I am currently working through the textbook "Complex Analysis" by Freitag and Busam.

Proposition III.2.1 (Convergence Theorem for Power Series) reads :

For each power series $$ a_{0} + a_{1} z + a_{2} z^{2} + ... $$

There exists a unique $r \in [0, \infty] := [0, \infty[ \quad \cup \quad \{\infty\}$

With the following property :

The series converges normally in the open disk $$ U_{r}(0) := \{ z \in \mathbb{C} ; \quad |z| < r\}$$

Having searched elsewhere I can't find anything about normal convergence in relation to power series. From what I understand, normal convergence relates to a series of functions, what series of functions is the book referring to ?

In addition, normal convergence seems to be quite a strong condition, since it implies uniform absolute convergence. However, I thought the ratio test was the tool most often used to find the radius of convergence, and this only implies absolute convergence. How is the ratio test then enough to find the ratio of convergence ?

The book gives the following definition for normal convergence :

A series $ f_{0} + f_{1} + f_{2} +...$ of functions $$ f_{n} : D \rightarrow \mathbb{C}, D \in \mathbb{C}, n\in \mathbb{N}_{0}$$ is called normally convergent in D, if for every point $a \in D$ there is a neighbourhood U and a sequence $(M_{n})_{n \geq 0}$ of non-negative real numbers, such that: $$ |f_{n}(z)| \leq M_{n} \text{ for all } z \in U \cap D, \quad \text{all} \quad n \in \mathbb{N}_{0}, \\ \text{and} \sum^{\infty}_{n = 0} M_{n} \text{ converges} $$


The functions are $f_n(z)=a_nz^n$. With this notation, $\sum_{n=0}^\infty a_nz^n=\sum_{n=0}^\infty f_n(z)$. This is the series that converges normally.

Also, you are right that the ratio test only implies absolute convergence, not normal convergence. At first, the radius of convergence only gives a domain (a disc with the radius of convergence $r$ as its radius) on which the series converges absolutely. But you can then show that in addition, on all smaller discs with radius $0<\rho<r$, there is a geometric series $\sum_n M\left(\frac{\rho}{r}\right)^n$ independent of $z$ which dominates the power series. And thus it also converges normally (since for each $z\in U_r(0)$ there is a $\rho<r$ such that $U_\rho(0)$ is a neighborhood of $z$).

Also keep in mind though, that normal convergence does not imply uniform convergence. Consider the geometric series $\frac{1}{1-x}=\sum_n x^n$. Its partial sums are bounded on $U_1(0)$, but its limit $\frac{1}{1-x}$ is not, so the convergence can't be uniform. The convergence really is just locally uniform, not globally.

  • $\begingroup$ Can I also just check that when you say "on all smaller discs ... there is a geometric series ...", you are referring to a single geometric series ? $\endgroup$ – Mr Lolo Jun 25 '20 at 16:42
  • $\begingroup$ (Deleted an earlier comment after working out where I went wrong, makes sense to me now, thanks!) $\endgroup$ – Mr Lolo Jun 25 '20 at 16:56
  • $\begingroup$ It's a different series for every disc, but on any single disc it is one single series dominating the power series in every point. Its terms are of the form $M\left(\frac{\rho}{r}\right)^n$, where $M$ only depends on the disc we're working with, so the whole thing depends only on the disc (since $r$ is constant and $\rho$ also only depends on the disc). $\endgroup$ – Vercassivelaunos Jun 25 '20 at 17:04

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