Let $k = \lim_{n \to \infty} a_n = \infty$. Construct the set $A$ such that every $m \in \mathbb{N}$ is divided by $k$:
$$A = \left\{\frac{1}{k},\frac{2}{k},\frac{3}{k},\frac{4}{k},\frac{5}{k},\frac{6}{k},\ldots, \frac{k}{k} \right\}$$
Is $\frac{\sqrt{2}}{2} \in A?$ Are other irrationals in $[0,1]$ also in $A$? I believe the answer is no, since $\sqrt{2}$ cannot be written as a quotient, and as a result is not contained in $A$. On the other hand, if some $m$ is large enough, $\frac{m}{k}$ gets arbitrarily close to $\frac{\sqrt{2}}{2}$.
This question is meaningless, because no such $k$ exists. There is no largest natural number, since for any $n\in\mathbb{N}$, $n+1$ is a larger natural number.