When I was working on a solution to this question I came across this type of series, $$ g(x) = \sum_{n=1}^\infty \frac{x-a_n}{(x-a_n)^2 + b_n^2} $$ where $a_n,b_n$ represent the real and imaginary parts of the zeroes of an entire function with growth order $\le 1$, which I believe is equivalent to $a_n^2 + b_n^2 \in o(n^{2})$. (I am also assuming $b_n\ne0$ for all $n$.) Under this assumption, I strongly suspect that $g(x)$ must be negative somewhere in $\mathbb{R}$, because each summand individually is negative for $x$ far enough negative, and it seems like in order to cancel out this negative part, the points $a_n+i b_n$ would have to be too densely spaced to satisfy $a_n^2+b_n^2 \in o(n^2)$, however I can't seem to prove this suspicion.
Is this true? If not, what is a counterexample?