Different proofs that $\lim_{n\to\infty}\sin n$ does not exist

Question

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.

Answer 1

$$ \left|\sin(n+1)-\sin(n-1)\right| = 2\sin(1)\left|\cos(n)\right| $$ 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.

Answer 2

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.

Answer 3

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 $\sin n_i\in A$.

Answer 4

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.

Answer 5

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.