A result about sequence satisfying $u_{n+1}-u_n=o(1)$

Question

I'm asked to prove that if $u_{n+1}-u_n=o\left(1\right)$ then $\displaystyle \frac{u_n}{n}\underset{n \rightarrow +\infty}{\rightarrow}0$.

I know that I can write that for all $\epsilon>0$, it exists $N$ such that if $n \geq N$ then $\left|u_{n+1}-u_n\right|<\epsilon'$. I'v tried to choose $\epsilon'=n\epsilon$ then I have $$\left|\frac{u_{n+1}}{n}-\frac{u_n}{n}\right|<\epsilon$$

But I cannot conclude, any hint ?

Answer 1

Let $(v_n)=(u_{n+1}-u_n)$, since $\lim\limits_{n\rightarrow +\infty}{v_n}=0$, by Cesaro's theorem you have that $$ \lim\limits_{n\rightarrow +\infty}\frac{1}{n}\sum_{k=1}^n{v_k}=0 $$ Moreover $$ \sum_{k=1}^n{v_k}=\sum_{k=1}^n{(u_{k+1}-u_k)}=u_{n+1}-u_1 $$ Finally, $$ \frac{u_n}{n}=\frac{u_{n+1}-u_1}{n}+\frac{u_1}{n}-\frac{u_{n+1}-u_n}{n}\underset{n\rightarrow +\infty}{\longrightarrow}0 $$

Answer 2

Hint. $\frac{u_n}{n} = \frac{u_0}{n} + \frac{1}{n} \cdot \sum_{j=1}^{n} (u_j - u_{j-1})$. Also, if a sequence goes to zero, what can you say about the mean value of it?

Answer 3

You don't need any fancy theorems for this. Simply observe that for $n>N$, we have

\begin{align} |u_{n}-u_N| & =|(u_{n}-u_{n-1})+(u_{n-1}-u_{n-2})+\cdots(u_{N+1}-u_N)| \\ & \le|u_{n}-u_{n-1}|+|u_{n-1}-u_{n-2}|+\cdots|u_{N+1}-u_N|\\ & \le \epsilon'(n-N) \end{align}

So $|u_n|\le |u_N|+\epsilon'(n-N)$, and $\left|\frac{u_n}{n}\right|\le \left|\frac{u_N}{n}\right|+\epsilon'$, which is $\le 2\epsilon'$ if $n\ge\frac{|u_n|}{\epsilon'}$.