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?
Best Answer
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).
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.