I imagine this to be a very classical question in complex analysis:


Consider the Hadamard product


$$g(\mu) = \prod\_{n=1}^{\infty}E\_1(\mu z\_n),$$
where $E\_1(z):=(1-z)e^z$ is the first elementary factor
for some sequence $z\_n \to 0$ fast enough, such that $g$ is entire. In fact, choosing $ (z\_m)\_{m \in \mathbb N}$ to be square-summable is sufficient.


I wonder if there are any estimates on the Taylor coefficients of $g$ known?


I can only think of the following very pedestrian approach:
Using the elementary inequality $$\left\lvert (1-w)e^w \right\rvert \le e^{\vert w \vert^2/2},$$


we can estimate
$$\vert g(\mu) \vert \le e^{\frac{\vert \mu^2\vert}{2} \sum\_{m=0}^{\infty} \vert z\_m \vert^2}=:h(\mu).$$


This implies for the power series $g(\mu) = \sum\_n a\_n \mu^n$ that
$\vert a\_n \vert \le r^{-n} \sup\_{\mu = r} h(\mu)$ for any $r.$


Optimizing over $r$ we find that
$$\vert a\_n \vert \le \frac{ e^{n/2}\Vert (z\_m)\_{m \in \mathbb N} \Vert^n\_{\ell^2}}{n^{n/2}}.$$


I am wondering if there is a systematic way to improve these estimates of $\vert a\_n\vert$ (apart from some obvious improvements such as $g'(0)=0$.



Taylor coefficients of Hadamard product
I doubt this bound can be improved much, at least for even $n$. Indeed: set $z\_1 = z\_2 = \ldots = z\_k = k^{-1/2}$ and $z\_{k+1} = z\_{k+2} = \ldots = 0$, so that the $\ell^2$ norm of $(z\_n)$ is $1$. (Intuitively, this is the worst-case scenario.) Then
$$ g(\mu) = (E\_1(k^{-1/2} \mu))^k = (1 - k^{-1/2} z)^k e^{z \sqrt k} .$$
As $k \to \infty$, the above functions converge to $e^{-z^2/2}$ locally uniformly on $\mathbb C$, and so the $n$th Taylor coefficient of $g$ converges to the $n$th Taylor coefficient of $e^{-z^2/2}$, which is equal to
$$ \frac{1}{2^{n/2} (n/2)!} \approx \frac{e^{n/2}}{n^{(n+1)/2} \sqrt{\pi}} $$
when $n$ is even. Compared to your "pedestrian" bound, this is off by a factor $n^{-1/2}$.


My guess would be that things are not different for odd $n$, but I did not think about it.


