I want to use Fourier Series to prove this formula: $$ \frac{\pi}{4}=\sum\limits_{n=1}^\infty\frac{\sin{(2n-1)x}}{2n-1} \quad\quad \text{if}\quad 0<x<\pi$$ To do this, my thought was to let $f(x)=x$ over $[0,\pi]$ and then use an odd periodic extension to define it over the interval $[0,2\pi]$. Doing this, and calculating the coefficients, I found: $$ a_n=0 \quad\quad\quad b_n=-\frac{2}{n}\cos{n\pi}=(-1)^{n+1}\frac{2}{n} $$ Since $f(x)$ is a polynomial and our interval is bounded, we know that $f$ is of bounded variation and so by Jordan's Theorem we know that we have convergence of our series to our function. So I arrive at: $$ x = 2\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}\sin{nx}}{n} $$ I thought that from here I could just let $x=\frac{\pi}{4}$ and the formula would pop right out but I'm not so sure because then all my $\sin$ terms would actually be evaluated. I also thought maybe I overcomplicated things and could just find the Fourier Series of $f(x) = \frac{\pi}{4}$ but that would just be itself. The other thing I thought to do was maybe to extend it as an even function over the interval $[0,2\pi]$ instead, but I don't really see why I would do this when the formula I'm looking to derive only involves signs.
Any hints would be much appreciated!