Convergence of sequence of means: Define $A_n=\frac{a_1+a_2+...a_n}{n}$, Find $\lim_{n\to\infty} \sqrt{n}(A_{n+1}-A_{n})$ 
Let $a_n$ be a sequence of real numbers such that $|a_n| \le 1$. 
  Define $A_n=\frac{a_1+a_2+...a_n}{n}$, Find
  $$\lim_{n \rightarrow \infty} \sqrt{n}(A_{n+1}-A_{n})$$

I was thinking of using Stolz Cesaro lemma, but that needs to show that $A_n$ is convergent which means that $a_n$ has to be convergent.
But I have no clue how to approach this one.
 A: *

*First: no, $(a_n)_n$ does not have to be convergent. Convergence implies convergence in Césaro mean, but the converse is not true.

*Yet (and second), indeed, even in spite of that first bound we cannot claim that $(A_n)_n$, the sequence of Césaro means, does converge. This is not true in general.

*Third... well, let's see. 
$$\begin{align}
\sqrt{n}(A_{n+1}-A_n) &= \sqrt{n}\left(\frac{1}{n+1}\sum_{k=1}^{n+1} a_k-\frac{1}{n}\sum_{k=1}^{n} a_k \right)
= \sqrt{n}\left(\frac{a_{n+1}}{n+1} - \frac{1}{n(n+1)} \sum_{k=1}^{n+1} a_k \right)\\
&=\frac{\sqrt{n}}{n+1}a_{n+1} - \frac{\sqrt{n}}{n(n+1)} \sum_{k=1}^{n+1} a_k
\end{align}$$
and since $(a_n)_n$ is bounded,
$$
\lim_{n\to\infty}\frac{\sqrt{n}}{n+1}a_{n+1} = 0, \qquad \lim_{n\to\infty}\frac{\sqrt{n}}{n(n+1)} \sum_{k=1}^{n+1} a_k = 0
$$
the second recalling that $\frac{\sqrt{n}}{n(n+1)} \sum_{k=1}^{n+1} \lvert a_k\rvert  \leq \frac{\sqrt{n}}{n}$.
Thus, despite our first and second point,
$$
\boxed{\lim_{n\to\infty} \sqrt{n}(A_{n+1}-A_n) = 0\,.}
$$
A: Solution
Notice that
\begin{align*}0 \leq|\sqrt{n}(A_{n+1}-A_n)|&=\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+1}|+|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, by the squeeze theorem, we may conclude that $$|\sqrt{n}(A_{n+1}-A_n)|\to 0,~~~(n \to \infty).$$
But $$-|\sqrt{n}(A_{n+1}-A_n)|\leq \sqrt{n}(A_{n+1}-A_n)\leq |\sqrt{n}(A_{n+1}-A_n)|,$$
by the squeeze theorem again, $$\lim_{n \to \infty}\sqrt{n}(A_{n+1}-A_n)=0.$$
