For
$$f(t)=t,\quad t\in[0,2\pi) \tag{1}$$
we find the fourier series
$$\hat{\hat{f}}(x)=\pi -2 \sum_{k=1}^\infty \frac{\sin(kx)}{k}\tag{2}$$
I want to calculate the value of
$$\sum_{n=1}^\infty\frac{(-1)^n}{2n-1}\tag{3}$$
using (2).
Now for $x=\pi/2$ we get
$$\hat{\hat{f}}(\pi/2)=\pi -2 \sum_{k=1}^\infty \frac{\sin(\frac{\pi}{2}k)}{k}$=\pi -2 \sum_{k=1}^\infty \frac{(-1)^k}{k} \tag{3}$$
So that looks a bit like (3). I should now be able to somehow reform the sum or "choose" k s.t. I actualyl get (3). But I just don't see it.