real analysis - $lim_{nrightarrowinfty}frac{a_{n}}{n}=1$ implies that $lim_{nrightarrowinfty}sup_{0leq kleq n}frac{|a_{k}-k|}{n}=0$

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

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.

Answer

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 ksuch 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 kwe 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 $\lim_{n}\sup_{0\leq k\leq n}\left|\frac{a_{k}-k}{n}\right|=0$.