Suppose that $f$ is holomorphic on all of $\mathbb{C}$ and that $$\lim_{n \rightarrow \infty}(\frac{\partial}{\partial z})^{n}f(z)$$
exists, uniformly on compact sets, and that this limit is not identically zero. Then the limit function $F$ must be a very particular kind of entire function. Can you say what kind ?
$\text{Conjecture}$
Functions of an order that are at most $1$, satisfy the following condition:
$$\lim_{n \rightarrow \infty}(\frac{\partial}{\partial z})^{n}f(z) = \Psi$$
In order to prove our Conjecture, one must construct an entire function that is at most of order of 1, and show that the $\lim_{n \rightarrow \infty}f(z)$ exists, and converages to some constant $\Psi$ the construction of our entire function can be seen in $\text{Lemma (1.1)}$
$\text{Lemma (1.1)}$
For arbitrary positive numbers $p$ and $\sigma$ one can construct an entire function of order $p$ and type $\sigma$ using in $(1.2)$
$(1.2)$
$$f(z)=\sum_{n=1}^{\infty}(ep\sigma/n)^{n/p}z^{n}$$
$\text{Lemma (1.2)}$
To attack our Conjecture, we substitute $(1.2)$ into $(1.1)$ the we have the following in $(1.3)$
$(1.3)$
$$\lim_{n \rightarrow \infty}(\frac{\partial}{\partial z})^{n}\sum_{n=1}^{\infty}(ep\sigma/n)^{n/p}z^{n}=\Psi$$
$\text{Remark}$
One performs the substitution to make clear one is trying to show the condition $\lim_{n \rightarrow \infty}(\frac{\partial}{\partial z})^{n}f(z) = \Psi$ is true
Since $(\frac{\partial }{\partial z})^{n}=(\frac{d}{dz})^{n}$, one can perform in $(1.4)$
$(1.4)$
$$\lim_{n \rightarrow \infty}(\frac{d}{d z})^{n}\sum_{n=1}^{\infty}(ep\sigma/n)^{n/p}z^{n}=\Psi.$$
$\text{Remark}$
At this stage of the proof, one wonders why didn't one just take $(\frac{\partial }{\partial z})^{n}\sum_{n=1}^{\infty}(ep\sigma/n)^{n/p}z^{n}$, one could that relies on a certain criteria being met which may not work for certain cases such as this one
$\text{Lemma 3}$
Changing $f(z)$ into a product representation, our recent developments now are as follows, in $(1.5)$
$(1.5)$
$$\lim_{n \rightarrow \infty}(\frac{d}{d z})^{n}\text{exp}\sum_{n=1}^{\infty}(ep\sigma/n)^{n/p}z^{n}=\Psi.$$
$$\lim_{n \rightarrow \infty}(\frac{d}{d z})^{n}\prod_{n=1}^{\infty} \exp (ep\sigma/n)^{n/p}z^{n}= \Psi.$$
$\text{Remark}$
The conservation, from sum to product was conducted by the following formula:
$$\exp \sum s_n = \prod e^{s_n}.$$
From $\text{Lemma 3}$ it seems like one can exploit the Logarithmic derivative of the product $\prod_{n=1}^{\infty} \exp (ep\sigma/n)^{n/p}z^{n}$, which at every point where $\prod_{n=1}^{\infty} \exp (ep\sigma/n)^{n/p}z^{n}$ is non-zero one would have $\lim_{n \rightarrow \infty} \sum_{n=1}^{\infty}\frac{a'n(z)}{1+a_{n}(z)}$. So far is my appoarch correct I feel what I have so far is not rigors, perhaps someone could help tighten my reasoning ?
Let $f_n = f^{(n)}$, and assume that $f_n$ converges locally uniformly to some $F$ (which is necessarily entire as well).
Then $f_n'$ converges locally uniformly (since $f_n' = f_{n+1}$) to the same function $F$. By general theory for uniform convergence, the limit of $f_n'$ is actually the derivative of $f$, i.e. $F = F'$.