Prove $\lim_{n\to\infty} \frac{a_n}{n}$ exists for positive sequence where $a_{n+m} \leq a_n + a_m$

I am only aware of proofs that use or essentially use the liminf and limsup. One way to see why those proofs work is to notice that $$ \frac{a_{m+n}}{m+n} \leq \frac{m}{m+n} \frac{a_m}{m} + \frac{n}{m+n} \frac{a_n}{n}. $$ So in effect, $a_n/n$ is bounded above by every "average" of the earlier terms. Now to prove the result, notice that $$ \frac{a_{kn}}{kn} \leq \frac{a_k}{k}, $$ and for any $k$, let $R(k) = \max_{m \leq k}a_m$. So for any $N = kn+r$, (e.g. $n = \lfloor N/k \rfloor$) $0 \leq r \leq k$, $$ \frac{a_N}{N} \leq \frac{kn}{N} \frac{a_{kn}}{kn} + \frac{r}{N}\frac{a_r}{r} \leq 1 \cdot \frac{a_k}{k} + \frac{R}{N}. $$ Now for a fixed $k$, if you let $N \to \infty$, you have that $$ \limsup_{N \to \infty} \frac{a_N}{N} \leq \frac{a_k}{k}. $$ Then taking a liminf with $k \to \infty$ will finish it off.

The canon is the following: writing $m=qn+r$ with $q=q(m)$ and $r=r(m)\in[0,n)$, we have $$ \frac{a_m}m=\frac{a_{qn+r}}{qn+r}\le\frac{qa_n+a_r}{qn+r}\to\frac{a_n}n $$ when $m\to\infty$, which implies that $$ \limsup_{m\to\infty}\frac{a_m}m\le\inf_n\frac{a_n}n\le\liminf_{m\to\infty}\frac{a_m}m $$ (you have to think that $n$ is fixed when letting $m\to\infty$).