I'm interested in finding an elementary proof for the following sum inequality: $$\sum_{k=1}^n \frac{\sin k}{k} \le \pi-1$$

If this inequality is easy to prove, then one may easily prove that the sum is bounded.


marked as duplicate by Guy Fsone, Rohan, JonMark Perry, Claude Leibovici, José Carlos Santos Dec 26 '17 at 22:48

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    $\begingroup$ We can write $1+\sum_{k=1}^n\frac{\sin k}k$ as $\int_0^1\sum_{k=1}^n\cos(kx)dx$. $\endgroup$ – Davide Giraudo Jun 23 '12 at 8:27
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    $\begingroup$ Possibly related: math.stackexchange.com/questions/13490/… $\endgroup$ – qoqosz Jun 23 '12 at 8:36
  • $\begingroup$ What is elementary? $\endgroup$ – Norbert Jun 23 '12 at 8:45
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    $\begingroup$ Abel summation? $\endgroup$ – zy_ Jun 23 '12 at 9:13
  • $\begingroup$ @Norbert: i thought of a trigonometric/geometrical approach. $\endgroup$ – user 1357113 Jun 23 '12 at 9:35

An approach of Abel Summation:


$$S_n=\sum_{i=1}^{n}\sin k,\quad S_0=0.$$


$$\sum_{k=1}^{n}\frac{\sin k}{k}=\sum_{k=1}^{n}\frac{S_k-S_{k-1}}{k}=\frac{S_n}{n}+\sum_{k=1}^{n-1}\frac{S_k}{k(k+1)}.$$

We have

$$S_n = \sum_{k=1}^{n}\sin k = \mathrm{Im}\left(\sum_{k=1}^{n}e^{ik}\right) = \mathrm{Im}\left(e^{i}\frac{1-e^{in}}{1-e^i}\right).$$

Hence $$|S_n| \leq \left|\frac{1-e^{in}}{1-e^i}\right| \leq \left|\frac{2}{1-e^i}\right| \approx 2.09.$$


$$\left|\sum_{k=1}^{n}\frac{\sin k}{k}\right| = \left|\frac{S_n}{n} + \sum_{k=1}^{n-1}\frac{S_k}{k(k+1)}\right| \leq 2.09 \left(\frac{1}{n} + \sum_{k=1}^{n-1}\frac{1}{k(k+1)}\right) = 2.09,$$

a sharper bound than $\pi-1$.

  • $\begingroup$ @y zhao: you applied here a nice trick. A good lesson to learn for me. Thanks! $\endgroup$ – user 1357113 Jun 24 '12 at 14:01

Let's first observe that $\sum_{k=1}^\infty u^k/k=-\ln(1-u)$.

If we're concerned about the convergence radius, we can always replace $u$ with $ue^{-\epsilon}$ and let $\epsilon\rightarrow0$. The branch of $\ln$ we're using is the one defined on $\mathbb{C}\setminus(-\infty,0]$: i.e. $\ln(re^{i\theta})=\ln r+i\theta$ where $r>0$ and $\theta\in(-\pi,\pi)$.

Inserting $\sin x=(e^{ix}-e^{-ix})/2i$, we get $$\sum_{k=1}^\infty \frac{\sin kx}{k} =\sum_{k=1}^\infty \frac{e^{ikx}-e^{-ikx}}{2ki} =\frac{\ln(1-e^{-ix})-\ln(1-e^{ix})}{2i} $$ At this point, I have two alternative solutions. In either case, I assume $x\in[0,\pi)$ to help stay within the selected branch of the logarithm.

You can look at the triangle with corners $O=0$, $I=1$ and $A=1-e^{-ix}$: this has $IO=IA$ and $\angle OIA=x$, so $\angle AOI=\frac{\pi-x}{2}$. This makes the imaginary part of $\ln(1-e^{-ix})=\angle AOI=\frac{\pi-x}{2}$; for $\ln(1-e^{ix})$ it is $-\frac{\pi-x}{2}$. The real part of the logarithm cancels out, and what remains is $\frac{\pi-x}{2}$.

Alternatively, while ensuring we stay within the branch of the logarithm, we get $$\sum_{k=1}^\infty \frac{\sin kx}{k} =\frac{1}{2i}\ln\frac{1-e^{-ix}}{1-e^{ix}} =\frac{\ln(-e^{-ix})}{2i} =\frac{\ln(e^{i(\pi-x)})}{2i} =\frac{\pi-x}{2}. $$

Thus, not only is the sum less than $\pi-1$. It is exactly $\frac{\pi-1}{2}$. And the more general sum $$\sum_{k=1}^\infty \frac{\sin kx}{k} =\frac{\pi-x}{2} $$ for $x\in[0,\pi]$: if $x=\pi$, the sum becomes zero (either by limit or because all the terms are zero).

  • $\begingroup$ wow. Another awesome proof! $\endgroup$ – user 1357113 Aug 17 '12 at 6:21
  • $\begingroup$ Only if $x=\pi$ the sum is 0, it doesn't diverge. $\endgroup$ – J.R. Oct 21 '12 at 9:38
  • $\begingroup$ @IHaveAStupidQuestion: Of course, you're right! All the terms of the sum are zero, so the sum becomes zero. Don't know what I was thinking. Will correct the answer. $\endgroup$ – Einar Rødland Oct 21 '12 at 12:32

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