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  • $\begingroup$ I remembered now from class, that at point of discontinuity of the function, the series converges to the average of the function at that point. But not sure this helps with your function, since at zero it blows up. But see Fouriercoscoefficient-for-generalized-function-with-singularity for related discuassion $\endgroup$ Commented Mar 1, 2024 at 23:45
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    $\begingroup$ I'm currently teaching EE261, The Fourier transform and its applications at Stanford (using Mathematica) and can confirm that such series converge to the average of limits on either side of a discontinuity. The square wave function is the clearest case. $\endgroup$ Commented Mar 2, 2024 at 0:24
  • $\begingroup$ @Nasser Hi, the "Fourier series coefficient" of $f(x)=(2-2\cos{x})^{-1}$ is $\tilde{f}_n = -|n|/2$. This is not a conventional Fourier series because $\sum_{n=-\infty}^{\infty} |n|\exp{\mathrm{i}nx}$ is nowhere convergent. The proper way is to understand it as the Fourier transformation between a distribution on $\mathbb{R}/2\pi\mathbb{Z}$ and a distribution on $\mathbb{Z}$. I'm surprised that Mathematica can give results in this vastly generalized sense. $\endgroup$ Commented Mar 2, 2024 at 0:28
  • $\begingroup$ @Nasser: The link refers to this actual page. What was your intention? $\endgroup$ Commented Mar 2, 2024 at 9:45
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    $\begingroup$ @user64494: Can you stop editing OP question if you do not understand the topic? Leave the title as it was typed by OP. $\endgroup$ Commented Mar 3, 2024 at 8:50