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

Martin Sleziak
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Tao Hacker
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    Right away, sire! – The Chaz 2.0 Jun 27 '11 at 18:22
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    @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
  • I removed the [number-theory] tag, but both [calculus] and [analysis] still seem a bit redundant. Someone should look into that. – Asaf Karagila Jun 27 '11 at 18:31
  • @Arturo: why don't you think the limit tag is appropriate here? – davidlowryduda Jun 27 '11 at 18:41
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    @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
  • @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 – davidlowryduda Jun 27 '11 at 18:45
  • This is a Hard Problem, isn't it? – TonyK Jun 27 '11 at 19:42
  • @TonyK: This problem and Diophantine approximation theory are related. I never studied the Diophantine approximation theory, so to ask next. – Tao Hacker Jun 27 '11 at 19:47
  • @Tao: Is it really homework? – TonyK Jun 27 '11 at 20:00
  • @Tao: At least tell us where did you find it? In which book, or forum, or course notes? – Beni Bogosel Jun 27 '11 at 20:19
  • This 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! – TonyK Jun 27 '11 at 20:26
  • @Beni Bogosel: If you understand Chinese, you can see link. In fact, I have asked on link. Sorry, I am a Chinese. My English is not very good. – Tao Hacker Jun 27 '11 at 20:37
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    @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
  • @Corel: Well, I did say a sophisticated troll...But in any case, your updated answer has proved me wrong! – TonyK Jun 27 '11 at 21:45

3 Answers3

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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 $$\sqrt[n]{|\sin n|} \rightarrow 1.$$

Corey
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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
  • @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) – Beni Bogosel Jun 27 '11 at 20:32
  • I'll turn my hint into a full solution. I haven't worked out the full solution yet so it might take me a minute. – Corey Jun 27 '11 at 20:57
  • Yes, this looks good indeed. – Aryabhata Jun 27 '11 at 22:55
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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

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

Beni Bogosel
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0

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...

davidlowryduda
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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? – davidlowryduda Jun 27 '11 at 18:33
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    @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 believe that $n$ is an integer in the question. Elseway, its trivial, as you point it. – Beni Bogosel Jun 27 '11 at 18:35
  • $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. – Beni Bogosel Jun 27 '11 at 18:36
  • @Beni: I deleted my answer. – The Chaz 2.0 Jun 27 '11 at 19:01
  • @Chaz: I understand. :) – Beni Bogosel Jun 27 '11 at 19:06
  • Incidentally, 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}]. – mac Jun 27 '11 at 19:06
  • @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. – Tao Hacker Jun 27 '11 at 19:16
  • @Tao: yeah, I know. But Mathematica gets this wrong, which is amusing/disconcerting/unusual. – mac Jun 27 '11 at 19:18
  • @mac:I am a Chinese, my English is not very good. If you understand Chinese, you can see link. Thank you. – Tao Hacker Jun 27 '11 at 19:22
  • @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. – Corey Jun 27 '11 at 19:25
  • Corey: 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. – mac Jun 27 '11 at 19:34
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    I don't see how this answers the question at all. – TonyK Jun 27 '11 at 20:38