# $\lim_{n\to \infty} \sqrt n(A_{n+1}-A_n)$ when $A_n = \frac{1}{n}(a_1+a_2+\cdots+a_n)$ and $|a_n| \leq 1$ [duplicate]

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 $\lim_{n\to \infty} \sqrt n(A_{n+1}-A_n)$ is

A. $0$

B. $-1$

C. $1$

D. None of them

ATTEMPT

I tok sequence $a_n=1$ and answer came out to be 0. But how do i choose between A and D.

Thanks

## marked as duplicate by StubbornAtom, Lord Shark the Unknown, José Carlos Santos calculus 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 10 '18 at 15:07

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• First derive the identity $(n+1)A_{n+1} - nA_n = a_{n+1}$ and rearrange it to get $\sqrt{n}(A_{n+1} - A_n) = \frac{a_{n+1} - A_{n+1}}{\sqrt{n}}$. Now use $a_n \leq 1$ to bound $A_n$ and use this to conclude. – Winther Dec 31 '15 at 5:32

## 2 Answers

We have \begin{align*} |A_{n+1} - A_n| &= \left|\frac{1}{n+1}(a_1+\dots+a_n+a_{n+1}) - \frac{1}{n}(a_1+\dots+a_n)\right| \\ &=\left|\frac{a_{n+1}}{n+1} - \left(\frac{1}{n}-\frac{1}{n+1}\right)(a_1+\dots+a_n)\right| \\ &\le\left|\frac{a_{n+1}}{n+1}\right| + \left(\frac{1}{n}-\frac{1}{n+1}\right)(|a_1|+\dots+|a_n|)\\ &\le\frac{1}{n+1} + \frac{1}{n(n+1)}(n)\\ &=\frac{2}{n+1} \end{align*} and hence $$\left|\sqrt{n}(A_{n+1}-A_n)\right|\le\frac{2\sqrt{n}}{n+1}\rightarrow 0$$ as $n\rightarrow\infty$.

$$|\sqrt{n}(A_{n+1}-A_n)|=\sqrt{n}\frac{|n(a_1+...+a_n+a_{n+1})-(n+1)(a_1+...+a_n)|}{n(n+1)}\\=\frac{|a_1+...+a_n-na_{n+1}|}{\sqrt{n}(n+1)}\leq\frac{2n}{\sqrt{n}(n+1)}=\frac{2\sqrt{n}}{n+1}\rightarrow 0, n\rightarrow \infty$$