I am having trouble calculating the following limit: $$\lim_{n \to \infty} \sqrt[n]{|\sin n|}\ .$$
Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$
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12Right away, sire! – 2011-06-27
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3@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. – 2011-06-27
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0Do you know that the limit exists? – 2011-06-27
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0I removed the [number-theory] tag, but both [calculus] and [analysis] still seem a bit redundant. Someone should look into that. – 2011-06-27
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0@Arturo: why don't you think the limit tag is appropriate here? – 2011-06-27
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3@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). – 2011-06-27
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0@Arturo: I see the wisdom in that. Funny enough, I only did it because I was already adding the limit tag, and I thought of leading by example. ;p – 2011-06-27
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0This is a Hard Problem, isn't it? – 2011-06-27
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0@TonyK: This problem and Diophantine approximation theory are related. I never studied the Diophantine approximation theory, so to ask next. – 2011-06-27
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0@Tao: Is it really homework? – 2011-06-27
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0@Tao: At least tell us where did you find it? In which book, or forum, or course notes? – 2011-06-27
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0This looks like a sophisticated troll! First, ask a deep question (possibly unanswered?); then accept the first answer that is not obviously wrong. I think we should keep an eye on this Tao Hacker person! – 2011-06-27
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0@Beni Bogosel: If you understand Chinese, you can see [link](http://www.douban.com/group/topic/20269517/). In fact, I have asked on [link](http://mathoverflow.net/questions/68934/calculate-lim-n-to-infty-sqrtn-sin-n-closed). Sorry, I am a Chinese. My English is not very good. – 2011-06-27
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1@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. – 2011-06-27
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0@Corel: Well, I did say a *sophisticated* troll...But in any case, your updated answer has proved me wrong! – 2011-06-27
3 Answers
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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0Sorry if I am missing something, but are you claming $\liminf \sin n \gt 0$? – 2011-06-27
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0I 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. – 2011-06-27
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0@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. – 2011-06-27
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0@Corey: Could you give more details please? How do you go from that inequality to the $\sin $ inequality? – 2011-06-27
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0@gary: What do you do with the subsequence $\sin n_k$ which converges to $0$? Indeed, $1$ is the limsup of the sequence considered, but we are looking for the limit (if it exists) – 2011-06-27
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0I'll turn my hint into a full solution. I haven't worked out the full solution yet so it might take me a minute. – 2011-06-27
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0Yes, this looks good indeed. – 2011-06-27
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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2Or any subsequence so that $|\sin n_k|$ is bounded away from $0$ would do. – 2011-06-27
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1The point is, does the limit exist? – 2011-06-27
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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0Depends on whether he means $n$ to be a real or $n$ is a natural number. – 2011-06-27
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0Why was my comment removed? Deleted answers, delete the comments attached. I think I get it :) – 2011-06-27
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0I 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? – 2011-06-27
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3@mixedmath: I think the use of the variable $n$ is meant to make you think of limits along $\mathbb{N}$. – 2011-06-27
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0I believe that $n$ is an integer in the question. Elseway, its trivial, as you point it. – 2011-06-27
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0$mixedmath: I commented on Chaz's answer. Sorry about that. I thought he changed the text of its answer after seeing it is wrong, and someone deleted my comment. – 2011-06-27
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0@Beni: I deleted my answer. – 2011-06-27
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0@Chaz: I understand. :) – 2011-06-27
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0Incidentally, Mathematica seems to get this wrong: Limit[Abs[Sin[x]]^(1/x),x->Infinity] returns 1. (And Wolfram Alpha gives the same answer). It does seem that Limit[ ..., x->Infinity] takes limits along $\mathbb{R}_+$ rather than $\mathbb{N}$ since Limit[Sin[2 Pi x], x->Infinity] gives Interval[{-1,1}]. – 2011-06-27
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0@mac:Let $F=\sqrt[x]{|\sin x|}$, if $x=2k\pi$, $F=0$. If $x=(2k+1/2)\pi$, $F=1$. So $\lim_{x \to \infty} \sqrt[x]{|\sin x|}$ does not exist. – 2011-06-27
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0@Tao: yeah, I know. But Mathematica gets this wrong, which is amusing/disconcerting/unusual. – 2011-06-27
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0@mac:I am a Chinese, my English is not very good. If you understand Chinese, you can see [link](http://www.douban.com/group/topic/20269517/). Thank you. – 2011-06-27
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0@mac: If Mathematica is getting it wrong, it is because it is computing the limit numerically. Since $m\{x\in [n\pi,(n+1)\pi] | \sqrt[x]{|\sin x|} < 1-\epsilon\}\rightarrow 0$ as $n\rightarrow \infty$, it is easy to mistake this limit for 1. – 2011-06-27
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0Corey: hm, I thought Mathematica was meant to find limits symbolically, or give up. For example, it fails to evaluate Limit[If[x - Floor[x] < 1/x, 0, 1], x -> Infinity] even though $m\{x\in [n,n+1]\colon f(x)<1\}\to 0$ as $n\to \infty$ for this function $f$. I believe the incorrect evaluation of the limit of $|\sin x|^{1/x}$ is a bug. – 2011-06-27
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1I don't see how this answers the question at all. – 2011-06-27