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Let $\{x_n\}$ be a sequence of real numbers with the property that for all positive integers $n$ and $m$, $|x_n -x_m|\lt \frac 1{mn}$.

To show $\{x_n\}$ is a cauchy sequence.

Attempt:

A sequence $\{x_n\}$ is said to be cauchy if for each $\epsilon \gt0$ there exist $n_0 \in \mathbb N$ such that $|x_n -x_m|\lt \epsilon $ for all $n,m \ge n_0. $

Now, $|x_n -x_m|\lt \frac 1{mn}\lt \frac 1{n} \lt \epsilon$ if $n\gt \frac{1}{\epsilon} $,

choose $n_0 \gt \frac{1}{\epsilon} $, using Archimedian property and hence we are done .

Am I right ?

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  • $\begingroup$ Looks correct indeed. $\endgroup$ – DonAntonio Feb 16 '16 at 18:23
  • $\begingroup$ It just works fine. $\endgroup$ – Crostul Feb 16 '16 at 18:23
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    $\begingroup$ Another way is using that every convergent sequence is Cauchy: Since $|x_n-x_m|<\frac{1}{mn}$, then $|x_n-x_1|<\frac{1}{n}$ for all $n$. Letting $n\to\infty$, the sequence is convergent and converges to $x_1$. BUT if you replace $x_1$ with any $x_m$, the conclusion is the same. That is, $x_1=x_2=x_3=...$, i.e. $\{x_n\}$ is constant $\endgroup$ – sinbadh Feb 16 '16 at 18:26
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    $\begingroup$ On the other hand, only one observation: since $m,n\ge 1$, then it must be $\frac{1}{mn}\le\frac{1}{n}$. It is a small observation and don't changes your idea. $\endgroup$ – sinbadh Feb 16 '16 at 18:27

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