Question regarding infinite Blaschke product According to Gamelin's $\textit{Complex Analysis}$, a finite Blaschke product is a rational function of the form $B(z)= e^{i \varphi} (\frac{z-a_1}{1-\bar{a_1} z} \cdots \frac{z-a_n}{1-\bar{a_n} z})$ where $a_1, ..., a_n \in \mathbb{D}$ and $0 \leq \varphi \leq 2\pi$. Similarly, I would guess that an infinite Blaschke product would be of the form $e^{i \varphi} \prod_{n=1}^\infty\frac{z-a_n}{1-\bar{a_n} z}$. I believe this is supposed to satisfy what is known as the Blaschke condition, i.e. $\sum_{n=1}^\infty (1-|a_n|) < \infty$, but how is that so? Can this be verified using the log function on the infinite product?
 A: Actually, the infinite Blaschke product, for $|a_n|\le1$ and $|z|<1$, is defined as
$$
e^{i\varphi}\prod_{n=1}^\infty\frac{|a_n|}{a_n}\frac{z-a_n}{\overline{a}_n z-1}\tag{1}
$$
The factor of $\;{-}\dfrac{|a_n|}{a_n}$ simply rotates $\dfrac{z-a_n}{1-\overline{a}_n z}$, which, for finite products, is incorporated into $e^{i\varphi}$. However, for infinite products, it is needed for convergence.
First, note that
$$
\begin{align}
\frac{|a_n|}{a_n}\frac{z-a_n}{\overline{a}_n z-1}
&=|a_n|\frac{z-a_n}{|a_n|^2 z-a_n}\\
&=(1-(1-|a_n|))\left(1+\frac{z(1-|a_n|^2)}{|a_n|^2 z-a_n}\right)\\
&=(1-(1-|a_n|))\left(1+\frac{z(1+|a_n|)}{|a_n|^2\left(z-\frac{1}{\overline{a}_n}\right)}(1-|a_n|)\right)\tag{2}
\end{align}
$$
where
$$
\begin{align}
\left|\frac{z(1+|a_n|)}{|a_n|^2\left(z-\frac{1}{\overline{a}_n}\right)}\right|
&\le\frac{1+|a_n|}{|a_n|^2}\frac{|z|}{1-|z|}\\
&\le6\frac{|z|}{1-|z|}\tag{3}
\end{align}
$$
when $|a_n|\ge\frac12$.
Equations $(2)$ and $(3)$ say that the infinite product in $(1)$ converges absolutely when $|z|<1$ and
$$
\sum_{n=1}^\infty(1-|a_n|)\tag{4}
$$
converges. That is, the infinite product $\prod\limits_{n=1}^\infty(1+z_n)$ converges absolutely when $\sum\limits_{n=1}^\infty|z_n|$ converges.
