The product $\prod_{n = 1}^{\infty} F_n(z)$ converges uniformly? Suppose we are given $\{F_n\}$ a sequence of holomorphic functions on the open set $\Omega.$ And there exists constants $c_n > 0$ such that $\sum c_n < \infty$ and $|F_n(z) - 1| \leq c_n$ for all $z \in \Omega.$ I get that the product $\prod_{n = 1}^{\infty} F_n(z)$ converges but how do we know that this converges uniformly? I tried to show that for big enough $N$ we can have $\prod_{n = 1}^{N} F_n(z)$ we can have the product approach some small $\epsilon$ as we know that $c_n\rightarrow \infty...$
 A: If $\prod F_n(z)$ converges pointwise, then $F_n(z)$ eventually avoids the branch cut for the principal value of $\log$ along the negative real axis: $-\infty < x \leqslant 0$.
We have
$$F_n(z) = 1 + (F_n(z) - 1) = 1 + G_n(z),$$
where $|G_n(z)| \leqslant c_n$.
Since $\sum c_n$ converges, $|G_n(z)| \leqslant c_n < 1/2$ for all $n$ greater than some $N$, and
$$|\log(1 + G_n(z))| \leqslant |G_n(z)| + \frac{1}{2} |G_n(z)|^2 + \ldots \\ \leqslant \frac{|G_n(z)|}{1 - |G_n(z)|} \\ \leqslant 2 |G_n(z)| \\ \leqslant 2c_n.$$
Hence, the series $\sum\log(1 + G_n(z))$ converges uniformly by the Weierstrass test and, consequently, so does the product $\prod F_n(z) = \prod (1 + G_n(z)) = \exp(\,\sum \log(1 + G_n(z))\, )$ since the exponential function is continuous.
A: If $\sum_{n=1}^\infty c_n<\infty$, then there exists an $N\in\mathbb N$, such that $c_n<\frac{1}{2}$, for $n\ge N$.
If $|F_n(z)-1|\le c_n<\frac{1}{2}$, then the range of $F_n$ lies in the disk $|z-1|<\frac{1}{2}$, where complex logarithm is definable. Let's $\,\mathrm{Log}\,z\,$ be the main branch of logarithm, with $\,\mathrm{Log}\,1=0$. 
We define $G_n(z)=\mathrm{Log}\,F_n(z)$, for $n\ge N$.
Also, if $|z-1|<\frac{1}{2}$, then
$$
\mathrm{Log}\,z=\mathrm{Log}\,\left(1-(1-z)\right)=\sum_{n=1}^\infty \frac{(1-z)^n}{n}
$$ 
and hence
$$
|\mathrm{Log}\,z|\le \sum_{n=1}^\infty \frac{|z-1|^n}{n}=-\log (1-|z-1|)=\log\left(\frac{1}{1-|z-1|}\right)=\log\left(1+\frac{|z-1|}{1-|z-1|}\right)\le 
\frac{|z-1|}{1-|z-1|}\le 2|z-1|$$ 
Thus
$$
F_1(z)\cdots F_n(z)=F_1(z)\cdots F_{N}(z)F_{N+1}(z)\cdots F_n(z)
=F_1(z)\cdots F_{N-1}(z)\exp\big(G_{N}(z)+\cdots+G_n(z)\big)
$$
It suffices to show that the sum
$$
\sum_{n=N}^\infty G_n(z)\tag{1}
$$
converges uniformly in $\Omega$.
We have that, for $n\ge N$ and $z\in \Omega$,
$$
|G_n(z)|=|\mathrm{Log}\,F_n(z)|\le 2 |F_n(z)-1|\le 2c_n,
$$
and hence the sum $(1)$ converges uniformly in $\Omega$, and so does $\prod F_n(z)$.
