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I am trying to show that $$\lim_{n\to \infty}\sin 2^n$$ diverges for $n \in \mathbb N$

I could show that assuming the limit converges, say to $L$ then

$$L=\lim_{n\to \infty}\sin 2^{n+1}$$ $$=2\lim_{n\to \infty}\sin 2^n\lim_{n\to \infty}\cos 2^n$$ $$=2L\lim_{n\to \infty}\cos 2^n$$

It cannot be $$\lim_{n\to \infty}\cos 2^n=\frac{1}{2}$$ since it implies $$\frac{1}{2}=\lim_{n\to \infty}\cos 2^{n+1}$$ $$=2(\lim_{n\to \infty}\cos 2^n)^2-1$$ $$=-\frac{1}{2}$$ So either $$\lim_{n\to \infty}\sin 2^n=0$$ or it diverges.

For the sequence to converge, necessarily it must be that $2^n$ gets arbitrarily close to $m\pi$ for some integer $m(n)$ as $n$ goes to infinity, which seems counterintuitive. But I couldn't prove it, so anyone has some good idea?

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    Suppose $0 < \lvert \sin 2^n\rvert < \frac{1}{2}$. Then what follows about $\lvert \sin 2^{n+1}\rvert$? – 2017-01-12
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    @Ian Miller$2^{-n}$ is never a multiple of $\pi$ and never equal zero, but it would converge to 0. – 2017-01-12

4 Answers 4

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One can see with the double angle formula that

$$\sin2^{n+1}=2\sin2^n\cos2^n=\pm2\sin2^n\sqrt{1-\sin^22^n}\approx\pm2\sin2^n$$

So if $\sin2^n$ gets close to $0$, $\sin2^{n+1}$ will double and get farther from $0$. Indeed, for if the limit to exist, the double angle theorem says it must be $\sqrt3/2$, which is a contradiction to your statement.

$$L=2L\sqrt{1-L^2}$$

edit:

To be more specific, if $0<|\sin2^n|<\frac12$, then it follows from the above that

$$\sqrt3|\sin2^n|<|\sin2^{n+1}|<2|\sin2^n|$$

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    L=0 satisfies your equation so It doesn't seem to disprove $\lim_{n\to \infty}\sin 2^n=0$ . – 2017-01-12
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    @user406323 but it is repelling. As hard as you try to get close, the double angle formula pushes it away from $0$. – 2017-01-12
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    So if $\lvert \sin 2^n\rvert$ is smaller then a certain number say $\frac{1}{2}$ then you get $\lvert \sin 2^{n+1}\rvert > \sqrt3 \lvert \sin 2^n\rvert$ which means it 'repels' until its absolute value is greater than $\frac{1}{2}$ again. – 2017-01-12
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    @user406323 Yes, that is basically how you do it. :D – 2017-01-12
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    @user406323 In case you couldn't figure out the bounds, I placed them in my edit. – 2017-01-12
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    Assuming that for some $0<\epsilon<1$ there exists $N\in \mathbb N$ such that $n\ge N$ implies $\lvert \sin 2^n\rvert<\epsilon$, you get $\lvert \sin 2^{n+k}\rvert>(2\sqrt{1-{\epsilon}^2})^k\lvert \sin 2^n\rvert>1$ for some big $k$, which is a contradiction. Therefore $\lim_{n\to \infty}\sin 2^n$ does not converge to 0 and diverges. Thanks, got it solved. – 2017-01-12
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    @user406323 Ok, glad to help! :D And just as the bell rings for me to leave the computer :-P – 2017-01-12
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    @user406323: Your inequality doesn't hold if $\sin2^N=0$. That's the nontrivial case. – 2017-01-12
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    @user406323 yes, you must first prove that we don't have $\sin2^n=0$ for natural $n$, but that is trivial. – 2017-01-12
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    @SimpleArt: If it is trivial that $\sin2^n\neq0$ for $n\in\Bbb N$, then please give a proof of it. – 2017-01-12
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    @JohnBentin [Niven's theorem](https://en.m.wikipedia.org/wiki/Niven's_theorem). – 2017-01-12
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    I think that you are referring to Niven's proof of Lambert's theorem, which can be found on the Wikipedia page for the proof that $\pi$ is irrational. However, neither Niven's nor any of the other proofs given there could be called trivial. – 2017-01-12
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    @JohnBentin I don't see how that is within the context of the problem anymore though. I think if someone wanted a proof of that, that it should be in a different question, and that referencing such a fact should be reasonable. – 2017-01-12
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If, for some $a$, $$ \lim\limits_{n \to \infty} \sin 2^n =\sin a $$ Introduce $$ b := \pi/2 -a $$ Suppose, with both $0 <|\epsilon_1| <|\epsilon$ and $0 <|\epsilon_2| < |\epsilon|$, $$ 2^n =\pm a \pm 2b +\epsilon_1 \quad (\mathrm{mod}\; 2\pi) \\ 2^{n+1} =\pm a \pm 2b +\epsilon_2 \quad (\mathrm{mod}\; 2\pi) $$ (They may be different numbers, and not necessarily same sign, so there are 4 signs.) Then by your construction, if we subtract them, either: $$ 2^n =\pm 2a \pm 2b +2|\epsilon_3|\quad (\mathrm{mod}\; 2\pi) \\ $$ where $0 <|\epsilon_3| <|\epsilon$. Make $\epsilon$ so small that this is absurd, unless $a =0$ or $a =\pi/2$.

