Show that the power series $\sum_{n=0}^\infty \frac{z^n}{n!}$ converges uniformly for all $z$.

Question

Show that the power series $\sum_{n=0}^\infty \frac{z^n}{n!}$ converges uniformly for all $z$.

I know that the definition for uniform convergence is: We say ${𝑓_𝑛}$ converges to 𝑓 uniformly on a subset 𝑆 of Ω iff for every $𝜖>0$ there is an $𝑛_0∈ℕ$ so that if $𝑛>𝑛_0$ then $|𝑓_𝑛(𝑧)−𝑓(𝑧)|<𝜖$ for all $𝑧∈𝑆$.

BUT I'm not sure if I can use it the same for a power series or if there are different rules with that. Can I?
Also, what would the f(z) be for this? I get that $f_n(z)=\frac{z^n}{n!}$, but how do I get $f(z)$?

Answer 1

When considering series of functions such as $z \mapsto \sum_{n=0}^\infty g_n(z)$, you take $f_n$ as the partial sum : $$ f_n(z) = \sum_{k=0}^n g_k(z). $$

However, the exponential series does not converge uniformaly on all $\mathbb C$. It does however uniformly converges on every bounded subset of $\mathbb C$. Indeed, if $|z| \leq M$ :

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

and the series $\sum_n \frac{M^n}{n!}$ converges (its sum is $e^M$).

Is that clearer?