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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...2012-04-25
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    @Fabian the lim sup of any bounded sequence of real numbers exists.2012-04-25
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    @countinghaus: i don't see a limsup... (maybe there is some problem with the display of the page on my computer)2012-04-25
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    You might use the result [here](http://www.cs.berkeley.edu/~wkahan/Math55/p249n17.pdf) 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 $1/(n+1)$.2012-04-25
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    Interesting: the limsup is shown as lim on my computer???? Does somebody else with the mathjax setting svg have the same problem?2012-04-25
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    Is it not true that $\limsup \cos(an) = 1$ for all $a$, rational or irrational? And isn't that easier by far to prove than $\pi$ is irrational?2012-04-25
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    @Fabian, yes, there is a bug with `\limsup` and `\liminf` in SVG mode. It will be fixed in the net release of MathJax.2012-04-26

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

For every irrational $\alpha$, the set $\{a+b\alpha: a,b\in \mathbb{Z}\}$ is dense in $\mathbb R$

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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?