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Why doesn't $d(x_n,x_{n+1})\rightarrow 0$ as $n\rightarrow\infty$ imply ${x_n}$ is Cauchy?

If we have a sequence $(x_n)_n$ that satisfies this rule:

$\forall \varepsilon > 0, \exists n_0 : \forall n > n_0 : |x_{n+1} - x_n| < \epsilon$

Does is always converge? The criterion is less strict than that of a Cauchy sequence, but I fail to find an example of a sequence which doesn't converge.

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marked as duplicate by David Mitra, Aryabhata, Davide Giraudo, Hans Lundmark, Kannappan Sampath Feb 20 '12 at 20:36

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.

Revise your way to write $\forall \varepsilon > 0, \exists n_0 : \forall n > n_0 : |x_{n+1} - x_n| < \epsilon$. Where is the second 'such that' should be a 'implies' . I would write $\forall \epsilon>0\;\;\exists n_{0}\in\mathbb{N}:(n>n_{0})\Rightarrow |x_{n+1}-x_{n}|<\epsilon$ – Paulo Henrique Feb 20 '12 at 20:05
up vote 9 down vote accepted

What about $s_n=\sum_{i=1}^n\frac 1 i$ ?

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Take $x_n=\sqrt n$. Then $$|x_{n+1}-x_n|=(\sqrt{n+1}-\sqrt n)\frac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}+\sqrt n}=\frac 1{\sqrt{n+1}+\sqrt n}\le \frac 1{\sqrt{n+1}}$$ so the rule is satisfied (show it with $\varepsilon$). But the sequence doesn't converge (in $\mathbb R$).

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