Suppose we have a sequence of entire functions $f\_n$ such that $$\text{$f\_n(z)\to0$ for each natural $z$}\tag{1}$$
(as $n\to\infty$).


Is it possible to give general additional conditions on the sequence $(f\_n)$ ensuring that (1) implies  

$$\text{$f\_n(z)\to0$ for each complex $z$?}\tag{2}$$


As a minimum, I would like such general additional conditions to hold and be easily verifiable when
$$f\_n(z)=-1+\frac1{n^z}\sum\_{k=0}^\infty k^z\,\frac{n^k}{k!}\,e^{-n};$$
cf. [this answer](https://mathoverflow.net/a/404024/36721).



On convergence of entire functions
Edited.


1. For your general question,
one sufficient condition is that your functions are of exponential type $<\pi$.
This is best possible since anything convergent to $\sin\pi z$ would be a counterexample, and $\sin\pi z$ has exponential type exactly $\pi$.


One can slightly improve this. For example when all $L^2(R)$ norms of your functions are bounded, independently of $n$. Then one can apply the Sampling theorem (which is usually credited to Nyquist in the West and Kotelnikov in the former Soviet Union, but in fact goes back to Cauchy), which estimates the
$L^2(R)$ norm in terms of $\ell^2(Z)$ norm, provided that the exponential
type is $\leq\pi$.


2. However your specific function does not have finite exponential type,
so these general results will not work, and you need to use some specific properties of your function. You say that yout functions satisfy $f\_n(z)\to 0$ for every integer $z$. Why is this so, and why your proof does not work for non-integer $z$?


