I'm struggling with a question on Fourier series. I need to expand the function which is $0$ for all $0<x<\pi/2$ and $\sin x$ for all $\pi/2 < x < \pi$ into a series of sines.
I'm trying to tackle the problem as follows:
First, I compute the integral $$b_n=\frac{2}{\pi}\int_0^{\pi}f(x)sin(nx)dx=0+\int_{\pi/2}^{\pi}\sin x \sin nx \, dx,$$which evaluates to $$\frac{2}{\pi} \frac{\cos\left(\frac{n\pi}{2}\right)}{n^2-1}.$$
For odd $n$ the result would be $0$; for even $n$, the result will be $\pm1$. So by substituting $k=2n$, we get rid of the odd/even problem and $\cos(k\pi)$ is now no more than $(-1)^k$: $$\frac{2}{\pi} \frac{(-1)^k}{(2k)^2-1}.$$ I assume we could now write our function as follows: $$f\sim\frac{2}{\pi}\sum_{k=1}^{\infty} \frac{(-1)^k}{(2k)^2-1}\sin(kx).$$
The following question is to verify that $$\sum_{j=0}^{\infty}\frac{(-1)^j(2j+1)}{4(2j+1)^2-1}=\frac{\pi}{8\sqrt{2}}$$ using our Fourier series expansion of $f$. I've been trying lots of different techniques, but none eventually result in proving the identity. For one, I don't quite recognize how the factor $(2j+1)$ might be brought into the series expansion. Any help would be extremely welcome!