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I think the answer should be $1$, but am having some difficulties proving it. I can't seem to show that, for any $\delta$ and $n > m$, $|n - k(2\pi)| < \delta$. Is there another approach to this or is there something I'm missing?

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I can understand that you have some difficulty proving your assessment. Maybe it is wrong and the limit does in fact not exist... – Fabian Apr 25 '12 at 18:18
@Fabian the lim sup of any bounded sequence of real numbers exists. – user29743 Apr 25 '12 at 18:22
@countinghaus: i don't see a limsup... (maybe there is some problem with the display of the page on my computer) – Fabian Apr 25 '12 at 18:24
You might use the result here that given an irrational $x$ and a positive integer $n$, there exists at least one positive integer $j\le n$ for which $jx$ and the integer $m$ nearest to $jx$ differ in magnitude by less than $1/(n+1)$. – David Mitra Apr 25 '12 at 18:24
Interesting: the limsup is shown as lim on my computer???? Does somebody else with the mathjax setting svg have the same problem? – Fabian Apr 25 '12 at 18:25
up vote 2 down vote accepted

No, that's exactly how you should show it. You can get what you want by using this question:

how to show this is a dense set?

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You are on the right track. If $|n-2\pi k|<\delta$ then $|\frac{n}{k}-2\pi|<\frac \delta k$. So $\frac{n}{k}$ must be a "good" approximation for $2\pi$ to even have a chance.

Then it depends on what you know about rational approximations of irrational numbers. Do you know about continued fractions?

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