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?
What is $\limsup\limits_{n\to\infty} \cos (n)$, when $n$ is a natural number?
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real-analysis
analysis
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0I 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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1@Fabian the lim sup of any bounded sequence of real numbers exists. – 2012-04-25
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0@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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0You 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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0Interesting: 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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0Is 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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0@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
2 Answers
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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?