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.



Distribution mod 1 of exponential growth sequences
I think the answer is yes. For any natural number $n$, let $P(n)$ denote the assertion that there exists an interval $I\_n$ of length $\sqrt{\varepsilon}/t\_n$ such that $\| \xi t\_m \| \geq \varepsilon$ for all $\xi \in I\_n$ and $m \leq n$. If $\varepsilon$ is small enough, it appears that $P(n)$ implies $P(n+A)$ for some suitably large constant $A$ (depending on $C,r$, and with $\varepsilon$ sufficiently small depending on $A,C,r$), basically by taking $I\_{n+A}$ to be a random subinterval of $I\_n$ and using the union bound to upper bound the probability that this fails to work. By completeness we can find a $\xi$ inside an infinite nested sequence of the $I\_n$ which should then do the job.


