3
$\begingroup$

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?

  • 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 2

2

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 $\mathbb R$

1

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?