Compute: $\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)$ 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 $\displaystyle\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)$ .

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 $|\frac1{\sqrt n(n+1)}(na_{n+1}-a_1-\dotso-a_n)|\le\frac{2n}{\sqrt n(n+1)}\to0$.