Let $\alpha \in \mathbb{R}$ be a fixed (positive) number. For each $k \in \mathbb{N}$ we choose $\varepsilon_k >0$ with the property that $\lim_k \varepsilon_k =0$.

If $\alpha \in \mathbb{R} \setminus \mathbb{Q}$, we know that there exist infinitely many $q_k \in \mathbb{Z}$ such that $$ \frac{1}{2} - \varepsilon_k \leq \{q_k \alpha \} \leq \frac{1}{2}, $$ since $\{q \alpha\}$ is dense in $[0,1)$. Here $\{ z \}$ denotes the fractional part of a number $z$, i.e. $\{z\} = z -[z]$. Can we also choose $q_k$ such that $q_k = O(\varepsilon_k^{-1})$? In other words, can we select $q_k$ so that $\limsup_k \varepsilon_k q_k < \infty$?

It seems to me that the answer is not trivial, since we are, roughly speaking, trying to single out a *slow* sequence of approximation. Maybe this holds only for some particular (kind of) irrational number $\alpha$.

**Edit.** It seems that there is a positive result if one requires the weaker condition $k/\varepsilon_k \geq q_k$, see here. Of course $q_k$ can be much larger than we need.

a priori, the largeness of the integer $q_k$? In some sense there are too few integers below $2^k$... $\endgroup$ – Siminore May 8 '16 at 11:32