& \frac{2}{\pi}\cdot \frac{\pi}{4}\\
&=& \frac{1}{2}.
\end{eqnarray*}$$
The evaluation of the second sum is similar except that instead of substituting in $x = \pi$ you substitute in $x = \pi/2$.
