Prove that the sequence {$n^{\frac{1}{2}}$} diverges where $n\in \mathbb N$

Question

My attempt :

I have tried to show by contradiction: let $L$ be the limit of the sequence. Then
$-e<n^{\frac{1}{2}}- L <e$ for all $n>N$ and $e> 0$ which implies that
$ (L-e)^2< n<(L+e)^2$ . However since $(L+e)^2 $ is a real number we can always choose infinitely many natural numbers which are greater than it and for which the equation doesn't hold. Am I correct?

Answer 1

Almost right but you cannot square an inequality involving negative numbers. So forget about the left hand in equality and look at $n^{1/2} <L+\epsilon$. You have to first prove that $L \geq 0$ so that you can square this in equality and get a contradiction as you have done.

Answer 2

Your idea is correct, but it is more natural to note that $\left(\sqrt n\right)_{n\in\mathbb N}$ is unbounded, whereas every convergent sequence is bounded.