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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.

xotix
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    in $(3)$ (the last one, you have two equations numbered $(3)$), when $k$ is even, then $k/2$ is an integer, hence $\sin(\pi k/2) = 0$, thus you are left with only odd $k$-s which is what is needed in $(3)$. – Hayk Jan 31 '19 at 14:40
  • Yeah I got that but you just helped me see that I still have the $(-1)'^n$ since the sinus is not odd. Thanks think I can solve it now. – xotix Jan 31 '19 at 15:13

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