Put
$$
\alpha=\inf_{k\geq1}\frac{x_k}{k}.
$$
Fix $k\geq1$ and write $n=qk+r$ with $0\leq r<k$. Repeated <subadditivity> gives
$$
x_n\leq qx_k+x_r,
$$
so
$$
\frac{x_n}{n}\leq\frac{qk}{n}\frac{x_k}{k}+\frac{x_r}{n}.
$$
The finitely many values $x_0,\ldots,x_{k-1}$ are bounded, hence
$$
\limsup_{n\to\infty}\frac{x_n}{n}\leq\frac{x_k}{k}.
$$
Taking the infimum over $k$ gives $\limsup x_n/n\leq\alpha$, while the definition of $\alpha$ gives $x_n/n\geq\alpha$ for every $n$. Therefore
$$
\boxed{\lim_{n\to\infty}\frac{x_n}{n}=\inf_{k\geq1}\frac{x_k}{k}.}
$$
This is <Fekete lemma>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 214 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-214/1/a/solution