I'm trying to prove the following:
If $$\displaystyle\lim_{n\rightarrow\infty}\frac{a_{n}}{n}=1$$ then $$\displaystyle\lim_{n\rightarrow\infty}\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{n}=0.$$
So, the hypotesis $\displaystyle\lim_{n\rightarrow\infty}\frac{a_{n}}{n}=1$ implies that $\forall\epsilon>0,\exists N:=N(\epsilon)\in\mathbb{N}$ such that $\forall n\geq N$ implies $\frac{|a_{n}-n|}{n}<\epsilon.$
Fixed $\epsilon>0,$ we have $\displaystyle\sup_{n\geq N}\frac{|a_{n}-n|}{n}<\epsilon.$
Now, for such $N$ I'd like to prove that $\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{n}<\epsilon$ $\forall n\geq N,$ but I'm stuck in this.
I think that if $n\geq N$ then $$\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{n}\leq\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{N}.$$
Then,for example, if $\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{N}=\frac{|a_{N}-N|}{N}$ it's done because of the hypotesis. If $\sup_{0\leq k\leq n}|a_{k}-k|$ is achieved in $0\leq k<n$ then $\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{N}<\epsilon$ only for large enough $N.$
Because of the above I think such limit have sense and it's equal to $0$ but I don't get how to prove it exactly.
Any kind of help is thanked in advanced.
Let $\varepsilon>0$ be arbitary. Then there exists $N_{1}\in\mathbb{N}$ such that $\left|\frac{a_{n}-n}{n}\right|<\varepsilon$ whenever $n\geq N_{1}$. Let $M=\max_{0\leq k<N_{1}}|a_{k}-k|$. Choose $N_{2}\in\mathbb{N}$ such that $\frac{M}{N_{2}}<\varepsilon$. Let $N=\max(N_{1},N_{2})$.
Let $n\geq N$ be arbitrary. Let $0\leq k\leq n$. If $0\leq k<N_{1}$, we have that $\left|\frac{a_{k}-k}{n}\right|\leq\frac{M}{N}<\varepsilon$. If $N_{1}\leq k\leq n$, then $\left|\frac{a_{k}-k}{n}\right|\leq\left|\frac{a_{k}-k}{k}\right|<\varepsilon$ because $k\geq N_{1}$. It follows that $\sup_{0\leq k\leq n}\left|\frac{a_{k}-k}{n}\right|<\varepsilon$. Hence $\lim_{n}\sup_{0\leq k\leq n}\left|\frac{a_{k}-k}{n}\right|=0$.