# Convergence of sequence of means: Define $A_n=\frac{a_1+a_2+…a_n}{n}$, Find $\lim_{n\to\infty} \sqrt{n}(A_{n+1}-A_{n})$ [duplicate]

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.

## marked as duplicate by StubbornAtom, Lord Shark the Unknown, 5xum, Christopher, José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 12 '18 at 14:07

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

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

• Great! This fact did not strike me – Legend Killer Jul 31 '18 at 4:41
• @LegendKiller There may be more direct, but here I favored the "heuristic" kind of approach. (After the fact, once can more directly bound directly $\lvert A_{n+1} - A_n\rvert$ and save a few lines, I am sure.) – Clement C. Jul 31 '18 at 4:42