# Show $\{x_n = \sqrt{n}\}$ is not Cauchy sequence

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?

• 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 – Stan Tendijck Mar 11 at 19:32

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.

• 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". – Jair Taylor Mar 11 at 19:57
• @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. – szw1710 Mar 11 at 23:43
• 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. – Jair Taylor Mar 11 at 23:45
• More importantly, this answer relies on the completeness of (and construction of) $\mathbb{R}$ which is not trivial. But of course it is correct. – Jair Taylor Mar 11 at 23:47

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.

• so if I chose n to be 1/4 or 1/8 would either of those work? – user597188 Mar 11 at 20:36
• Uh, what? $n$ is an integer. And, for the anti-Cauchy criterion we're trying to prove, it has to be greater than $m$. – jmerry Mar 11 at 21:00
• sorry I meant like n/2$\sqrt{m}$<1/4 or 1/8 – user597188 Mar 11 at 21:07
• 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. – jmerry Mar 11 at 21:11
• 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? – user597188 Mar 11 at 21:18

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.