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.

$f(x) = |\sin(x)| \quad \Rightarrow\quad f(x) = \left\{ 
  \begin{array}{l l}
    -\sin(x) & \quad \forall x \in [- \pi, 0\space]\\
   \sin(x) & \quad \forall x \in [\space 0,\pi\space ]\\
  \end{array} \right.$
The Fourier coefficients associated are 
$$a_n= \frac{1}{\pi}\int_{-\pi}^\pi f(x) \cos(nx)\, dx = \frac{1}{\pi} \left[\int_{-\pi}^0 -\sin (x) \cos(nx)\, dx + \int_{0}^\pi \sin(x) \cos(nx)\, dx\right], \quad n \ge 0$$
$$b_n= \frac{1}{\pi}\int_{-\pi}^\pi f(x) \sin(nx)\, dx = \frac{1}{\pi} \left[\int_{-\pi}^0 -\sin (x) \sin(nx)\, dx + \int_{0}^\pi \sin(x) \sin(nx)\, dx\right], \quad n \ge 1$$
All functions are integrable so we can go on and compute the expressions for $a_n$ and $b_n$.
$$a_n  = \cfrac{2 (\cos(\pi n)+1)}{\pi(1-n^2)}$$
$$b_n = 0$$
The $b_n = 0$ can be deemed obvious since the function $f(x) = |\sin(x)|$ is an even function. and $a_n$ could have been calculated as $\displaystyle a_n= \frac{2}{\pi}\int_{0}^\pi f(x) \cos(nx)\, dx $ only because the function is even.
The Fourier Series is $$\cfrac {a_0}{2} + \sum^{\infty}_{n=1}\left [ a_n \cos(nx) + b_n \sin (nx) \right ]$$
$$= \cfrac {2}{\pi}\left ( 1 + \sum^{\infty}_{n=1} \cfrac{(\cos(\pi n)+1)}{(1-n^2)}\cos(nx)\right )$$
$$= \cfrac {2}{\pi}\left ( 1 + \sum^{\infty}_{n=1} \cfrac{((-1)^n+1)}{(1-n^2)}\cos(nx)\right )$$
$$= \cfrac {2}{\pi}\left ( 1 + \sum^{\infty}_{n=1} \cfrac{2}{(1-4n^2)}\cos(2nx)\right )$$
Since for an odd $n$, $((-1)^n+1) = 0$ and for an even $n$, $((-1)^n+1) = 2$
At this point we can't just assume the function is equal to its Fourier Series, it has to satisfy certain conditions. See Convergence of Fourier series.
Without wasting time, (you still have to prove that it satisfies those conditions) we assume the Fourier Series converges to our function i.e
$$f(x) = |\sin(x)| = \cfrac {2}{\pi}\left ( 1 + \sum^{\infty}_{n=1} \cfrac{2}{(1-4n^2)}\cos(2nx)\right )$$
Note that $x=0$ gives $\cos(2nx) = 1$ then 
$$f(0) = |\sin(0)| = \cfrac {2}{\pi}\left ( 1 + 2\sum^{\infty}_{n=1} \cfrac{1}{(1-4n^2)}\right ) =0$$ which implies that
$$\sum^{\infty}_{n=1} \cfrac{1}{(1-4n^2)} = \cfrac {-1}{2}$$ and $$\boxed {\displaystyle\sum^{\infty}_{n=1} \cfrac{1}{(4n^2 -1)}= -\sum^{\infty}_{n=1} \cfrac{1}{(1-4n^2)} = \cfrac {1}{2}}$$
Observe again that when $x = \cfrac \pi 2$, $\cos (2nx) = cos(n \pi) = (-1)^n$, thus
$$f \left (\cfrac \pi 2 \right) = \left |\sin \left (\cfrac \pi 2\right )\right | = \cfrac {2}{\pi}\left ( 1 + 2\sum^{\infty}_{n=1} \cfrac{(-1)^n}{(1-4n^2)}\right ) =1$$ which implies that
$$\sum^{\infty}_{n=1} \cfrac{(-1)^n}{(1-4n^2)} = \cfrac {1}{4}(\pi -2)$$ and $$\boxed {\displaystyle\sum^{\infty}_{n=1} \cfrac{(-1)^n}{(4n^2 -1)}= -\sum^{\infty}_{n=1} \cfrac{(-1)^n}{(1-4n^2)} = \cfrac {1}{4}(2-\pi)}$$
