I am looking for a different proof that ##(S_n) = \frac{1}{n}## is cauchy.(adsbygoogle = window.adsbygoogle || []).push({});

The regular proof goes like this (concisely):

##\left|\frac{1}{n} - \frac{1}{m} \right| \leqslant \left|\frac{m}{nm}\right| \ (etc...) \ <\epsilon ##

but I was thinking about an alternative proof. Is my proof correct:

let ##\epsilon > 0## by Archimedian property ##\exists N \ s.t. \frac{1}{N}<\epsilon##

This is equivalent to ##\frac{1}{N}<\frac{\epsilon}{2}## "may be ommited"

Now ##\forall n, m \geqslant N## we have by ##\Delta## ineq.

##\left|\frac{1}{n} - \frac{1}{m} \right| \leqslant \left|\frac{1}{n} \right| + \left|\frac{1}{m} \right|\leqslant \frac{1}{N}+\frac{1}{N} < ε##

What do you guys think? Thanks...

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# Cauchy Sequence

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