How to prove $\lim\limits_{n \to \infty}\sin 2^n \neq 0$? We can prove $\{\sin n\}$ diverge.
We can also prove $\{\sin n^2\}$ diverge.
But can we prove $\{\sin 2^n\}$ diverge?
I tried the same methods like the former two, but failed. Can anyone help?
 A: The easiest way is to look at the sequence $2^n$ mod $2\pi$ since knowing a number mod $2\pi$ tells you its sine. Let $S$ be the set of limit points of the sequence $2^n$ mod $2\pi$ (looking at this as a circle). We will be able to see that, since $\pi$ is irrational (and, in particular, is not of the form $2^n/m$ for any integers $n$ and $m$), the sines fail to converge. We may immediately observe that $\sin(2^n)$ converges if and only if the image $\sin[S]$ consists of only a single point, since all points in $S$ are approached arbitrarily close infinitely often and, within any radius and after sufficiently long, the sequence will remain within that radius of an accumulation point.
First, note that $S$ is not empty because the circle is compact. Let $\theta\in S$ be any accumulation point. Note that $2\theta\in S$ as well. However, unless $\theta = 0$ or $\theta=\pi$ it is not the case that $\sin(\theta)$ and $\sin(2\theta)$ and $\sin(4\theta)$ are all equal. Thus, $S$ must be a subset of $\{0,\pi\}$. In any case, $0$ must be an accumulation point.
However, then we run into an issue: $0$ is not a power of $2$ mod $2\pi$ because $\pi$ is irrational. Thus, we find powers of $2$ which are arbitrarily close to $0$, but are not equal to $0$ - but this is a problem because, whenever have a $x$ close within $\pi/6$ of $0$, we can certainly find a number of the form $2^nx$ in the interval $[\pi/6,\pi/3]$ or $[-\pi/3,-\pi/6]$ just by doubling $x$ until we hit one of these intervals. Thus, powers in these intervals occur infinitely often, but then there must be an accumulation point therein, so the sequence fails to converge!
(Note: If $1/\pi$ is a normal number, as is conjectured, then $S$ is actually just the whole circle - and, what is more, $2^n$ is equidistributed mod $2\pi$)

Another, somewhat more high-powered way to finish the proof, is to note that if $z$ is irrational, then the binary expansion of $z$ contains at least $n+1$ distinct subsequences in $\{0,1\}^n$ infinitely often. You can apply this to $1/\pi$ to see that $S$ is actually infinite - and the proof of the lemma used here is a fun property to work out.
A: Note that $\sin(2^n)\neq 0$ and that $\lvert \sin(2^{n+1})\rvert \leq \lvert \sin(2^n) \rvert$ implies that $2 \lvert \cos(2^n) \rvert \leq 1$. This, in turn, implies $\lvert \sin(2^n) \rvert \geq \tfrac12 \sqrt3$. So either the sequence has a tail that is strictly increasing in norm or it must exceed $\tfrac12\sqrt3$ infinitely often. Either way, it cannot converge to $0$.
A: If $\lim_{n\rightarrow\infty}\sin(2^n)=0$ then for every $\epsilon\in\Bbb R\setminus\{0\}$ we should find a $N\in\Bbb N$ such that $$n\geq N\Rightarrow |\sin(2^n)|<\epsilon$$
Let, in particular, $\epsilon<1$ and $y=\log_2\left(2^{N}\cdot2\pi+\frac{\pi}{2}\right)$. Then we have both:


*

*$y>N$, since $y=\log_2(2^N\cdot 2\pi+\frac{\pi}{2})>\log_2(2^N\cdot 2\pi)>\log_2(2^N)=N$

*$\sin(2^y)=1$
