Zeros of $f_n(z)=1+\frac{1}{z}+\frac{1}{2!z^2}+...+\frac{1}{n!z^n}$ are in $B_{\varepsilon}(0)$ I want to prove that for every $\varepsilon >0$ there is a $N\in\mathbb{N}$ so that for every $n\ge N$ all zeros of $$f_n(z)=1+\frac{1}{z}+\frac{1}{2!z^2}+...+\frac{1}{n!z^n}$$  are in $B_{\varepsilon}(0).$
This should be done by using Rouché's theorem (symmetric version) I think.
Has anyone an idea how I can apply that theorem in an useful way?

First note that the problem is equivalent to proving that for every $\epsilon>0$, $\exists N$ such that for $\forall n\geq N$,$f_n(\dfrac{1}{z})=P_n(z)$ has all zeroes outside $B(0,\epsilon)$. Assume on the contrary the existence of such an $\epsilon.$ Then for each $P_n$, $\exists z_n \in B(0,\epsilon)$ such that $P_n(z_n)=0.$ Observe that $P_n \to f$, where $f(z)=e^z$. Now, replacing $z_n$ by a convergent subsequence if necessary, we assume that $z_n \to z$ and obtain $P_n(z_n) \to f(z)$, i.e, $e^z=0$, which is the  required contradiction.
