Intuitively, $\sqrt{n}$ is not convergent. However, $|\sqrt{n+p}-\sqrt{n}|<\epsilon, \forall \epsilon>0, p\geq 1$

Question

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

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?

Answer 1

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}} $.

Answer 2

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.

Answer 3

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.