If there is a $m$ such that ${a}_{m}=\mathrm{\infty}$, then, by subadditivity, we have ${a}_{n}=\mathrm{\infty}$ for all $n>m$. Then, both sides of the equality are $\mathrm{\infty}$, and the theorem holds.
So, we suppose that ${a}_{n}\in \text{\mathbf{R}}$ for all $n$. Let $L={inf}_{n}\frac{{a}_{n}}{n}$ and let $B$ be any number greater than $L$. Choose $k\ge 1$ such that
For $n>k$, we have, by the division algorithm^{} there are integers ${p}_{n}$ and ${q}_{n}$ such that $n={p}_{n}k+{q}_{n}$, and $0\le {q}_{n}\le k1$.
Applying the definition of subadditivity many times we obtain:

$${a}_{n}={a}_{{p}_{n}k+{q}_{n}}\le {a}_{{p}_{n}k}+{a}_{{q}_{n}}\le {p}_{n}{a}_{k}+{a}_{{q}_{n}}$$ 

So, dividing by $n$ we obtain:

$$\frac{{a}_{n}}{n}\le \frac{{p}_{n}k}{n}\frac{{a}_{k}}{k}+\frac{{a}_{{q}_{n}}}{n}$$ 

When $n$ goes to infinity, $\frac{{p}_{n}k}{n}$ converges to $1$ and $\frac{{a}_{{q}_{n}}}{n}$ converges to zero, because the numerator is bounded by the maximum of ${a}_{i}$ with $0\le i\le k1$. So, we have, for all $B>L$:
Finally, let $B$ go to $L$ and we obtain

$$L=\underset{n}{inf}\frac{{a}_{n}}{n}=\underset{n}{lim}\frac{{a}_{n}}{n}$$ 
