[Math] Square root of a divergent series diverges.

Question

If we let $\sum a_n$ converges, we cannot say anything about convergence of $\sum \sqrt{a_n}$ The counter examples being $a_n = \frac{1}{n^2}$ and $a_n = \frac{1}{n^4}$

But what about if the origional series diverges. I think it is the case that the square root of a divergent series is divergent, just dont know how to prove. Going along with the same idea, how would I prove the same for the series of $\sqrt{a_n}/n$ i think that diverges as well, where $a_n$ converges

edit: where $a_n$ converges

Answer 1

(i) If $a_n \not\to 0$, then $\sqrt{a_n} \not\to 0$.

(ii) If $a_n \to 0$, then $\sqrt{a_n} > a_n$ eventually.