Show that $\lim_{n \to \infty}\frac {a_n} {n}=1 \implies \lim_{n \to \infty }\sup_{0\le k \le n} \frac {|a_k - k|} {n }=0$

Let $\varepsilon>0$ be given, then there exists $N_{1}\in\mathbb{N}$ such that $\left|\frac{a_{n}}{n}-1\right|<\varepsilon$ whenever $n\geq N_{1}$. Choose $N_{2}\in\mathbb{N}$ such that $\frac{|a_{k}-k|}{\varepsilon}\leq N_{2}$, for $k=0,\ldots,N_{1}-1$. Let $N=\max(N_{1},N_{2})$. Let $n\geq N$ and $0\leq k\leq n$ be arbitrary. If $k<N_{1}$, we have \begin{eqnarray*} |\frac{a_{k}-k}{n}| & \leq & \frac{\varepsilon N_{2}}{N}\\ & \leq & \varepsilon. \end{eqnarray*} If $k\geq N_{1}$, we have \begin{eqnarray*} |\frac{a_{k}-k}{n}| & \leq & \frac{k\varepsilon}{n}\\ & \leq & \varepsilon. \end{eqnarray*} This shows that $\sup_{0\leq k\leq n}|\frac{a_{k}-k}{n}|\leq\varepsilon$ whenever $n\geq N$. Hence $\lim_{n\rightarrow\infty}\sup_{0\leq k\leq n}|\frac{a_{k}-k}{n}|=0$.

Hint: Choose $N$ such that $|a_k-k |<k\epsilon$ for $k >N$. Observe that the sup over $k \leq n$ is the maximum of sup over $k \leq N$ and the sup over $N<k\leq n$.