# proving existence of a sequence such that the limit exists?

Can anyone prove the existence of a sequence $(n_{k})_{k\in \mathbb{N}}$ of distinct positive integers such that the limit: $\lim_{k\rightarrow \infty }\sin(n_{k})$ exists in $\mathbb{R}$

I can definitely construct a sequence $(n_{k})_{k\in \mathbb{N}}$ such that $\frac{1}{2}\leq \sin(n_{k})\leq 1$, but this doesn't imply that this sequence is convergent. Any suggestions?

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the question as phrased doesn't make much sense: suppose $a_n \to a$, then $\sin(a_n) \to \sin(a)$. – Kris Jul 19 '12 at 17:35
I assume that the $n_k$ themselves have to be divergent or some similar restriction, otherwise as Kris said this is trivial. – Robert Mastragostino Jul 19 '12 at 17:40
@Kris: I fixed the typo. the terms of the sequence are positive integers – C. Lambda Jul 19 '12 at 17:53
I think one can prove that the set $\{\exp(in)\}_{n=1}^\infty$ is dense on the unit circle. This would imply the existance of such a sequence. Or, you can just copy the proof from here: math.stackexchange.com/questions/4764/… – Per Alexandersson Jul 19 '12 at 18:08
Yes, see en.wikipedia.org/wiki/Irrational_rotation – Per Alexandersson Jul 19 '12 at 18:09

Every bounded sequence has convergent subsequence, see Bolzano-Weierstrass theorem.

If you apply this to the sequence $(\sin n)_{n=0}^\infty$, you get the desired result.

(Or you can mimic the proof of Bolzano-Weierstrass theorem, if you prefer.)

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I turn my comment into an answer: It follows directly from http://en.wikipedia.org/wiki/Irrational_rotation

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As @MartinSleziak noted, existence of such a sequence is easy. Slightly less obvious is how to construct the sequence more-or-less explicitly. Suppose $p$ and $q$ be positive integers such that $\left| \dfrac{p}{q} - \pi\right| < \dfrac{\epsilon}{q}$. Then $|p - q \pi| < \epsilon$ so $|\sin(p)| < \epsilon$. So we could use a sequence of good rational approximations of $\pi$. For example, we could take $p_k/q_k$ to be the convergents of the continued fraction of $\pi$, which have $|p_k/q_k - \pi| < 1/q_k^2$, and thus $|\sin(p_k)| < 1/q_k$.

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A couple of things to think about:

1) Let $\phi: \mathbb{R} \to \mathbb{R} / 2\pi\mathbb{Z}$ be the projection map. Then $\phi(\mathbb{N})$ is dense in $\mathbb{R} / 2\pi\mathbb{Z}$.

2) Let $n_k$ be an increasing sequence of (distinct) positive integers defined by picking $n_k$ such that $n_k$ $($ mod $2\pi ) < \frac{1}{k}$.

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