I'm not quite sure what you're asking...but...first, in the world of complex variables trigonometric functions and exponential functions all unify. Thus by moving to complexes we have a nice uniform way of dealing with sine and cosine.
If $c_n=a_n+ib_n$, then
$$ c_ne^{int}+c_{-n}e^{-int} = c_ne^{int}+\bar{c_{n}}e^{-int} =$$ $$ (a_n+ib_n)(\cos(nt)+i\sin(nt))+(a_n-ib_n)(\cos(nt)-i\sin(nt)) = 2a_n\cos(nt)-2b_n\sin(nt) $$
So dealing with trig functions or exponentials is just a matter of notation/taste.
Now to address your question as to what these series are for...
Recall that an analytic function is equal to its Taylor series:
$$ f(x) = \sum\limits_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k $$
If we stop this sum, say at $N$, we have
$$ f(x) \approx \sum\limits_{k=0}^N \frac{f^{(k)}(a)}{k!}(x-a)^k $$
which is a polynomial approximation of $f(x)$. Polynomials are nice. Taylor polynomials give us easily understood approximations of our function.
Now Fourier series do essentially the same thing for periodic functions. If we stop the summation at some $N$, we get a Fourier polynomial. This is a nice approximation made up of "easy" to understand trig functions.
From a theoretic viewpoint, Taylor series are wonderful because you can treat analytic functions sort of like polynomials. Fourier series allow one to treat nice periodic functions sort of like trig functions.
$f(x) = |\sin(x)| \quad \Rightarrow\quad f(x) = \left\{
\begin{array}{l l}
-\sin(x) & \quad \forall x \in [- \pi, 0\space]\\
\sin(x) & \quad \forall x \in [\space 0,\pi\space ]\\
\end{array} \right.$
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 $$\boxed {\displaystyle\sum^{\infty}_{n=1} \cfrac{(-1)^n}{(4n^2 -1)}= -\sum^{\infty}_{n=1} \cfrac{(-1)^n}{(1-4n^2)} = \cfrac {1}{4}(2-\pi)}$$
Best Answer
$$f(x)=|x| \sim \frac{\pi}{2} - \sum_{n=1}^\infty \frac{4}{\pi (2n-1)^2}\cos (2n-1) x.$$
Evaluating this at $x=0$:
$$\frac{\pi}{2} = \sum_{n=1}^\infty \frac{4}{\pi (2n-1)^2}$$
$$\sum_{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$$
$$\sum_{n=0}^\infty \frac{1}{(2n-1)^2}=1+\frac{\pi^2}{8}$$