I have stumbled into this problem, without a given answer.
Let $f$ be an entire function such that $F(z) = \lim\limits_{n\to\infty} f^{(n)}(z)$ exists $\forall z \in \mathbb{C}$ with local uniform convergence.
- What can you say about the function $F$?
- What can you say about the function $f$?
I have sort of convinced myself that $F(z) =Ce^z$ and thus $f(z)=F(z)$ but i am very doubtful about this and even if it is correct I have no idea how to prove it, and there is probably more information you have to provide about the given functions. Such that $F$ is analytic implies that $f$ is analytic.
As already worked out in the comments, $F(z) = \lim_{n\to\infty} f^{(n)}(z)$ (locally uniformly) implies that $$ F'(z) = \lim_{n\to\infty} f^{(n+1)}(z) = F(z) $$ so that $F(z) = Ce^z$ for some constant $C \in \Bbb C$.
Then $g(z) = f(z) - Ce^z$ satisfies $$ \lim_{n\to\infty} g^{(n)}(z) = \lim_{n\to\infty} f^{(n)}(z) - Ce^z = F(z) - Ce^z = 0 $$ so that it remains to characterize all entire functions $g$ with the property that $$ \lim_{n\to\infty} g^{(n)}(z) = 0 $$ locally uniformly in $\Bbb C$. Writing $g$ as a power series $g(z) = \sum_{k=0}^\infty \frac{b_k}{k!} z^k$ we have the necessary condition $$ \lim_{k\to\infty} b_k = \lim_{k\to\infty} g^{(k)}(0) = 0 \,. $$ That condition is also sufficient: If $b_k \to 0$ then for $|z| \le R$ $$ \left| g^{(n)}(z) \right| = \left| \sum_{k=0}^\infty \frac{b_{k+n} }{k!} z^k\right| \le \sum_{k=0}^\infty \frac{|b_{k+n}| }{k!} R^k \, . $$ Given $\epsilon > 0$ we can choose $N$ such that $|b_n| < \epsilon e^{-R}$ for $n > N$, which implies that $$ \left| g^{(n)}(z) \right| \le \epsilon e^{-R} \sum_{k=0}^\infty \frac{1}{k!} R^k = \epsilon $$ for $n > N$ and $|z| \le R$.
Summarizing the results: If $f$ is a an entire function then $(f^{(n)})$ converges locally uniformly in $\Bbb C$ if and only if $$ f(z) = Ce^z + \sum_{k=0}^\infty \frac{b_k}{k!} z^k $$ for some $C \in \Bbb C$ and some sequence $(b_k)$ of complex numbers converging to zero. In that case the limit function is $F(z) = Ce^z $.