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This question already has an answer here:

Let $(x_n)$ be a real sequence with the property that for all $n \in \mathbb{N}$, $$|x_{n + 1} - x_n| < \frac{1}{2^n}$$ I want to show, using the definition of a Cauchy sequence, that $(x_n)$ must be Cauchy.

I have found that the property implies that for any $(m, n) \in \mathbb{R}^2$, assuming without loss of generality that $m > n$, it must be true that $$|x_n - x_m| \leq \sum\limits_{i = n}^m \frac{1}{2^i}$$

How can I proceed from there ? Is this even the right way to approach this problem?

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marked as duplicate by Martin R, Chinnapparaj R, Cameron Buie, mrtaurho, Adrian Keister Apr 12 at 13:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ What is $k$, and how you got this inequality? Substituting $n = m + 1$ you get $x_n - x_m| < 0$ that can't be right. $\endgroup$ – mihaild Apr 12 at 11:22
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    $\begingroup$ Also: math.stackexchange.com/q/683023/42969 $\endgroup$ – Martin R Apr 12 at 11:26
  • $\begingroup$ Fixed my question, thank you. I get this result by repeatedly applying the triangle inequality. I will go and check out this possible duplicate. $\endgroup$ – oranji Apr 12 at 11:27
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Here is a slightly different argument which is not presented on the linked duplicates:

  • $s_n = \sum_{k=1}^{n}(x_k - x_{k-1}) = x_n - x_0$ is (absolutely) convergent
  • $\Rightarrow x_n = s_n + x_0$ is convergent
  • convergent sequences are Cauchy-sequences
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    $\begingroup$ You can always add another answer to the possible duplicate target (and vote to close). That is preferred (as I understand it) because the various solutions are kept in a single place. $\endgroup$ – Martin R Apr 12 at 11:36
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    $\begingroup$ Btw, this only “hides” the triangle-inequality argument in the conclusion $\sum_{k=1}^{n}|x_k - x_{k-1}|$ convergent $\implies \sum_{k=1}^{n}(x_k - x_{k-1})$ convergent (and uses the Cauchy criterion, which is not needed to answer the original question). $\endgroup$ – Martin R Apr 12 at 11:46
  • $\begingroup$ @MartinR : I wouldn't call it "hiding". It is just using known basic facts and, that way, avoiding to repeat some standard calculations. And btw., I wrote "... slightly different ..." in my introductory sentence. :-) $\endgroup$ – trancelocation Apr 12 at 12:02

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