$f(x) = |\sin(x)| \quad \Rightarrow\quad f(x) = \left\{
\begin{array}{l l}
-\sin(x) & \quad \forall x \in [- \pi, 0\space]\\
\sin(x) & \quad \forall x \in [\space 0,\pi\space ]\\
\end{array} \right.$
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
Yes. Fourier series can help. It is equivalent to finding the limiting function of the Fourier series
$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^3}}}} $ .
Note that$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {nt} \right)}}{n}} $ converges to $(\pi - t)/2$ for $0 < t < 2 \pi $ . Observe that
${\mathop{\rm Im}\nolimits} Log(1 - {e^{i2t}}) = - 2\sum\limits_{n = 1}^\infty {\frac{{\sin (2nt)}}{{2n}}} $ .
We can deduce this from $Log(1 - {z^2}) = - 2\sum\limits_{n = 1}^\infty {\frac{{{z^{2n}}}}{{2n}}} $.
Hence $\sum\limits_{n = 1}^\infty {\frac{{\sin (2nt)}}{{2n}}} = - \frac{1}{2}{\mathop{\rm Im}\nolimits} Log(1 - {e^{i2t}}) = \frac{1}{2}\frac{1}{2}(\pi - 2t) = \frac{1}{4}(\pi - 2t)$ for $ 0 < t < \pi $.
Therefore, $\sum\limits_{n = 1}^\infty {\frac{{\sin ((2n - 1)t)}}{{2n - 1}}} = \sum\limits_{n = 1}^\infty {\frac{{\sin (nt)}}{n}} - \sum\limits_{n = 1}^\infty {\frac{{\sin (2nt)}}{{2n}}} = \frac{1}{2}(\pi - t) - \frac{1}{4}(\pi - 2t) = \frac{\pi }{4}$ for $0 < t < \pi$ . We may integrate the above Fourier series term by term to give the integral of the function on the right:
$ - \sum\limits_{n = 1}^\infty {\frac{{\cos \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^2}}}} + \sum\limits_{n = 1}^\infty {\frac{1}{{{{(2n - 1)}^2}}}} = \frac{\pi }{4}t$
I.e.,
$\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^2}}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{{{(2n - 1)}^2}}}} - \frac{\pi }{4}t = \frac{{{\pi ^2}}}{8} - \frac{\pi }{4}t$ .
Integrating again gives:
$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^3}}}} = \frac{{{\pi ^2}}}{8}t - \frac{\pi }{8}{t^2}$
Now for $x \ge 1$ , $ t = \pi/x \le \pi $.
Substituting this value of $t$ in the above equation gives:
$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {(2n - 1){\textstyle{\pi \over x}}} \right)}}{{{{(2n - 1)}^3}}}} = \frac{{{\pi ^3}}}{8}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)$ ,
which is equivalent to your equation for $x \ge 1$.