How can I compute this limit : $$\lim_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {\left\lvert \sin k\right\rvert} $$ Thank you.


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    $\begingroup$ What are your thoughts and approach? $\endgroup$ – Jaideep Khare Mar 22 '17 at 17:39
  • $\begingroup$ Think of which techniques you have learned in your course. Does any of those ring a bell? $\endgroup$ – mathreadler Mar 22 '17 at 17:43
  • $\begingroup$ I tried to apply Stolz theorem but i have not got any result. $\endgroup$ – Gustave Mar 22 '17 at 17:43
  • $\begingroup$ Equidistributed sequence, ergodic process, Monte-Carlo integration - did you see any of those terms in your course? Is it analysis or statistic course? $\endgroup$ – A.Γ. Mar 22 '17 at 18:40
  • $\begingroup$ No..it is just an analysis exercise. $\endgroup$ – Gustave Mar 22 '17 at 18:47

By Weyl's equidistribution theorem the sequence $\{e^{in}\}_{n\geq 0}$ is dense in the unit circle and much more: it is equidistributed. In particular, since $\sin(k)=\text{Im}\,e^{ik}$, the limit $\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\left|\sin(k)\right|$ is the average value of the function $\left|\sin(x)\right|$, i.e. $$ \frac{1}{\pi}\int_{0}^{\pi}\sin(x)\,dx = \color{red}{\frac{2}{\pi}}.$$

  • $\begingroup$ Thanks Jack D'Aurizio. Can we solve it without this theorem? $\endgroup$ – Gustave Mar 22 '17 at 21:20
  • $\begingroup$ @Gustav: I believe we cannot. As a matter of fact, we may build sequences that are dense in $[-1,1]$ such that $\lim_{n\to +\infty}\sum_{k=1}^{n}a_k$ does not exist or differs from the expected value given by the associated integral. $\endgroup$ – Jack D'Aurizio Mar 22 '17 at 21:33
  • $\begingroup$ However, we may prove the above result by exploiting the convergents of the continued fraction of $pi$, Lagrange's theorem and the Lipschitz-continuity of the sine function: $\left|\sin(k)\right|$ is an almost-periodic sequence with respect to the periods $7,106,113,33102,\ldots$. But that is not really different than going through the equidistribution theorem. $\endgroup$ – Jack D'Aurizio Mar 22 '17 at 21:37
  • $\begingroup$ +1. A 'naive' application of Stolz-Ces$\mathrm{\grave{a}}$ro fails !!!. It "is not regular enough to apply standard manipulations" as you pointed out in the above comments. $\endgroup$ – Felix Marin Mar 23 '17 at 6:26

Hint: $|\sin k|=\sin (k \bmod \pi)$ The values of $k \bmod \pi$ will bounce around in the interval $[0,\pi)$ so you are asked for the average value of $\sin x$ over this interval. What integral can you do to get this?

  • $\begingroup$ Really Ross Millikan, i didn't understand your question, can you aske it in more simple way? Thanks. $\endgroup$ – Gustave Mar 22 '17 at 18:22
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    $\begingroup$ Important to note this is guaranteed because integers are not fractions of any irrational (which $\pi$ is), if the k were a fraction of $\pi$ the answer might be different. $\endgroup$ – mathreadler Mar 22 '17 at 18:26
  • $\begingroup$ How would you find the average value of $\sin x$ over $[0,\pi)$? $\endgroup$ – Ross Millikan Mar 22 '17 at 18:33
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    $\begingroup$ Bouncing around or density is not enough: we need equidistribution, that is stronger than density. $\endgroup$ – Jack D'Aurizio Mar 22 '17 at 20:16
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    $\begingroup$ @mathreadler: what does it mean compute the density? If it stands for know the distribution, well, of course, a weighted integral solves the problem. $\endgroup$ – Jack D'Aurizio Mar 23 '17 at 14:13

EDIT as Jack points out below in comments I forgot the absolute value. I am sure this answer can be fixed by various means, but it would take some time, won't be pretty and probably add to any confusion already present. I should probably remove it and make a question of how to solve it using a similar technique.

Here is another approach, although probably not easier than the integral: use the addition formula

$$\sin(v+w) = \sin(v)\cos(w)+\cos(v)\sin(w)$$

a lot.

$$\sin(n+1) = \sin(n)\cos(1)+\cos(n)\sin(1)$$ $$\cos(n+1) = \cos(n)\cos(1)+\sin(n)\sin(1)$$

If you have learned to calculate recurrence relations with matrices for example a course in linear algebra, now is the time to put that to use.

  • $\begingroup$ Sorry, but I don't see the relation here. $\endgroup$ – Gustave Mar 22 '17 at 18:55
  • $\begingroup$ Maybe you can come back and look at it some time after you done some linear algebra. $\endgroup$ – mathreadler Mar 22 '17 at 19:02
  • $\begingroup$ Sorry mathreadler, can you give me a hint? $\endgroup$ – Gustave Mar 22 '17 at 19:04
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    $\begingroup$ How do you plan to deal with the involved absolute values? $\endgroup$ – Jack D'Aurizio Mar 22 '17 at 20:15
  • $\begingroup$ You are right @JackD'Aurizio and I can't find a way to patch it. Maybe in combination with Fourier series estimation of $|\sin(x)|$ could do it but that would be even more horribly overkill and not make for a very clean answer. $\endgroup$ – mathreadler Mar 23 '17 at 13:16

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