Let $f(n)$ be a sequence of positive integers such that the limit $\alpha=\lim_{n\to \infty} f(n)/n$ exists and is a finite number greater than 1. Is it true that the range $R$ of this sequence has positive density in the integers in the sense that $\lim \inf_N|R\cap \{1,...,N\}|/N>0$?
Can this lim inf be bounded below in terms of $\alpha$? Is it equal to $1/\alpha$? Any intuition appreciated!
Without assuming that $f$ is injective, the answer is certainly no. For a counterexample, define $f(n) = \lfloor \sqrt{2n} \rfloor^2$. Then the range of $f$ is contained in the set of perfect squares and thus has density $0$. On the other hand, $f(n) = (\sqrt{2n}+O(1))^2 = 2n + O(\sqrt n)$ and therefore $\lim_{n\to\infty} f(n)/n = 2$.
(In this counterexample one can change the $2$ to any positive number $\alpha$, including $\alpha\in(0,1]$. There are lots of other variants as well, such as letting $f(n)$ be the smallest prime exceeding $2n$.)