I am interested in approximating the function $1-e^{-z}$. I want to approximate it by a family of functions $f_n$ that are entire and $f_n(2\pi k i) = 0$ for all $2\pi k i$ with $|k|\le n$ and $f_n(2\pi k i) \ne 0$ for all $2\pi k i$ with $|k|> n$. Is this possible?
When I say that the sequence $f_n$ approximates $1-e^{-z}$, I mean that there exists a sequence of functions as described and an open set $D$ such that $\lim_{n\to\infty}f_n(z) = 1-e^{-z}$ uniformly in $D$.
So far I have succeeded only on the second part.
We have that for any $z$ such that the left hand side makes sense, it holds that $$ \frac{1}{1-e^{-z}} = \frac12+\frac1z + \sum_{k\ge0}\frac{2z}{z^2+4\pi^2 k^2}.$$ So if we define $$ f_n(z) = \frac12+\frac1z + \sum_{k=0}^n\frac{2z}{z^2+4\pi^2 k^2} $$ then clearly $$ \lim_{n\to\infty} f_n(z) = \frac{1}{1-e^{-z}}. $$
This implies that $$ \lim_{n\to\infty} \frac{1}{f_n(z)} = 1-e^{-z}. $$
However these functions have poles which do not seem to escape to infinity as $n\to\infty$, so I cannot really use them.
EDIT: As WimC noted, given my requirements, $1-e^{-z}$ is a good approximation of itself, so I changed it to what I actually need.
EDIT2: The solution to this is given by the Weierstraß product theorem, as suggested by Daniel Fischer. We get $$ \lim_{n\to\infty} e^{-\frac{s}{2}}s \prod_{k=1}^n\left(1+\frac{s^2}{4\pi^2 k^2}\right) = 1-e^{-s}. $$