Editing Uniform convergence (section)

=== Exponential function ===
The series expansion of the [[exponential function]] can be shown to be uniformly convergent on any bounded subset <math>S \subset \C</math> using the [[Weierstrass M-test]].

'''Theorem (Weierstrass M-test).''' ''Let <math>(f_n)</math> be a sequence of functions <math>f_n:E\to \C</math> and let <math>M_n </math> be a sequence of positive real numbers such that <math>|f_n(x)|\le M_n</math> for all <math>x\in E</math> and <math>n=1,2, 3, \ldots</math> If <math display="inline">\sum_n M_n</math> converges, then <math display="inline">\sum_n f_n</math> converges absolutely and uniformly on <math>E</math>.''

The complex exponential function can be expressed as the series:

:<math>\sum_{n=0}^{\infty}\frac{z^n}{n!}.</math>

Any bounded subset is a subset of some disc <math>D_R</math> of radius <math>R,</math> centered on the origin in the [[complex plane]]. The Weierstrass M-test requires us to find an upper bound <math>M_n</math> on the terms of the series, with <math>M_n</math> independent of the position in the disc:

:<math>\left| \frac{z^n}{n!} \right|\le M_n, \forall z\in D_R.</math>

To do this, we notice

:<math>\left| \frac{z^n}{n!}\right| \le \frac{|z|^n}{n!} \le \frac{R^n}{n!}</math>

and take <math>M_n=\tfrac{R^n}{n!}.</math>

If <math>\sum_{n=0}^{\infty}M_n</math> is convergent, then the M-test asserts that the original series is uniformly convergent.

The [[ratio test]] can be used here:

:<math>\lim_{n \to \infty}\frac{M_{n+1}}{M_n}=\lim_{n \to \infty}\frac{R^{n+1}}{R^n}\frac{n!}{(n+1)!}=\lim_{n \to \infty}\frac{R}{n+1}=0</math>

which means the series over <math>M_n</math> is convergent. Thus the original series converges uniformly for all <math>z\in D_R,</math> and since <math>S\subset D_R</math>, the series is also uniformly convergent on <math>S.</math>