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.
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, $$ \boxed{\lim_{n\to\infty} \sqrt{n}(A_{n+1}-A_n) = 0\,.} $$
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, $$\lim_{n \to \infty}\sqrt{n}(A_{n+1}-A_n)=0.$$