For $a=\pi/2$, we know $2^n$ is $2k\pi \pm \pi/2 +\epsilon_4$. But then $2^{n+1} =4k\pi +\pi +2\epsilon_4$ (or minor case, which is similar). Make $\epsilon$ so small that this is absurd.

It remains to deal with $a=0$. Here, similarly, $2^n =k\pi +\epsilon_5$. If $\epsilon_5 \neq 0$, then times 2 so many times, that $\epsilon_5$ exceeds $\epsilon$.

That $2^n =k\pi$, for some $k$, is outrageous.

(A bit hard to write very clearly, but draw a graph and you will get it. If someone finds a better way to put it, edit by all means.)

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    Why is it outrageous that $2^n$ could be $k\pi$ for some $k\in\Bbb N$ ? – 2017-01-12
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    Because $\pi$ is irrational ~ – 2017-01-12
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    But the limit of a rational number can be irrational. – 2017-01-12
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    In that line, I deal with the remaining case that exactly $2^n =k\pi$, which cannot be. (Arguably, my approach is too tedious) – 2017-01-12
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    @SimpleArt .He is NOT saying that $\pi$ is not the limit of a sequence of rationals whose numerators are powers of $2.$ He is saying that $2^n\ne k \pi$ for any integers $n,k.$ – 2017-01-12
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    The proof that $\pi$ is irrational is way beyond the stage of analysis that the OP's exercise is at. The task here is to show the weaker result that $\pi$ is not of the form $2^m/(2n+1)$, for any integers $m$ and $n$. – 2017-01-12
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    Sorry, do I need to complete my sketch very thoroughly > – 2017-01-12
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$\pi\not\in \mathbb Q.$ Let $2^n=k_n\pi +d_n$ where $k_n\in \mathbb Z$ and $0<|d_n|< \pi /2.$

Suppose $\lim_{n\to \infty}\sin 2^n=0.$ Then $\lim_{n\to \infty} |d_n|=0.$

If $n\geq n_0\implies |d_n|<\pi /8,$ consider $j\in \mathbb N$ such that $2^j|d_{n_0}|<\pi /2<2^{j+1}|d_{n_0}|.$ Then $$2^{j-1}|d_{n_0}|=(1/2)\cdot 2^j|d_{n_0}|<\pi /4$$ so we have$$|d_{n_0+j-1}|=2^{j-1}|d_{n_0}|.$$

Then $$\pi /8<(1/4)\cdot 2^{j+1}|d_{n_0}|=2^{j-1}|d_{n_0}|=|d_{n_0+j-1}|<\pi/8$$ a contradiction.

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    Why is $\pi\not\in\Bbb Q$ ? – 2017-01-12
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    @JohnBentin Because it is transcendental :-) – 2017-01-12
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    But the point is to provide a proof at the level of the OP's exercise, not to deduce the result as an easy consequence of an advanced result. The fact that $\pi$ is not an odd fraction of a power of $2$ is all that's needed; and this may be a lot easier to show than the irrationaity of $\pi$ (or, a fortiori, its transcendence). – 2017-01-12
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Hint:

$$2^n=2\pi\frac{2^n}{2\pi}$$ so that the quadrant of the angle $2^n$ is given by the first two bits of the fractional part of $\dfrac{2^n}{2\pi}$, which are nothing but the $n^{th}$ and $n+1^{th}$ bits of the fractional part of $\dfrac1{2\pi}$, a transcendental number.

If you can show that all bit pairs occur infinitely many times, you are done.