Let $t_n$ be a sequence of real numbers and $C,r>1.$ Suppose that for every $n\geq 1$ we have $\frac{1}{C}r^n\leq t_n \leq Cr^n.$ Does there exist a real number $\xi$ and an $\varepsilon>0$ such that $|| \xi t_n ||\geq \varepsilon$ for every $n\geq1$?

Here $|| x ||$ denotes the distance between $x$ and the nearest integer.

If $r/C^2>1$, then the sequence is lacunary and the answer is yes (by a result discovered independently by Khintchine, Pollington and De Mathan). This is not my area so I'm neither familiar with the literature nor adept at such arguments. Basic Mathscinetting turned up intersting realted results but nothing *I* could use to answer the above question.