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Possible Duplicate:
Sine function dense in $[-1,1]$

Does there exist a subsequence $n_k$ where $1\leq k < \infty $ of the sequence of natural numbers, such that the sequence $\sin n_k$ is convergent?

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    Question and title don't match.2012-06-17

2 Answers 2

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Yes, in fact, given any $x$, $-1\le x\le1$, there's a subsequence such that $\sin n_k$ converges to $x$. In other words, $\sin n$ is dense in $[-1,1]$.

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    Well, yes... But I was stumped on the proof.2012-06-17
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    Then maybe you should edit your question so that it asks what you really want to ask. Anyway, I imagine a web search for "sine" and "dense" will turn up proofs galore.2012-06-17
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    Hint: if $\alpha$ is irrational, then $\{{n\alpha\}}$, the set of fractional parts of $n\alpha$, as $n$ runs through the natural numbers, is dense in $[0,1]$.2012-06-17
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    @GerryMyerson and $\{n\alpha\}$ is uniformly distributed between $0$ and $1$, as $n\to\infty$.2012-06-17
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    @Frank, I know that, but it's overkill for this problem.2012-06-17
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Every bounded real sequence has a convergent subsequence.

This is known as Bolzano-Weierstrass theorem.

This answers your original question. However, as you can see from other answers and comments, about this particular sequence you can show even more than that.