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This question already has an answer here:

In this question it was proved that limit $$ \lim_{x\to\infty}\sin x $$ doesn't exists. What about $$ \lim_{n\to\infty}\sin n? $$ I asking about usual limit, where $n$ is integer. I know that this limit doesn't exists, I want to see different proves. My first idea was to use Kronecker's approximation theorem, but it's not very suited for such simple question.

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marked as duplicate by Najib Idrissi, user147263, Henning Makholm, Community Jul 16 '15 at 3:33

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  • $\begingroup$ use exactly the same idea . $\endgroup$ – Dave Nguyen Jul 15 '15 at 16:31
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    $\begingroup$ @DaveNguyen, ha-ha. I cannot use $2\pi n$ and $\pi/2$. And $\pi$ is irrational $\endgroup$ – Michael Galuza Jul 15 '15 at 16:32
  • $\begingroup$ It is utterly unnecessary to make a new question asking for new proofs when there's already a question about this fact. Questions can receive multiple answers on this site. $\endgroup$ – Najib Idrissi Jul 15 '15 at 17:45
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$$ \left|\sin(n+1)-\sin(n-1)\right| = 2\sin(1)\left|\cos(n)\right| $$ hence $\{\sin n\}_{n\in\mathbb{N}}$ cannot be a Cauchy sequence, since $\cos(n)$ is bounded away from zero for infinitely many values of $n$. As an alternative, since $\sin n=\text{Im}(e^{in})$ and the length of circumference of radius one is $2\pi\approx 6$, among: $$\sin(n),\quad \sin(n+1),\quad\ldots\quad \sin(n+7)$$ there are two numbers whose difference is at least one, so, again, $\{\sin n\}_{n\in\mathbb{N}}$ cannot be a Cauchy sequence.

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    $\begingroup$ Second part of your answer is very-very elegant and clear, thanks. $\endgroup$ – Michael Galuza Jul 15 '15 at 17:02
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That doesn't exist either.

No matter how large $n$ is, there will always be a larger $n'$ such that $n'$ is between $2\pi k-1$ and $2\pi k$ for some $k\in\mathbb N$. Thus $\sin n' <0 $ infinitely often, which means that a limit, if one exists, cannot be positive.

But then $n'+2$ is between $2\pi k+1$ and $2\pi k+2$, and therefore $\sin (n'+2) > \sin 1 \approx 0.84$. So the limit, if it exist, cannot be less than $\sin 1$. Since every real number is either positive or less than $\sin 1$ (and some are both), there is no number that can possibly be the limit.

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Expanding on the idea of Kronecker's approximation theorem: The map $T(x) = x + 1$ is uniquely ergodic on $[0, 2\pi]$, so $f(x) = \sin x$ has \begin{align*} \frac{1}{N}\sum_{n\leq N} \chi_A (\sin T^n 0) = \frac{1}{N} \sum_{n\leq N} \chi_A (\sin n) \to \operatorname{vol}(A) \end{align*} as $N\to\infty$ for any measurable $A\subset [0, 2\pi]$, where $\chi_A$ is the characteristic function for $A$. In particular, for any $A$ of positive measure, there exists an infinite sequence $n_1, n_2, \dots$ such that $\sin n_i\in A$.

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  • $\begingroup$ Very cool and very heavy for such problem) I wanted to see exactly such proofs. It's sad that we cannot use algebra for proof. Or...?:) $\endgroup$ – Michael Galuza Jul 15 '15 at 16:54
  • $\begingroup$ Oh, sure. But you said above that you were looking for different proofs of the result. You might be able to put something together using the various addition formulas for $\sin$, but ergodicity is the underlying "reason" why the limit doesn't exist (and, in fact, fails to exist as spectacularly as possible: the limits of convergent subsequences fill [-1, 1]). The same sort of reasoning applies to any function with irrational period. $\endgroup$ – anomaly Jul 15 '15 at 16:57
  • $\begingroup$ I thought about it, and I feel that ergodicity is a "reason", but I should to prove Kronecker's theorem first (it's h/w for first year students). And, may be exists other "underlying reasons") $\endgroup$ – Michael Galuza Jul 15 '15 at 17:00
  • $\begingroup$ About the other "underlying reasons": Of course, that's why I put 'reason' in quotes. :) $\endgroup$ – anomaly Jul 16 '15 at 4:13
  • $\begingroup$ Could you give the source on the result you prove. I mean the convergence for $\sum \chi_{A}(T^n f(x))$ $\endgroup$ – Ice sea Oct 9 '17 at 11:41
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If $|\sin(n+1) - \sin(n)\,|$ is very small, then either $n \approx 2k\pi + \frac\pi2 - \frac12$ or $n \approx 2k\pi - \frac\pi2 - \frac12$, where $k$ is an integer. Then either $\sin(n+1) \approx \sin(n) \approx 0.878$ or $\sin(n+1) \approx \sin(n) \approx -0.878$.

Using the same value of $n$ for the moment, consider $|\sin(n+2) - \sin(n+1)\,|$. We have $n + 2 \approx 2k\pi + \frac\pi2 + \frac32$ or $n +2 \approx 2k\pi - \frac\pi2 + \frac32$, so either $\sin(n+2) \approx 0.071$ or $\sin(n+2) \approx -0.071$, and $|\sin(n+2) - \sin(n+1)\,| \approx 0.807.$

That is, whenever you have two successive terms that are close together, the next term is significantly different. The sequence therefore cannot converge.

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  • $\begingroup$ It's Henning Makholm's words and idea $\endgroup$ – Michael Galuza Jul 15 '15 at 17:16
  • $\begingroup$ @MichaelGaluza They are related but not the same. Henning shows the value is negative infinitely often; I don't. Henning uses $\sin(1)$, I use $\sin\left(\frac\pi2-\frac12\right)$. But if you are saying Henning's answer is better, I have no argument with you: I think I rather prefer his answer too, and would not have written a different one if he had posted before I started writing mine. $\endgroup$ – David K Jul 15 '15 at 19:21
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I got another idea on how to prove that $\lim_{n\to\infty}\sin(n)$ doesn't exists, I'll do this via contradiction.

So first let's assume the limit exists, $$ \lim_{n\to\infty}\sin(n)=a\Rightarrow\lim_{n\to\infty}\sin(2n)=a\in[-1,1] $$ Now we use the representation $$ \sin(2x)=2\sin(x)\cos(x)\Rightarrow \lim_{n\to\infty}\cos(n)=\frac{1}{2}=\lim_{n\to\infty}\cos(2n) $$ Now we use another representation $$ \cos(2x)={\cos(x)}^2-{\sin(x)}^2\Leftrightarrow\cos(2n)-{\cos(n)}^2=-{\sin(n)}^2 $$ And now we take the limit and we get, since all limits exist and everything is continuous $$ \lim_{n\to\infty}(\cos(2n)-{\cos(n)}^2)=\lim_{n\to\infty}(-{\sin(n)}^2)\\ \Leftrightarrow\ (\frac{1}{2}-\frac{1}{4})=\frac{1}{4}=-a^2\Leftrightarrow a^2=-\frac{1}{4} $$ which is a contradiction.

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  • $\begingroup$ @DavidK the existence of $\lim_{n\to\infty}\cos(n)$ is not an assumption but a an implication since both sequences are linked through the representation $\cos(n)=\frac{\sin(2n)}{2\sin(n)}$. Or am I mistaken? $\endgroup$ – user190080 Jul 15 '15 at 19:47
  • $\begingroup$ On further thought, I think you're right about that. $\endgroup$ – David K Jul 15 '15 at 19:49

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