Subadditive sequences and the limit $\sup_{n\to\infty} a_n / n$

$a_n$ is sub-additive, i.e. $a_{m+n}\leq a_n + a_m$, the limit exists and is equal to $\inf_{n\in\mathbb{N}^\ast} a_n/n$,

Now we fix $m$ and take $n = mq + r$ with $r, q\in\mathbb{N}$ and $r < m$. Therefore, $$a_n \leq q\cdot a_m + a_r \, \, \Rightarrow\,\, \frac{a_n}{n} \leq \frac{q}{n} a_m + \frac{a_r}{n} \, .$$ Note that $m$ is fixed, so if we pass $n\nearrow +\infty$, we have $q\to +\infty$ and $$ \frac{q}{n} = \frac{q}{mq + r} = \frac{1}{m + \frac{r}{q}} \xrightarrow{ n\to+\infty } \frac{1}{m} \,\, , \quad \Big\vert \frac{a_r}{n} \Big\vert \leq \max\big\{ a_1, a_2, \ldots, a_m \big\}\cdot \frac{1}{n}\xrightarrow{n\nearrow+\infty} 0 \, . $$ It follows that $\limsup_{n\to+\infty}a_n\big/n \leq a_m\big/m$ and thus $\limsup_{n\to+\infty}a_n\big/n \leq \inf_{m} a_m\big/m$. On the other side, $$\text{$ \dfrac{a_n}{n} \geq \inf_{m} \dfrac{a_m}{m}$ implies $\liminf_{n\to+\infty}\dfrac{a_n}{n} \geq \inf_{m} \dfrac{a_m}{m}$ .} $$ Hence $$\lim_{n\to+\infty} \frac{a_n}{n} = \inf_{n} \frac{a_n}{n} $$

Q.E.D.

a variant of this result is given in [Szegö and Polya, Problems and Theorems in Analysis, Vol1, Chapter 3] : assume for all indices $m$ and $n$ $$ a_m+a_n - 1 < a_{m+n} < a_m+a_n +1 $$ Then the following limit exists $$ \lim \frac{a_n}{n} = \omega $$ and moreover verifies $$ \omega \,n -1 < a_n < \omega\, n +1 $$ All this follows because $a_n+1$ and $1-a_n$ are both subadditive.