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We know that any sequence on $S^1$ must have a converging extracted subsequence, as $S^1$ is compact. Now, consider the sequence $a_n=(\cos(n),\sin(n))$. Could you find explicitly a subset of the natural numbers such that the corresponding subsequence converges? I don't even know whether it is possible to work it out, or whether there exists a nice representation of the solution. I thought about it and I found out a proof, but later I noticed that it actually contains an error. I'd prefer not to write here the proof, at least in this first time, since I could influence your reasoning. Thank you in advance,


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up vote 5 down vote accepted

Consider the continued fraction for $2\pi$. Let $\frac{h_n}{k_n}$ be the $n^{th}$ convergent. Then the numbers $h_n$ will give you a convergent subsequence which converges to $(1,0)$.

Since $$|2\pi -\frac{h_n}{k_n}|\leq \frac{1}{k_n k_{n+1}},$$ (see this theorem) it follows that $$|2k_n \pi -h_n|\leq \frac{1}{k_n+1}.$$ Since $k_{n+1}\rightarrow \infty$, and both $\sin$ and $\cos$ are continuous, it follows that $$\lim_{n\rightarrow \infty} (\cos (h_n), \sin (h_n))=(1,0).$$

Hope that helps,

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Let me see if I get you: you would finish the proof with $$\lim_{n\to\infty}\cos(h_n)=\lim_{n\to\infty}\cos(|2\pi k_n -h_n|)=\cos(\lim_{n\to\infty}|2\pi k_n -h_n|)$$ and then proving that $\lim_{n\to\infty}|2\pi k_n -h_n|=0$ as $$|2\pi k_n -h_n|\le\frac{1}{k_{n+1}}$$. I am sorry about my huge ignorance about continued fractions and maths in general but... why should $k_n$ tend to infinity? Thanks a lot again, bye – fatoddsun Jun 1 '11 at 8:13
@fatoddsun: I suggest taking at look at: The short answer to why $k_n\rightarrow \infty$ is because $k_n <k_n+1$ and they are integers. A more precise answer is that $k_n$ satisfies the recurrence relation $$k_n =a_nk_{n-1}+k_{n-2}$$ for all $n$, where $a_n\geq 1$. (When we write a continued fraction down, we write it in the short form $[a_0;a_1,a_2,a_3,\cdots $ which refers to $$a_0 +\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\cdots}}}.$$ The coefficient $a_0$ is the only one that can be zero for an irrational) – Eric Naslund Jun 2 '11 at 17:26
Thank you very much indeed! – fatoddsun Jun 5 '11 at 9:56

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