Question: Show that for all $r > 0$, there exists $N \in \mathbb{N}$ such that for $n \geq N$, the polynomial $ 1 + z + \frac{z^2}{2!} + \ldots + \frac{z^n}{n!} $ has all its zeros outside the closed disc $\overline{D(0, r)}$.
I have some doubts in this solution,
Solution:
As $\exp$ is nowhere zero on $\overline{D(0; r)}$, define $\varepsilon = \min\{|\exp(z)| : z \in \overline{D(0; r)}\} > 0$.
From uniform convergence of $1 + z + \frac{z^2}{2!} + \ldots + \frac{z^n}{n!}$ on $\overline{D(0, r)}$, one obtains that there exists $N \in \mathbb{N}$ such that, for all $n \geq N$ and $z \in \overline{D(0; r)}$,
$ || {1 + z + \frac{z^2}{2!} + \ldots + \frac{z^n}{n!}} | - \exp(z)| \leq | {1 + z + \frac{z^2}{2!} + \ldots + \frac{z^n}{n!}} - \exp(z) | < \frac{\varepsilon}{2}. $
This implies that, for all $n \geq N$ and $z \in \overline{D(0, r)}$,
$ | {1 + z + \frac{z^2}{2!} + \ldots + \frac{z^n}{n!}} | > |\exp(z)| - \frac{\varepsilon}{2} > 0. $
Therefore, it has all zeros outside the closed disc $\overline{D(0, r)}$
Doubts:
How can the uniform convergence of this polynomial on the closed disk can be proved?
Is $\frac{\varepsilon}{2}$ stating some logic or I can take it like $\frac{\varepsilon}{5}$, $\frac{\varepsilon}{4}$, etc.