I was hoping someone could explain to me how to prove a sequence is Cauchy. I've been given two definitions of a Cauchy sequence:

$\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $n,m> N$ $\Rightarrow |a_n - a_m| ≤ \epsilon$

and equivalently $\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $n> N$ $\Rightarrow |a_{n+p} - a_n| ≤ \epsilon$, $\forall p \in \mathbb{N}$

I understand that proving a sequence is Cauchy also proves it is convergent and the usefulness of this property, however, it was never explicitly explained how to prove a sequence is Cauchy using either of these two definitions. I'd appreciate it if someone could explain me how to prove a sequence is Cauchy perhaps $a_n = \sqrt{n+1} - \sqrt{n}$ ? or another example just for me to grasp the concept.

  • $\begingroup$ I have shown an example of how to use the definition of a Cauchy sequence. I changed the sequence to an easier one (to be honest because the one you suggested looked like a mess). In your particular example you should just prove directly that the limit is zero. $\endgroup$ – I. Cavey Dec 27 '15 at 18:33
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    $\begingroup$ As with all epsilon-N proofs, there's no magic method, although the example given by @I.Cavey is a good one. In my experience, it is typically some trickery due to the fact that n,m>0. As a side note, proving that a series is convergent (by typical epsilon-N definition of convergence) also suffices since convergent implies cauchy (and vice versa.) $\endgroup$ – Andres Mejia Dec 27 '15 at 20:14
  • $\begingroup$ Thank you all for your help, I think I've got it now $\endgroup$ – CrossGuard Dec 27 '15 at 21:57

As an easier example of how to apply the definition of a Cauchy sequence, define the sequence $\{\frac{1}{n}\}$. Given any $\epsilon>0$, you would like to find an $N$ such that for any $n,m>N$, $\left|\frac{1}{n}-\frac{1}{m}\right|<\epsilon$. This would certainly be the case if $\frac{1}{n},\frac{1}{m}<\frac{\epsilon}{2}$ since $\left|\frac{1}{n}-\frac{1}{m}\right|\leq\left|\frac{1}{n}\right|+\left|\frac{1}{m}\right|$. Therefore you need to force $n,m>\frac{2}{\epsilon}$. Conveniently enough, you are allowed to choose whichever $N$ you like, so choosing $N>\frac{2}{\epsilon}$ will do the trick. The above constitutes the work you do beforehand, now the proof.

Claim: The sequence $\{\frac{1}{n}\}$ is Cauchy.

Proof: Let $\epsilon>0$ be given and let $N>\frac{2}{\epsilon}$. Then for any $n,m>N$, one has $0<\frac{1}{n},\frac{1}{m}<\frac{\epsilon}{2}$. Therefore, $\epsilon>\frac{1}{n}+\frac{1}{m}=\left|\frac{1}{n}\right|+\left|\frac{1}{m}\right|\geq\left|\frac{1}{n}-\frac{1}{m}\right|$. Thus, the sequence is Cauchy as was to be shown.

  • $\begingroup$ Everything you wrote is correct, but I think your point would be better illustrated by $a_n = \frac{(-1)^n}{n}$. $\endgroup$ – Pedro M. Dec 27 '15 at 20:13

For the particular example you chose, it is very easy to show directly that it converges to zero, because $$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}}.$$ Nevertheless, this same identity allows you to show that it is Cauchy, since $$\left|\frac{1}{\sqrt{m+1} + \sqrt{m}} - \frac{1}{\sqrt{n+1} + \sqrt{n}}\right| \leq \left| \frac{1}{2\sqrt{m}} - \frac{1}{2\sqrt{n}} \right| \leq \frac{1}{2\sqrt{\min(m,n)}}< \epsilon$$ whenever $\min(m,n) \geq N > \frac{1}{4\epsilon^2}$.


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