If the sum of $f(n)$ diverge, then does the sum of $\sqrt{f(n)}$ diverge? Lets say that $$\sum_{n=a}^\infty f(n)$$ diverges.
Does $$\sum_{n=a}^\infty \sqrt{f(n)}$$
necessarily diverge?
 A: Suppose $\sum f(n)$ diverges. If $f(n)$ is unbounded or does not vanish, the result is immediate, so assume it is bounded and vanishing.
Observe that there exists some $k>0$ such that $f(n)<1$ for all $n>k$. This implies that $\sqrt{f(n)}>f(n)$ for any $n>k$. Since we know $\sum f(n)$ diverges, so must this series whose tail is composed of larger terms.
EDIT NOTE: This argument assumes that all terms stay in $\mathbb{R}$
A: For the case when $f(n)\ge 0\forall n$, simply observe that $$\sum_{n\ge 0}f(n)\le \left(\sum_{n\ge 0}\sqrt{f(n)}\right)^2$$ which implies divergence of $\sum_{n\ge 0}\sqrt{f(n)}$ whence $\sum_{n\ge 0}f(n)$ diverges.
A: There are two cases:
1) $f:\mathbb{N} \to \mathbb{C}$. Then the answer is no. Let $f(n)=1/n$ for $n$ odd and $f(n)=1/n-\epsilon i/n^2$. The sum of $f(n)$ obviously diverges but the sum of the square roots (being careful to use one branch of the root) will become, as $\epsilon \to 0$, $\sum \frac{(-1)^n}{\sqrt{n}}$ which is an alternating series.
2) $f:\mathbb{N} \to \mathbb{R}$. Then the answer is yes. The fact that $\sum f(n)$ diverges means that at least one of $A=\sum_{I_1}f(n)$ or $B=\sum_{I_2}f(n)$, where $A,B$ are the sums of the positive or negative terms of $f(n)$ respectively, diverge. WLOG $B$ diverges.
The sum $C=\sum_{I_2}\sqrt{f(n)}=\sum_{I_2}\sqrt{|f(n)|}i$ will then also diverge. Indeed, suppose it converges. Then eventually $|f(n)| \to 0$ for $n \in I_2$. Then $\sum_{I_2}f(n)<\sum_{I_2}\sqrt{|f(n)|}$ converges, contradiction (where in this sum it is implicit we have taken it over all sufficiently large $n$).
