# 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 [closed]

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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## closed as off-topic by 900 sit-ups a day, anorton, Tomás, Adam Hughes, RecklessReckonerAug 1 at 3:14

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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

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.