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Question: Show that for a sequence $\{x_m\}$ of real numbers to be a Cauchy sequence, it is necessary, but not sufficient that $|x_{m+1}-x_m|$ converges to zero.

This is how I proved that it is necessary: for $\{x_m\}$ to be cauchy, for all $\epsilon \gt 0$ there exists $n_o \in \mathbb{N}$ such that $|x_a-x_b| \lt \epsilon$ for all $a,b \ge n_o$. Now we can choose $a=m+1$ & $b=m$ such that both $a,b \ge n_o$. and by increasing or decreasing $m$ we can always ensure that for all $\epsilon$, $|x_{m+1}-x_m| \lt \epsilon $. Is this proof correct?

How to prove that it is not sufficient condition?

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I know this question is very simple but even then please guide me if I am going in the right direction. – Mathy Jun 24 '13 at 2:14
See this post for examples showing the condition is not sufficient. – David Mitra Jun 24 '13 at 2:20
It means the way I have proved necessary condition is also wrong? – Mathy Jun 24 '13 at 2:23
No, your proof of necessary is right. As for not sufficient, you need to give an example of a sequence which satisfies the condition, but does not satisfy the condition of being Cauchy. David's link provides good examples. – Calvin Lin Jun 24 '13 at 2:27
The right idea seems to be there, but your phrasing could be better. Say: "Let $\epsilon>0$. Choose $N$ so that $|x_a-x_b|<\epsilon$ for all $a,b>N$. Then in particular, for $m>N$, $|x_{m+1}-x_m|<\epsilon$. So..." – David Mitra Jun 24 '13 at 2:27

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