# Square root of a divergent series diverges.

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

• I think you want your counterexamples to be $1/n^2$ and $1/n$? Nov 21, 2014 at 23:12
• Well, if you use $1/n^2$ you get the series to of $1/n$ and if you use $1/n^4$ you get it to be of $1/n^2$. And since $1/n$ does not converge it wouldn't work as a counterexample Nov 21, 2014 at 23:15
• If $\sum a_n$ diverges, then $\sum\sqrt{a_n}$ diverges as well, but $\sum\sqrt{a_n}/n$ can converge or diverge. Nov 21, 2014 at 23:18
• could you explain why? Nov 21, 2014 at 23:19
• I suppose it depends which statement you're looking for a counterexample to! Nov 21, 2014 at 23:20

(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.
• (1) If $a_n \to 0$, then $0 \le a_n < 1$ for all $n > N$ for some $N$. (2) If $0 < a_n < 1$ then $\sqrt{a_n} > a_n$ (since also $0 < \sqrt{a_n} < 1$). Nov 22, 2014 at 0:06
• @martycohen: Technically, you only get $\ge$, not $>$, since $a_n = 0 \implies a_n \to 0$. Nov 22, 2014 at 1:22