If $a_n$is subadditive, $\frac{a_n}{n}$ is convergent.

We say that the sequence $a_n$ is subadditive, if $a_{m+n} \le a_m+a_n (\forall m,n \in \mathbb{N})$.

Prove, that in that case, $\frac{a_n}{n}$ is convergent(or maybe the limit is -$\infty$?).

What I tried: For $m=n=1$, we have:

$a_2 \le a_1+a_1 = 2a_1$. Then $a_4 \le 2a_2 \le 4a_1$.

In that logic, we can see, that $\frac{a_4}{4} \le \frac{a_2}{2} \le a_1$, and this goes until infinity. We have a monotonically decreasing sequence, but I don't know if anything comes from this.

Any ideas? Thanks!