Subadditive sequences and the limit $\sup_{n\to\infty} a_n / n$ Let $\{ a_n, n\geq 1\}$ be a sub-additive sequence, i.e. for any $m,p\geq 1$
$$ a_{m+p} \leq a_m + a_p. $$
.Prove that
$$\lim_{n\to\infty} \frac{a_n}{n} = \inf_n \frac{a_n}{n}.$$
I guess the first step is to state
$$a_{n} = a_{mp+q} \leq a_{mp} + a_q$$
but I'm not sure where to go from here....
 A: $a_n$ 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: 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.
