prove that if $\lim\limits_{n\to\infty}(a_{n+1}-a_{n})$ exist then $\lim\limits_{n\to\infty}\frac{a_{n}}{n}=\lim\limits_{n\to\infty}(a_{n+1}-a_{n})$

Question

prove that if $\lim\limits_{n\to\infty}(a_{n+1}-a_{n})$ exist then $\lim\limits_{n\to\infty}\frac{a_{n}}{n}=\lim\limits_{n\to\infty}(a_{n+1}-a_{n})$
So that's what I did so far Let $$B_{n}=\frac{(a_{2}-a_{1})+(a_{3}-a_{2})+...+(a_{n+1}-a_{n})}{n}=\frac{(a_{n+1}-a_{n})}{n}$$ and we know that if$\ \lim\limits_{n\to\infty}(a_{n+1}-a_{n})$ exist that's mean that $\lim\limits_{n\to\infty}B_{n}=\lim\limits_{n\to\infty}(a_{n+1}-a_{n})$
Thus we have $\lim\limits_{n\to\infty}\frac{(a_{n+1}-a_{n})}{n}=\lim\limits_{n\to\infty}(a_{n+1}-a_{n})$ But I don't know how to go from here

Answer 1

Let $L$ be the limit

$$L=\lim_{n\to\infty}(a_{n+1}-a_n)\tag 1$$

We assume that $L$ is finite.

From $(1)$, for all $\epsilon >0$, there exists an $N$ such that whenever $n>N$,

$$|a_{n+1}-a_{n}-L|<\epsilon/2 \tag2$$

With a given $\epsilon>0$ and corresponding $N$ fixed such that $(2)$ holds, we have (defining $a_0=0$)

$$\begin{align} \left|\frac{a_{n}}{n} -L\right|&=\left|\frac1n \sum_{k=1}^n(a_k-a_{k-1}-L)\right|\\\\ &\le\frac 1n \sum_{k=1}^{N}|a_k-a_{k-1}-L| +\frac1n \sum_{k=N+1}^n|a_k-a_{k-1}-L|\\\\ &\le \frac Nn \sup_{k\in[1,N]}|a_k-a_{k-1}-L|+\frac12 \epsilon\left(\frac{n-N}{n}\right)\tag 3 \end{align}$$

We can now take $n$ so that the first term on the right-hand side of $(3)$ is less than $\epsilon/2$. And we are done!