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Here is my question: Prove there is a sequence of increasing positive integers $n_i$ such that the limit of $\sin(n_i)$ exists as $i \to \infty $. The problem with this question is... I don't know how to start it! Doesn't the sine function fluctuate between -1 and 1? How would a limit exist if it keeps going back and forth? Thank you for any help you give me! :)

jimjim
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Mary A.
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4 Answers4

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Take $a_n = \sin (n)$ then since $[-1,1] \supset \bigcup_{n=0}^{\infty} a_n$ and $[0,1]$ is compact these points have an accumulation point in $[-1,1]$.

Hence you have found $a_{k_n} \rightarrow A$, to find an increasing sequence of indeces use the fact when you have $\phi :\mathbb{N}\rightarrow \mathbb{N}$ then also $a_{\phi(k_n)}\rightarrow A$.

clark
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  • You mean $[-1,1]$. Nice proof, anyhow! – Pedro Dec 08 '13 at 23:57
  • @PedroTamaroff yes indeed, thank you! – clark Dec 08 '13 at 23:58
  • Oh, so use sequential compactness, then? Thank you! – Mary A. Dec 09 '13 at 00:05
  • @MaryA. Yes sequential compactness will give you $a_{k_n}$ converging to a point in $[-1,1]$. Now you only have to rearrange the $k_n$ so they are increasing. And any such rearrangement gives the same limit. – clark Dec 09 '13 at 00:09
  • @MaryA. here see a link proving this fact by a very intuitive answer of qiaochu youan (it does not matter that the limit is zero) http://math.stackexchange.com/questions/185778/rearrangement-of-sequences-with-limit-0 – clark Dec 09 '13 at 00:19
  • All this was posted a long time ago but, first, the fact that $[0,1]$ is compact is not relevant (but the fact that $[-1,1]$ is compact is) and, second, the fact that $a_{k_n}\to A$ does not imply that $a_{\phi(k_n)}\to A$ (but it implies that $a_{k_{\phi(n)}}\to A$). – Did May 03 '17 at 07:03
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A consequence of Dirichlet's approximation theorem is that there exists infinitely many positive integers $n_k$ such that $\left|\pi-\dfrac{p_k}{n_k}\right|\leqslant\dfrac1{n_k^2}$ for some integer $p_k$. Then $|n_k\pi-p_k|\leqslant\dfrac1{n_k}$ hence $|\sin(p_k)|\leqslant\dfrac1{n_k}$. In particular, $\sin(p_k)\to0$ when $k\to\infty$.

Did
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The sequence whose $n$-th term is $\sin n$ is a bounded (by $\pm1$) sequence of real numbers. By the Bolzano-Weierstrass Theorem, it has a convergent subsequence. (You should certainly make your peace with this theorem before worrying about more advanced things like Dirichlet's Theorem or Thue-Roth-Siegel.)

Andreas Blass
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Hint: in the Reals, continuity is equivalent to sequential continuity, i.e., if $x_n \rightarrow x $ , then $f(x_n) \rightarrow f(x) $ . And $\sin x$ is continuous.

Edit: What I meant is we can use a sequence to approximate a multiple of $\pi$, and then use sequential continuity. This follows from Roth-Thue-Siegel. Equivalently, by Weil's equidistribution theorem http://en.wikipedia.org/wiki/Equidistribution_theorem on {$n\pi$}, the sequence $n\pi$ is equidistributed $mod1$ , so that it will be indefinitely-close to an integer infinitely often, say at $n_1,n_2,....$. Take these terms $n_i$ of the sequence that are within $<\epsilon$ of an integer for the actual sequence $n_i$. Then $n_i -k\pi <\epsilon$ , so that $Sinn_i \rightarrow Sink\pi=0$.

This is the specific argument, which apparently I have to provide, but others do not: the sequence $n \pi; n=1,2,3,...$ considered $Mod1$ , meaning we consider $nx$ minus the greates integer less than $nx$ , i.e., the decimal part of $nx$. These decimal parts are dense in $[0,1]$. This means, in particular, that for any $\epsilon/n$ , there is an integer $m$ with $m\pi-k < \epsilon $ for some integer $k$. Since $n\pi$ is dense ($Mod1$ in $[0,1]$) , we can find an integer $m_2>m$ , so that $|m_2\pi -k_2| <\epsilon/2$. The same can be done for any $\epsilon/n$ , i.e., we can find an increasing sequence of integers $m_i$ , so that $|m_ipi -k_i < \epsilon/I $ . The sequence $k_i$ is indefinitely -close to integer multiples of $\pi$. Then $sink_i \rightarrow sin\pi=0$.

user99680
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