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The question is pretty self explanatory, I'm studying Fourier Series with the book Mathematical Methods for Physicists written by Arfken and it does not explain that.

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    @IvanLerner Please consider accepting the answer below if it resolved your question satisfactorily.2013-11-27

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As mentioned in the comments, study the argument here carefully. I'll add a little play-by-play commentary to guide you.

Basically, the steps for the canonical example $f(x)=\operatorname{sign}(x)$ on $-\pi are:

  1. Compute the indicated Fourier series and consider $F_N(x)$, the $N$th partial Fourier sum of $f$.
  2. We want to find the value of the first positive local max of $F_N(x)$. Do this by standard calculus followed by some trigonometry to sum the resulting cosine series in terms of a single sine function.
  3. From there, it's easy to find the critical number of interest; call this $x^*$.
  4. Evaluating $F_N(x^*)$, the resulting sum can be massaged a bit and then recognized as a Riemann sum.
  5. The Riemann sum is one corresponding to ${1\over \pi}\int_0^\pi {\sin x\over x}\,dx$.
  6. Since $N$ is finite, we have $F_N(x^*)\approx {1\over \pi}\int_0^\pi {\sin x\over x}\,dx\approx 0.589\dots$.
  7. This last value is the value of that first positive local max and thus the overshoot above the true value at $x^*$ (which is of course 1/2) is about $0.0589-0.5=0.089$, i.e., about 8.9% of the unit jump here.

I hope that sketch is helpful.