Suppose $\theta >0$ is irrational. Is it true that for $n\in\mathbb N$ $$ \cos 2\pi n\theta > 1-\varepsilon \tag{1} $$ holds infinitely often?
The question is inspired by a series candidate (posted by another user) whose terms are of the form $(\exp(i2\pi n\theta)-1)^{-1}$. My intuition says the terms are norm-wise large infinitely often (for irrational $\theta$) so the series diverges. But if anything, I'm just practicing technique.
For divergence purposes, it's sufficient that there is some $\varepsilon_0$ for which (1) holds i.o. For example, can we make $\cos 2\pi n\theta > 0.5$ i.o? It holds that $\cos x >0.5$ if $x \in \left ( -\pi/3 + 2\pi k, \pi/3 + 2\pi k \right )$ for some $k\in \mathbb N$ ($x<0$ not relevant for now). With that in mind I get the inequalities $$ 6k-1 < 6n\theta < 6k+1 $$ and I'm stuck. Does this hold infinitely often? More specifically, does there exist a subsequence $m_n$ such that for every $n$ there exists $k_n$ satisfying $$ 6k_n -1< 6m_n\theta < 6k_n+1? $$
I vaguely remember problems of this sort, but I can't recall relevant keywords to find such problems.
If $\theta$ is irrational then $ n\theta$ is uniformly distributed on $[0,1]$. Hence $2\pi n\theta$ is uniformly distributed on $[0,2\pi]$ so, since $\cos $ is continuous the sequence is dense on $[-1,1]$. In particular, it gets as close to $1$ as possible.