Determining Fourier series for $\lvert \sin{x}\rvert$ for building sums

My math problem is a bit more tricky than it sounds in the caption. I have the following Task (which i in fact do not understand):

"Determine the Fourier series for $f(x)=\lvert \sin{x}\rvert$ in order to build the Sum for the series: $\frac{1}{1*3}+\frac{1}{3*5}+\frac{1}{5*7}+\dots$"

My approach: first, calculating the Fourier series. There is no period, Intervall or Point given. i think it must turn out to be something like this: $a_{n} = \frac{2}{\pi} \int_{0}^{\pi} |\sin x|\cos nx dx$ but what would be the next step to build the series?

and second: the other given series. I think its all about the uneven numbers, so i have in mind 2n-1 and 2n+1 are the two possible definitions. So it could be something like this:

$(\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots) = \sum\limits_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}$

But i cannot make the connection between this series, its sum and |sin x|.

despite i think the sum should be something around $\sum\limits_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}$ (but i cannot proof yet)

P.S.: i know the other Convergence of Fourier series for $|\sin{x}|$ -question here at stackexchange, but i think it doesn't fit into my problem. despite i don't understand their Explanation and there is no way shown, how to determine the solution by self.
You were on the right track. First, calculate the Fourier series of $f(x)$ (you can leave out the magnitude signs because $\sin x \ge 0$ for $0\le x \le \pi$ ): $$a_n=\frac{2}{\pi}\int_{0}^{\pi}\sin x\cos nx\;dx= \left\{ \begin{array}{l}-\frac{4}{\pi}\frac{1}{(n+1)(n-1)},\quad n \text{ even}\\ 0,\quad n \text{ odd} \end{array}\right .$$ For $a_0$ we get $a_0=\frac{2}{\pi}$. Therefore, the Fourier series is $$f(x) = \frac{2}{\pi}-\frac{4}{\pi}\sum_{n \text{ even}}^{\infty}\frac{1}{(n+1)(n-1)}\cos nx=\\ =\frac{2}{\pi}\left ( 1-2\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n-1)}\cos 2nx \right )\tag{1}$$
Now we can evaluate the series. Set $x=\pi$, then we have $f(\pi)=0$ and $\cos 2n\pi = 1$. Evaluating (1) with $x=\pi$ we get $$0 = 1 - 2\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n-1)}$$ which gives your desired result.
• You might note that $f$ is Lipschitz at $x=\pi$ so that the series converges to $f(\pi)$. – copper.hat May 10 '13 at 19:18