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Consider the sequence {$x_n$},$x_n$=$\sqrt{n}$

Show that $\forall \varepsilon > 0, \exists n_0 \in \Bbb N$ s.t. $\forall n \geq n_0$, |$x_{n+1}-x_n$|<$\varepsilon$.

This is what I have:

Let $\varepsilon>0$ and let $n_0$ = $(\frac{1}{2\varepsilon})^2, \forall n \geq n_0$. $|\sqrt{n+1}-\sqrt{n}|$. So, $|\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})}|$.

Then, $|\frac{1}{\sqrt{n+1}}$ + $\sqrt{n}|$ $\leq$ $|\frac{1}{\sqrt{n}+\sqrt{n}}|$=$|\frac{1}{2\sqrt{n}}|$= $\varepsilon$.

Therefore, $|x_{n+1}-x_n|<\varepsilon, \forall n \geq n_0$

After I did all this it got me thinking can you prove {$x_n$} is not cauchy? if so how?

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  • $\begingroup$ You made a typo, $|\frac{1}{\sqrt{n+1}} + \sqrt{n}|$ probably needs to be $|\frac{1}{\sqrt{n+1} + \sqrt{n}}|$. Your proof is correct. Note that for Cauchy you need that for any $m,n\geq n_0$: $|x_n-x_m|<\epsilon$ which is not the case $\endgroup$ – Stan Tendijck Mar 11 at 19:32
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Any Cauchy sequence converges. Since $\sqrt{n}\to+\infty$, it diverges, so it is not Cauchy.

Telling about the condition in question, we have $$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0,$$ so it holds trivially by definition of a limit.

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    $\begingroup$ You forgot "in a complete metric space" after "any Cauchy sequence converges." I am guessing that they have not proven that $\mathbb{R}$ is complete and they will have to prove the non-Cauchy property "by hand". $\endgroup$ – Jair Taylor Mar 11 at 19:57
  • $\begingroup$ @JairTaylor, If the metric os not specified, usually one thinks about the Eucldean one, which is, of course, complete. In OP's question we have $|x_{n+1}-x_n|$, which clearly indicates the Euclidean metric. Please show me, where OP specifies another metric. $\endgroup$ – szw1710 Mar 11 at 23:43
  • $\begingroup$ Of course, it is implied in this case the metric is Euclidean. But it is important to note that Cauchy sequences do not necessarily converge in other metrics. $\endgroup$ – Jair Taylor Mar 11 at 23:45
  • $\begingroup$ More importantly, this answer relies on the completeness of (and construction of) $\mathbb{R}$ which is not trivial. But of course it is correct. $\endgroup$ – Jair Taylor Mar 11 at 23:47
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You have a solid argument for the main part, aside from the typo $|\frac1{\sqrt{n+1}}+\sqrt{n}|$ in the second line (already noted in a comment).

After I did all this it got me thinking can you prove $\{x_n\}$ is not Cauchy? If so, how?

The definition of a Cauchy sequence refers not only to the difference between adjacent pairs of elements, but also the difference between pairs of elements any index distance apart - as long as they're both far enough out in the sequence.

Just eyeballing it, the square roots increase without bound; we should be able to find one that exceeds any particular target. So, then, given some $m$, can you find some larger $n$ with $\sqrt{n}-\sqrt{m} > 1$? A rule to do this would immediately contradict the Cauchy criterion for $\epsilon=1$, and show the sequence isn't Cauchy.

Don't worry about being terribly precise here. We don't need the smallest $n$, just anything that works.

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  • $\begingroup$ so if I chose n to be 1/4 or 1/8 would either of those work? $\endgroup$ – user597188 Mar 11 at 20:36
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    $\begingroup$ Uh, what? $n$ is an integer. And, for the anti-Cauchy criterion we're trying to prove, it has to be greater than $m$. $\endgroup$ – jmerry Mar 11 at 21:00
  • $\begingroup$ sorry I meant like n/2$\sqrt{m}$<1/4 or 1/8 $\endgroup$ – user597188 Mar 11 at 21:07
  • $\begingroup$ I still doubt that's what you mean; there's no reason to divide $n$ and $m$. Now, in what I wrote, the $1$ is arbitrary. We could replace it by $1/4$, or by $40$. But, whatever we do, it has to be a constant, not depending on $m$ at all. $\endgroup$ – jmerry Mar 11 at 21:11
  • $\begingroup$ So if I let $\epsilon$ =1/4 id have (m-n)/( $\sqrt{m}$ + $\sqrt{n}$) $\geq$ n/(2$\sqrt{m}$) then $\sqrt{m}$/2 - n/(2$\sqrt{m}$) > 1/2-n/(2$\sqrt{m}$) >1/2-1/4 > 1/4 which is what epsilon equals but how would I do it if I let epsilon=1? $\endgroup$ – user597188 Mar 11 at 21:18
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Let $\epsilon_0=\frac13$. For all $n\in\mathbb{N}$ and $p=n$, $$ |\sqrt{n+p}-\sqrt n|=(\sqrt{2}-1)\sqrt n\ge\sqrt 2-1>\epsilon_0. $$ Namely, $\sqrt{n}$ is not Cauchy.

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