For $s\in\mathbb{C}$, let $S(s)$ be a series that checks for the Hurwitz Theorem,
If $S(s)=0$, then by Hurwitz theorem, there exists a sequence $s_n$ so that $s_n \xrightarrow[n\to \infty]{} s$ verifying $S_n(s_n)=0$ for all $n$,
where $S_n$ is the truncation of the $n$-th order of $S$.
At each $n$, $|S_n(s_n)|^2=0$ implies that for some sequences $a_n(s)$ and $b_n(s)$, that both converge to reals: $$a_n(Re(s_n))/n^{\alpha_n}-b_n(Re(s_n))/n^{\alpha_n}=0$$where $\alpha_n$ is strictly positive for all $n$ and converges to a strictly positive real, which in turn implies: $$a_n(Re(s_n))=b_n(Re(s_n))$$ for all $n$,
My question is, in these conditions, does $S(s)=0$ imply that: $$\lim_{n\to\infty} a_n(Re(s_n))= \lim_{n\to\infty} b_n(Re(s_n))$$ Or is the actual necessary condition the way looser condition: $$\lim_{n\to\infty} a_n(Re(s_n))/n^{\alpha_n}-b_n(Re(s_n))/n^{\alpha_n}=0$$
?
