Let ${x_n}$ be a sequence such that there exist $A>0$ and $C\in (0,1)$ for which $|x_{n+1} -x_n|<AC^n$, for any $n\geq 1$. Show that $\{x_n\} $ is Cauchy. Is this conclusion still valid if we assume only $\lim_{n\to\infty} |x_{n+1}-x_n|=0? $
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Do you mean $AC^n$? – user259242 Nov 04 '16 at 12:59
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You need to show us your thoughts, your attempts. – GEdgar Nov 04 '16 at 13:01
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Yes i mean $ AC^n$. – chenar abdulla Nov 04 '16 at 13:02
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Okay, I have shown that $x_n$ is Cauchy. What now? – Wojowu Nov 04 '16 at 13:04
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@GEDgar, i think we have to show that AC^n is less than epsilon, then we can make it to be cauchy – chenar abdulla Nov 04 '16 at 13:05
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Look at the definition of "Cauchy". It is not just that $|x_{n+1}-x_n|$ is small. – GEdgar Nov 04 '16 at 13:08
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@GEDgar Yes i know. We can add x_m and subtract x_m – chenar abdulla Nov 04 '16 at 13:39
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For the last part, see: Is a bounded sequence such that $\lim(a_{n+1}-a_n)=0$ necessarily Cauchy? and An example of a bounded pseudo Cauchy sequence that diverges? and Pseudo-Cauchy sequence. – Martin Sleziak Nov 04 '16 at 15:56
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And the first part is rather similar to this: http://math.stackexchange.com/questions/661047/cauchy-sequences-involving-geometric-series or http://math.stackexchange.com/questions/572635/analysis-question-cauchy-sequence – Martin Sleziak Nov 04 '16 at 16:00
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Note, that for $m > n$, we have \begin{align*} \def\abs#1{\left|#1\right|}\abs{x_m - x_n}&= \abs{\sum_{k=n}^{m-1} x_{k+1} - x_k}\\ &\le \sum_{k=n}^{m-1}\abs{x_{k+1} - x_k}\\ &\le A \sum_{k=n}^{m-1} C^k\\ &= AC^n \sum_{k=0}^{m-n-1} C^k\\ &= AC^n \frac{C^{m-n} - 1}{C - 1}\\ &\le \frac{AC^n}{1-C}\\ &\to 0, n \to \infty. \end{align*} For your additional question, have a look at $x_n = \sum_{k=1}^n \frac 1k$.
martini
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