Given a pair of strictly increasing functions $f,g:\mathbb{N}\to \mathbb{N}$
define:


$P\_N(f,g)\doteq \left(z\in \mathbb{C}\mapsto \prod\_{i=1}^{f(N)}\left(1+\frac{z}{v\_i(N)}\right)\in \mathbb{C}\right),$


where $v\_i:\mathbb{N}\to \mathbb{C}, |v\_i(N)|\geq g(N) \mbox{ for } i=1,2,\ldots,f(N);$


and such that $\lim\_{N\to\infty} \sum\_{i=1}^{f(N)}\frac{1}{v\_i(N)}=1.$ 


Define $S$ to be the set of functions $\varphi:\mathbb{C}\to \mathbb{C}$ such that there exists a sequence $(P\_N(f,g))\_{N\in\mathbb{N}}$ with some $f,g:\mathbb{N}\to \mathbb{N}$ strictly increasing functions such that $P\_N(f,g)$ converges compactly to $\varphi$ as $N\to\infty.$


Question: Can you describe $S$?



Infinite product's question
$S$ is the set of all entire zero-free functions $F$ with $F(0)=F'(0)=1$. To approximate such an $F$, just cut off its Taylor series $F(z)=1+z+\sum\_{n\ge 2} a\_n z^n$ at high enough degree $N\_1=f(1)$. Make sure this polynomial $p$ approximates $F$ well enough on $|z|\le 1=g(1)$ (say) so that it will be zero-free there. Since $p(0)=p'(0)=1$ also, its factorization is of the required form. Continue in this way, with $f(2)>f(1)$ and $g(2)>g(1)$ etc.


Conversely, any locally uniform limit of $P\_n(f,g)$'s is of course zero-free and satisfies $F(0)=F'(0)=1$.


