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 $\mathbb{R}$