Lemma 2 (Fekete Lemma)

Let $(a_n)_{n\geqslant 1}$ be a sequence of real numbers. Suppose that

for all $n\geqslant 1$ and all $m\geqslant 1$. Then the sequence $(a_n/n)_{n\geqslant 1}$ converges and satisfies

lim$_{n\rightarrow+\infty}\frac{a_n}{n}=$ inf$_{n\geqslant 1}\frac{a_n}{n}$.

Put $a_0= 0$ ; the inequality $a_{n+m}\leqslant a_n+a_m$ remains valid for all $n\geqslant$ 0 and all $m\geqslant 0$. Fix an integer $m\geqslant$ 1. For every integer $n\geqslant 1$, let $q(n)$ and $r(n)$ be the integers such that $n=q(n)m+r(n)$ and $0\leqslant r(n)< m$ (E, III, p. 39, th. 1). The hypothesis then implies

Letting $n$ tend to $+\infty$, we deduce that lim sup$_n(a_n/n)\leqslant a_m/m$ since $q(n)/n\rightarrow 1/m$. Since this holds for every $m\geqslant 1$, we therefore have

lim sup$_{n\rightarrow+\infty}\frac{a_n}{n}\leqslant$ inf$_{m\geqslant 1}\frac{a_m}{m}\leqslant$ lim inf$_{n\rightarrow+\infty}\frac{a_n}{n}$.

These inequalities prove the convergence of the sequence $(a_n/n)_{n\geqslant 1}$ as well as the formula lim $a_n/n=$ inf$_{n\geqslant 1}a_n/n$.