How do I prove the sequence $s_{n}=\sqrt{n}$ diverges?

Question

I think this needs to be done as a counterexample:

Assume $s_{n}$ converges to $s$. Then $|s_{n} - s|<\epsilon$ for all $n>N$. And then I get stuck...

Answer 1

Well, if you are determined to use a counter example, then

$|s_{n} - s|<\epsilon$ so $s_n^2 - 2s_n*s + s^2 = n + s - 2s_n*s < \epsilon^2$

$|s_n| < |s - \epsilon|$ so $n + s \pm(s - \epsilon)*2*s < \epsilon^2$

$n <e^2 - s \mp (s - \epsilon)*2*s = V$ for some finite and constant V. So for all n > N then n < V which is ....

Answer 2

hint: Show that it is unbounded