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I’m attemping to solve this problem in a Mathematical Analysis workbook.

Prove:$$\boldsymbol{ {\left| \sum_{k=1}^{n} \frac{\sin kx}{k} \right|} \leq \operatorname{Si}(\pi) = {\int_0^{\pi} \frac{\sin x}{x} \,\mathrm dx} \approx 1.85}$$

Some ideas:

  • Directly compute the derivative of $\displaystyle \sum_{k=1}^{n} \frac{\sin kx}{k} $, the extrema points can be written analytically. In fact, they are $\dfrac{2m\pi}{n}$ and $\dfrac{(2m+1)\pi}{n+1}$, where $m$ is an arbitrary integer. From this perspective, the upper boundary is evidently the optimal one. And some tools from number theory can be used (for example, when $n$ is a prime number, $\{k/n|k\}$ forms a Complete residual lineage of $n$). But there are tough situations hard to treat.
  • Because the derivative of $\displaystyle \sum_{k=1}^{n} \frac{\sin kx}{k}$ is a cos series, which can be written in a simple form, we can evaluate $\displaystyle \sum_{k=1}^{n} \frac{\sin kx}{k}$ by the cos series’ integral. But it seems hard to cope with either.
  • Notice that the L.H.S. is a partial sum of the Fourier series of a linear function.
Rócherz
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Marian Xu
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  • What is $x$? Is it any arbitrarily fixed real number? – Surajit Mar 16 '25 at 12:23
  • Of course. Otherwise the inequality doesn't make sense. – Marian Xu Mar 16 '25 at 12:53
  • Which inequality are you talking about? What does $\mathrm{Si}(\pi)$ mean? Is it a function of $x$? – Surajit Mar 16 '25 at 13:25
  • It's only a notation of int_0^{x} frac{sinx}{x} – Marian Xu Mar 16 '25 at 15:33
  • You can use this derivation to get $\sum_{k=1}^{n} \frac{\sin kx}{k} = \text{Si}\left( \frac{(2n+1)x}{2} \right)$. Taking $|\cdot|$ gives $\left| \sum_{k=1}^{n} \frac{\sin kx}{k} \right| = \left| \text{Si}\left( \frac{(2n+1)x}{2} \right) \right|$ and using the fact that $|\text{Si}|$ is max at $\pi$ implies $\left| \sum_{k=1}^{n} \frac{\sin kx}{k} \right| = \left| \text{Si}\left( \frac{(2n+1)x}{2} \right) \right| \leq | \text{Si}( \pi ) | = \text{Si}( \pi )$. – The Art Of Repetition Mar 17 '25 at 17:53
  • Are you sure the form is right? – Marian Xu Mar 18 '25 at 04:41

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