Uniform Convergence of the Complex Exponential Function

Question

Here is an excerpt from a textbook I am reading:

The prime example of a power series is the complex exponential function, which is defined for $z \in \Bbb{C}$ by $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$ ...... the series above converges absolutely for every $z \in \Bbb{C}$. To see this, note that $\left|\frac{z^n}{n!} \right| = \frac{|z|^n}{n!}$, so $|e^z|$ can be compared to the series $\sum |z|^n/n! = e^{|z|} < \infty$. In fact, this estimate shows that the series defining $e^z$ is uniformly convergent in every disc in $\Bbb{C}$.

So, I see how the estimate shows that $\sum_{n=0}^\infty \frac{z^n}{n!}$ converges absolutely for a fixed $z \in \Bbb{C}$, but I don't see how uniform convergence follows. The author is clearly appealing to some theorem that I can't recall at the moment. Initially I thought it was the Weierstrass $M$-test, but this doesn't seem right; perhaps I am applying incorrectly.

Answer 1

The function is uniformly convergent in any disc, and in fact it is even uniformly convergent in any bounded subset of $\mathbb C$. Indeed, if $|z| \leq R$,

$$ \left|\frac{z^n}{n!}\right| \leq \frac{R^n}{n!} $$

and the series $\sum_n \frac{R^n}{n!}$ is convergent (of sum $e^R$).

Answer 2

Consider $a_n=\frac{1}{n!}$. First of all the radius of convergence, $R=\infty$, where $\frac{1}{R}=\limsup_{n\to \infty} |a_n|^{\frac{1}{n}}$. (in our case, we can easier calculate that as the limit of $|a_n/a_{n+1}|).$ Now, consider any closed disk $D(0,r)$, $r>0$, $z$ with $|z|\leq r$ and $r'>r$ (and less than $R$ in general). Then, eventually $|a_n|^{\frac{1}{n}} < \frac{1}{r'}$ ( by the definition of $R$. Follows that $|a_n z^n| < \big( \frac{|z|}{r'} \big)^n< \big( \frac{r}{r'} \big)^n$, eventually (where $\frac{r}{r'}<1).$ Now you just apply the M-test and gives you the uniform convergence on the closed disk $D(0,r)$.