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Let $(x_n)$ an increasing sequence and auditing, $$|x_{2^{n+1}} - x_{2^n}| \le \frac{1}{2^n}.$$ Prove that $(x_n)$ is convergent.

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This site isn't here to do your homework for you. Please refer to this to see how to properly ask for help: meta.math.stackexchange.com/questions/1803/… –  Tyler Nov 20 '11 at 1:24
"auditing" I guess is some foreign word literally translated to English? –  GEdgar Nov 20 '11 at 1:24
Please do not use display mode for math in the titles. And, please carefully read the comments here. –  cardinal Nov 20 '11 at 1:25
$\text{auditing} = \text{satisfying}$? –  cardinal Nov 20 '11 at 1:26
Welcome to MSE! I see that all of your questions have been downvoted... perhaps you should start thinking about being more friendly to other users when asking questions. One way to do this is to not ask questions on the imperative like in the books, and perhaps add a little information about what you've tried to do on your problem and where you got stuck. This will make people more willing to help. –  Patrick Da Silva Nov 20 '11 at 1:29

1 Answer 1

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Well clearly $x_{2^n}$ is convergent because it is a Cauchy sequence : $$ |x_{2^{n+k}} - x_{2^n} | \le \sum_{i=0}^{k-1} \frac 1{2^{n+i}} = \frac 1{2^n} \sum_{i=0}^{k-1} \frac 1{2^i} \le \frac 1{2^{n-1}} < \varepsilon $$ when $n$ is sufficiently large. Since for $m \in \mathbb N$, there exists $n$ such that $2^n \le m < 2^{n+1}$, you have $x_{2^n} \le x_{m} \le x_{2^{n+1}}$ so that $x_m$ has the same limit as $x_{2^n}$ by a sandwich argument.

Note : Be less rude when you ask questions. We're not robots.

Hope that helps,

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sorry for my rudeness ... i'm very sorry,will not happen again... tanks for the response! very nice –  FrConnection Nov 20 '11 at 1:38
I am not offended, but the community usually doesn't like it. I am more offended if the user that does this has over 50 reputation and keeps doing that ; that means he just doesn't learn... just get better =) –  Patrick Da Silva Nov 20 '11 at 7:43

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