Prove that the sequence $(a_n/n)$ converges 
Let $(a_n)$ be a positive and monotonically increasing sequence that
satisfies
$$ a_{n \cdot m} \geq n a_m \; \; \; \; \text{for} \; \; n,m \in
 \mathbb{N} $$
Suppose $\sup_{n \in \mathbb{N}} \left\{ \dfrac{a_n}{n} \right\} < +
 \infty $, Prove ${\bf carefully}$ that $\left( \dfrac{ a_n }{n}
 \right) $ ${\bf converges}$

Attempt:
By hypothesis, if $\alpha$ is the supremum of the sequence, then $\dfrac{a_n}{n} \leq \alpha $ so we observe that $(a_n/n)$ is bounded. If we can prove that is monotonic, then we are done. Let $b_n = a_n/n$. Then, we have
$$ \dfrac{b_{n+1} }{b_n} = \dfrac{n}{n+1} \cdot \frac{a_{n+1}}{a_n} > \dfrac{n}{n+1} $$
Which leads nowhere. But, if we use the property of the sequence with $m=1$ we see that
$$ a_n \geq n a_1 $$
And in particular with $n$ replaced by $n+1$ one sees that $a_{n+1} \geq (n+1) a_1 $ or that $\dfrac{ a_{n+1} }{n+1} \geq a_1 $. Now, Id be tempted to say that
$$ \dfrac{ a_{n+1} }{n+1} - \dfrac{ a_n }{n} \geq a_1 - a_1 = 0 $$
but unfortunaly the inequality is not always true. Am I on the right directio nto solve this problem?Any hint/suggestion is welcome!
 A: Here is an approach I think will lead to a solution:
As per the counterexample in the comment, trying to prove monotonicity will fail. But part of the counterexample is that in order to create the break in monotonicity, we had to increase the constant greatly, impacting the rest of the sequence. Also we have that $a_n$'s growth is at most linear, otherwise the $a_n/n$ sequence isn't bounded. 
So assume we have infinitely many breaks in monotonicity. Each break in monotonicity will impact the later terms by forcing the linear constant up. If there are infinitely many breaks then (I think) one can show that this lower bound for the linear constant is unbounded. 
A: Clearly $\limsup_{k \to \infty} \frac{a_k}{k} \le \sup_n \frac{a_n}{n}$. Fix $n \ge 1$. For $k \ge 1$, write $k = nq+r$ for $0 \le r \le n-1$. Then $\frac{a_k}{k} = \frac{a_{nq+r}}{nq+r} \ge \frac{a_{nq}}{nq+r} \ge \frac{qa_n}{qn+r}$. So, $\liminf_{k \to \infty} \frac{a_k}{k} \ge \frac{a_n}{n}$. It follows that $\liminf_{k \to \infty} \frac{a_k}{k} \ge \sup_n \frac{a_n}{n}$.
