 Solution

Set $L={\inf }_{k\ge 1}{x}_k/k$. It may be $-\infty$, but cannot be $+\infty$ because $x_1$ is finite. Fix $k$ and write $n=qk+r$, $0\le r<k$. Repeated use of the subadditive sequence inequality giveswhere $C_1=0$. Dividing by $n$ and letting $n\to \infty$ gives $\mathrm{limsup}{x}_n/n\le x_k/k$. Since this holds for every $k$, the upper limit of a sequence is at most $L$, while every ratio is at least $L$. If $L$ is finite these bounds agree. If $L=-\infty$, choose $k$ with $x_k/k$ below any prescribed real number; the same upper-limit bound proves convergence to $-\infty$. This establishes the extended-real Fekete lemma:A finite real limit of a sequence requires a lower linear bound. For example $x_n=-n^2$ satisfies the premise but has normalized limit of a sequence $-\infty$, so that qualification cannot be dropped.