How do you show that the exponential is not uniformly convergent on the entire complex plane? I want to show that the series $\sum_{z=0}^\infty\frac{z^n}{n!}$ is not uniformly convergent on the entire complex plane.  I understand that we just need to show that the supremum norm of the difference $\sup_{z \in \mathbb{C}} |\sum_{n=k}^\infty \frac{z^n}{n!}| $ does not go to $0$ as $k \to \infty$, but I'm not sure how to proceed rigorously. Of course, for every $k$ I can pick $z$ so that $|\frac{z^k}{k!}|$ is way bigger than $0$, but how can I be certain that the subsequent terms   are not going to cancel it out?
 A: Just restrict your attention to positive $z$.  Then all the terms will be positive and you don't have to worry about any cancellation.
Alternatively, actually, this is not even necessary: uniform convergence of a series implies that for any $\epsilon>0$, the terms must be eventually smaller than $\epsilon$ where the "eventually" is uniform.  The point is that the individual terms are just the differences between two consecutive partial sums which you know must be approaching the limit.  In detail, suppose that the series $\sum_{n=0}^\infty\frac{z^n}{n!}$ did converge uniformly.  Then you could find $N$ such that $\sum_{n=0}^k\frac{z^n}{n!}$ is within $\epsilon/2$ of $e^z$ for all $z$ and all $k\geq N$.  But this implies that for all $n>k$, $|\frac{z^n}{n!}|$ must be at most $\epsilon$, since it is the difference between two partial sums that are each within $\epsilon/2$ of $e^z$.  That is, $|\frac{z^n}{n!}|\leq \epsilon$ for all $z$ and all $n>N$.  But you know this is not possible since for any $n$ you can find $z$ such that $|\frac{z^n}{n!}|$ is large.
