# Find $\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)$ where $A_n = \frac{1}{n}(a_1 + a_2 + \cdots + a_n)$

Let $\{a_n\}, {n\geq 1}$, be a sequence of real numbers satisfying $|a_n|\leq 1$ for all $n$. Define $$A_n = \frac{1}{n}(a_1 + a_2 + \cdots + a_n),$$ for $n\geq 1$. Then find $\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)$ .

I proceed in this way $$\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)=\lim_{n \rightarrow \infty}\sqrt{n}\left[\frac{1}{n+1}(a_1 + a_2 + \cdots + a_n+a_{n+1})-\frac{1}{n}(a_1 + a_2 + \cdots + a_{n})\right]=\lim_{n \rightarrow \infty}\left[{(na_{n+1}-a_1 - a_2 - \cdots - a_n})\frac{1}{\sqrt{n}(n+1)}\right]$$ Please help me to complete from here

You almost solved the problem with your calculation. Now you just have to note that with $|a_n|\le1$ we have $|na_{n+1}-a_1-\dotso-a_n|\le2n$, so $|\frac1{\sqrt n(n+1)}(na_{n+1}-a_1-\dotso-a_n)|\le\frac{2n}{\sqrt n(n+1)}\to0$.
Notice that \begin{align*}|\sqrt{n}(A_{n+1}-A_n)-0|&=\frac{\sqrt{n}}{n(n+1)}|na_{n+1}-a_1-a_2-\cdots-a_n|\\&\leq \frac{\sqrt{n}}{n(n+1)}(n|a_n|+|a_1|+|a_2|+\cdots|a_n|)\\ &\leq \frac{\sqrt{n}}{n(n+1)}(n+1+1+\cdots+1)\\&=\frac{\sqrt{n}}{n(n+1)}(n+n)\\&=\frac{2\sqrt{n}} {n+1}.\end{align*} Since $\dfrac{2\sqrt{n}} {n+1} \to 0$ as $n \to \infty$. Thus, for any $\varepsilon>0$, there exists $N \in \mathbb{N_+}$ such that $$|\sqrt{n}(A_{n+1}-A_n)-0| \leq \dfrac{2\sqrt{n}} {n+1}<\varepsilon$$when $n>N.$ According to the $\varepsilon-N$ definition of the limit, we can conclude that $$\lim_{n \to \infty}\sqrt{n}(A_{n+1}-A_n)=0.$$