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So, we're given a function $f(x) = \begin{cases} 2, &-\pi < x\le 0 \\ 6, &0 < x\le\pi \end{cases}$, while $f(x+2π) = f(x)$ for any $x\in\Bbb R$.

Now, I've calculated the Fourier series of $f(x)$, which is $$F = 2+\sum_{n=0}^\infty\left(\frac{8\sin(\frac{πn}{2})}{πn}\cos(nx)+\frac{4(1-\cos(\frac{πn}{2}))}{πn}\sin(nx)\right)$$

We're then supposed to use this Fourier series $F$, to calculate this sum $\sum_{k=0}^∞ \frac{(-1)^k}{2k+1}$.

I've read some other replies here regarding the same subject, but I just cannot understand how to do this. Can someone help me understand what I'm supposed to do?

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    $\begingroup$ The DC term is usually the average which is $4$ here. So, without checking the details, something is off. The denominators should be of the form $2n+1$. $\endgroup$ – copper.hat Apr 7 at 7:44
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    $\begingroup$ I got the Fourier series as$$f(x)=4+\frac4\pi\sum_{n=1}^\infty\frac{1-\cos(n\pi)}n\sin(nx)$$Since $1-\cos(n\pi)=0$ for even $n$, we can also write$$f(x)=4+\frac8\pi\sum_{n=1}^\infty\frac{\sin((2n-1)x)}{2n-1}$$ $\endgroup$ – Shubham Johri Apr 7 at 7:47
  • $\begingroup$ @copper.hat $1-\cos(n\pi)=0$ for even $n$ $\endgroup$ – Shubham Johri Apr 7 at 7:49
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    $\begingroup$ @ShubhamJohri: Seems like an answer to me with the fact that the series converges pointwise except at transition points. $\endgroup$ – copper.hat Apr 7 at 7:52
  • $\begingroup$ Thanks for the corrections, will need to recheck my calculations. But what about using the Fourier series to find the ∑ sum? $\endgroup$ – sdds Apr 7 at 7:54
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I got the Fourier series as$$f(x)=4+\frac4\pi\sum_{n=1}^\infty\frac{1-\cos(n\pi)}n\sin(nx)$$Since $1-\cos(n\pi)=0$ for even $n$, we can also write$$f(x)=4+\frac8\pi\sum_{n=0}^\infty\frac{\sin\big((2n+1)x\big)}{2n+1}$$$x=\pi/2$ is a point of continuity, so the series converges to $f(\pi/2)=6$ at the point.$$\implies6=4+\frac8\pi\sum_{n=0}^\infty\frac{\sin\big((2n+1)\pi/2\big)}{2n+1}\\\implies\frac\pi4=\sum_{n=0}^\infty\frac{(-1)^{n}}{2n+1}$$

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