 Solution

This is the Fekete lemma, with its possible extended-real limit made explicit. Put $L={\inf }_{m\ge 1}{x}_m/m$. Since $x_n\le n{x}_1$, $L$ cannot be $+\infty$; it may be $-\infty$.

Fix $m\ge 1$ and write $n=qm+r$, with $0\le r<m$. Repeated subadditivity giveswhere for this calculation set $x_0=0$. The finitely many possible remainder terms are bounded. Since $q/n\to 1/m$,This holds for every $m$, whereas $x_n/n\ge L$ by the definition of an infimum. If $L$ is finite, the upper and lower bounds agree. If $L=-\infty$, choosing $m$ with arbitrarily negative $x_m/m$ shows that $x_n/n\to -\infty$. ThusA finite real limit is not guaranteed: $x_n=-n^2$ is a subadditive sequence and has $x_n/n=-n\to -\infty$. If the limit is intended to be real rather than extended real, one must add a linear lower bound.