Alternative Proof of Cauchy Sequence ##\left(S_n\right) = \frac{1}{n}##

[Bachelier]

I am looking for a different proof that $(S_n) = \frac{1}{n}$ is cauchy.

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...

 

[kevinferreira]

I think it's perfectly ok.

 

[jbunniii]

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

I'm not sure what this means, but the rest is fine.

Another way to prove it is to note that $(s_n)$ converges to $0$ (easy proof), and use the fact that any convergent sequence is Cauchy.

Proof: if $(x_n)$ is a sequence which converges to some number $L$, then given $\epsilon > 0$, there is some $N$ for which $|x_n - L| < \epsilon / 2$ for all $n \geq N$. Therefore, if $m \geq N$ and $n \geq N$, then $|x_n - x_m| = |x_n - L + L - x_m| \leq |x_n - L| + |x_m - L| \leq \epsilon$.

 

[Bachelier]

I'm not sure what this means, but the rest is fine.

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

Well since ε in this case can be so small, then we can use a larger N > 1/ε (actually twice larger) to end up with ε/2+ε/2 in the end instead of 2ε.