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$|\sqrt{n+p}-\sqrt{n}|<\epsilon$

Clearly, $|\sqrt{n+p}-\sqrt{n}|=\frac{p}{\sqrt{n+p}+\sqrt{n}} \leq p/\sqrt{n} \rightarrow 0$.

But by definition of a cauchy sequence, if we can choose $\exists N: l,m\geq N \Rightarrow |\sqrt{m}-\sqrt{l}|<\epsilon$, it is cauchy. If we take w.l.o.g. $m=n$ and $l=n+p=m+p$, $p\geq 0$, what is stopping this from being cauchy?

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    $\begingroup$ for any $n$, $\sqrt{n+p} - \sqrt{n} $ can be made as large as you like by making $p$ large. $\endgroup$ – Calvin Khor Apr 19 at 22:25
  • $\begingroup$ For fixed $p$, $\sqrt{n+p} - \sqrt{n}$ is $p$ times the difference quotient of the sequence $\sqrt{n}$ with $\Delta n = p$. So what you've found is that these differences go to $0$ with $n$, which is not surprising because the derivative of $\sqrt{x}$ is $1/ (2\sqrt{x})$ which also tends to $0$ with $x$. $\endgroup$ – Jair Taylor Apr 19 at 22:48
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Your proof assumes that $p$ is fixed.

Also, in $|\sqrt{n+p}-\sqrt{n}|=\frac{p}{\sqrt{n+p}+\sqrt{n}} \leq 1/\sqrt{n} \rightarrow 0 $, it is not true that $\frac{p}{\sqrt{n+p}+\sqrt{n}} \leq \frac1{2\sqrt{n}} $.

What is true is that $\frac{p}{\sqrt{n+p}+\sqrt{n}} \leq \frac{p}{\sqrt{n}} $.

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  • $\begingroup$ I meant $|\sqrt{n+p}-\sqrt{n}|=\frac{p}{\sqrt{n+p}+\sqrt{n}} \leq p/\sqrt{n} \rightarrow 0$, my bad. Doesnt this result then hold for all finite $p$, then and thus all numbers $m,l$ which are separated by some finite $p$? $\endgroup$ – BayesIsBae Apr 19 at 22:33
  • $\begingroup$ As I wrote, you assume that $p$ is fixed. If $p > \sqrt{n}$, the difference does not go to zero. $\endgroup$ – marty cohen Apr 19 at 22:35
  • $\begingroup$ ah ok, that makes sense now (specifically, with your $p\geq \sqrt{n}$ example)! $\endgroup$ – BayesIsBae Apr 19 at 22:37
  • $\begingroup$ For the sake of clarity to anyone who stumbles upon this question, I will state that my mistake seems to be that I mistook the nature of $l, m$. They are both fixed numbers (i.e., not some "function of n"). The sequence is called pseudocauchy math.stackexchange.com/questions/2791970/pseudo-cauchy-sequence, but as we can see - if we take fixed $l,m \geq N$ for any positive integer $N$, we can make the difference as large as we like. $\endgroup$ – BayesIsBae Apr 19 at 22:55
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The difference scales up as $p$ increases so you can't make two terms which are far apart but beyond a point $n$ arbitrarily close.

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In order for a sequence to be cauchy, you have to be able to pick $\ell$ and $m$ arbitrarily - it must always be true, for all $\ell$ and $m$ beyond $N$, that $|f(\ell) - f(m)| < \epsilon$, not just ones you choose according to some rule.

Since you've restricted your choice of $m$ based on your existing choice of $\ell$, you haven't satisfied the requirements of cauchy convergence.

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  • $\begingroup$ If one can choose $l,m$ arbitrarily, there is some separation and this holds for all separations, no? $\endgroup$ – BayesIsBae Apr 19 at 22:37
  • $\begingroup$ it "holds", but it doesn't mean anything: since we've already picked $\epsilon$ by the time we reach picking $\ell$ and $m$, it's possible to have $\ell$ and $m$ arbitrarily far apart to the point where the $\sqrt\ell - \sqrt m > \epsilon$. Or even $\gg$ or $\ggg$. $\endgroup$ – Dan Uznanski Apr 19 at 22:41

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