How to prove $f_n(z)=\sum_{j=0}^n\frac{z^j}{j!}$ doesn't converge uniformly to $e^z$ on $\Bbb{C}$?

Question

We showed in class that the functions $f_n(z)=\sum_{j=0}^n\frac{z^j}{j!}$ converge uniformly to $e^z = \sum_{n=0}^\infty\frac{z^n}{n!}$ in any disc $D(0;R)$ such that $R>0$. But does $(f_n)$ converge uniformly to the exponential function in all $\Bbb{C}$?

My first guess is "no", since $e^z$ is not bounded on $\Bbb{C}$, but I need some help (hints preferentially, not a complete proof please) on how to prove that. Thanks in advance!

Answer 1

Your guess is good, but the reason is not, since polynomials are also unbounded on $\mathbb C$.

In order to prove that the sequence does not prove uniformly, you need to prove that there exists some $\epsilon > 0$ such that for all $N\in\mathbb N$, there exists a $n>N$ and a $z\in\mathbb C$ such that $|f_n(z) - e^z| > \epsilon.$

In order to do that, think about the limit of $$\lim_{x\to\infty} \frac{f_n(x)}{e^x}$$ where the limit is taken over the reals.

1. First of all, what is this limit equal to?
2. Second of all, how can you use this?

Answer 2

Here is a hint: first think about the real exponential. In this case, given any polynomial $p(x)$, we can find values of $x$ for which $|e^x-p(x)|$ is arbitrarily large.