# Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$

I am having trouble calculating the following limit: $$\lim_{n \to \infty} \sqrt[n]{|\sin n|}\ .$$

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Right away, sire! – The Chaz 2.0 Jun 27 '11 at 18:22
@Tao: You might not know this as a new user, but questions in an imperative tone are strongly discouraged. Please ask the question and elaborate on what you have already tried, whether this is homework or not, etc. – InterestedGuest Jun 27 '11 at 18:25
Do you know that the limit exists? – mac Jun 27 '11 at 18:28
@mixedmath: I think [limit] is redundant, and subsumed with [calculus]. But that may just be me. By the way: it's probably best if instead of putting words like "I'm having trouble" into the OPs mouth, you allowed him to try to edit and rephrase the problem himself, once it has been politely pointed out that posting in the imperative is frowned upon (as was done here). – Arturo Magidin Jun 27 '11 at 18:43
@TonyK: Asking argumentative questions about 0/0 or .9999...=1 would make somebody a troll. I would agree that the question was poorly motivated, but it is a challenging and interesting problem. – Corey Jun 27 '11 at 21:42

Hint:

$\pi$ is not a Liouville number, so there exists $m\in\mathbb{N}$ such that for all $p,q\in\mathbb{Z}$ with $q>1$, we have $$\left| \pi - \frac{p}{q}\right| \geq \frac{1}{q^m}.$$ This should allow you to keep $\sin n$ away from 0.

Edit: Full Solution:

Let $m$ be as above. So for all $p,q\in\mathbb{Z}$ with $q>1$ we have $$\left| \pi - \frac{p}{q}\right| \geq \frac{1}{q^m}.$$ Now take $n\in\mathbb{N}$. Take $q_n$ so that $|q_n\pi-n|$ is minimized. Then we have $$\frac{\pi}{2} \geq \left| q_n\pi - n\right| \geq \frac{1}{q_n^{m-1}}.$$ Next we note that $|\sin n| = |\sin(q_n\pi-n)|$. Now since $\sin$ is increasing on $[0,\pi/2]$ we have $$|\sin(q_n\pi-n)|\geq \sin\frac{1}{q_n^{m-1}} \geq \frac{1}{2}\frac{1}{q_n^{m-1}}.$$ Such an estimate holds for each $n$, with $q_n\approx \frac{n}{\pi}$. So now we have $$\frac{1}{2}\frac{1}{q_n^{m-1}} \leq |\sin n| \leq 1.$$ Now take $n$th roots of everything and let $n\rightarrow\infty$. The LHS goes to 1 so $$\sqrt[n]{|\sin n|} \rightarrow 1.$$

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Sorry if I am missing something, but are you claming $\liminf \sin n \gt 0$? – Aryabhata Jun 27 '11 at 19:58
I don't think this is the claim. Rather, you want to use something like $|\sin n| \ge \tfrac12\mathrm{dist}(n,\pi\mathbb{Z})$ with the inequality Corey gave to get a positive lower bound on $|\sin n|^{1/n}$. Although I haven't worked this out in detail myself. – mac Jun 27 '11 at 20:04
@Aryabhata: That is not what I am claiming (as that statement is not true). Instead this claim can be used to give a bound of the type $\left|\sin n \right| \geq \frac{K}{n^m}$. Then after taking $n$th roots and letting $n\rightarrow \infty$ you have the result. – Corey Jun 27 '11 at 20:15
@Corey: Could you give more details please? How do you go from that inequality to the $\sin$ inequality? – Beni Bogosel Jun 27 '11 at 20:15
+1. Looks good to me. – TonyK Jun 27 '11 at 21:43

If the limit exists, then it should be equal to 1, because $|\sin n|$ is dense in $[0,1]$ and there exists a subsequence $|\sin n_k|$ converging to $1$. Then

$$\sqrt[n_k]{|\sin n_k|} \to 1.$$

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Or any subsequence so that $|\sin n_k|$ is bounded away from $0$ would do. – mac Jun 27 '11 at 19:22
The point is, does the limit exist? – TonyK Jun 27 '11 at 20:06

HINT:

There are arbitrarily large multiples of $\pi$, and there are arbitrarily large odd multiples of $\pi /2$.

Of course, this only happens if we consider real n as opposed to natural n...

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Depends on whether he means $n$ to be a real or $n$ is a natural number. – Thomas Andrews Jun 27 '11 at 18:29
Why was my comment removed? Deleted answers, delete the comments attached. I think I get it :) – Beni Bogosel Jun 27 '11 at 18:30
I am interpreting it as the standard limit, which I interpret to mean for all real numbers. It does matter. Do you think that limits in intro classes are restricted to integers? – mixedmath Jun 27 '11 at 18:33
@mixedmath: I think the use of the variable $n$ is meant to make you think of limits along $\mathbb{N}$. – mac Jun 27 '11 at 18:34
I don't see how this answers the question at all. – TonyK Jun 27 '11 at 20:38