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

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

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Downvoted for the usual reasons. – Did Jul 18 '12 at 7:58
In the event OP doesn't know what the usual reasons are, they have to do with phrasing questions as orders, and with not showing any evidence of having tried to do the problem. – Gerry Myerson Jul 18 '12 at 8:40
Dear Ranbhir, I think you are new to this community, it takes some time for you to get adjusted with the rules of this community . Being a member of this forum, I take the privilege to explain you these things. (1). Please phrase your questions with proper wording and REQUEST the users for the help. (2) Adding a simple THANK YOU, at the end will make your question appear interesting. (3) Please remove the habit of posting it as a homework assignment, People won't help you then. (4) Use the tag , <<Home Work >> if you want to hear hints.. Contd. – Iyengar Jul 18 '12 at 8:48
(5) I remember the Kinley mineral water bottle caption, it says : " We give back more than we take " , that should be your attitude. You shouldn't solely depend upon users for help. You must understand that each user will be having his own work and they are doing a free service for us, sparing their valuable time. Users like @GerryMyerson , and others may have busy schedules, some work as the professors in the university and are packed between their schedules. They can't help you from the scratch. Please do work on it as far as possible, and if you are unable to figure it out .. Contd. – Iyengar Jul 18 '12 at 8:52
Please post the question along with the effort you did. Ask your question precisely and then clearly state where you got an obstacle that is not letting you to proceed further. (6) You can also use the Chat rooms, and see if some user is free to assist you from elementary level, but I can't promise that you find such users. There may be , there may not be. (7) Try reposting your questions , slowly you get upvotes and slowly your position will improve, users start responding to you. Trust me, I was in a more worst state when I entered, there were many times where I got 9 down votes and closed – Iyengar Jul 18 '12 at 8:56

1 Answer 1

up vote 5 down vote accepted

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

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