Convergence of $\sin(nx)$

Question

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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?

Question and title don't match. – Arturo Magidin Jun 17 '12 at 02:30 — Arturo Magidin, Jun 17 '12 at 02:30

score 2 · Answer 1 · answered Jun 17 '12 at 01:47

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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 $[-1,1]$.

answered Jun 17 '12 at 01:47

Gerry Myerson

179,216

2

Well, yes... But I was stumped on the proof. – BSNL_IN Jun 17 '12 at 01:55
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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. – Gerry Myerson Jun 17 '12 at 02:02
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 $[0,1]$. – Gerry Myerson Jun 17 '12 at 02:04
@GerryMyerson and ${n\alpha}$ is uniformly distributed between $0$ and $1$, as $n\to\infty$. – Yai0Phah Jun 17 '12 at 08:05
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@Frank, I know that, but it's overkill for this problem. – Gerry Myerson Jun 17 '12 at 12:24

score 2 · Answer 2 · answered Jun 17 '12 at 06:47

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.

Convergence of $\sin(nx)$

2 Answers2