Some proofs about the infinite products; the series and the analyticity Definition A.1.4. The infinite product $\prod_{n=1}^{\infty}(1+a_{n}(x))$, where $x$ is a real or complex variable in a domain, is uniformly convergent if $p_{n}(x)=\prod_{m=k}^{n}(1+a_{n}(x))$ converges uniformly in that domain, for each $k$.
Theorem A.1.5. If the series $\sum_{n=1}^{\infty}|a_{n}(x)|$ converges uniformly in some region, then the product $\prod_{n=1}^{\infty}(1+a_{n}(x))$ also converges uniformly in that region.
Corollary A.1.6. If $a_{n}(x)$ is analytic in some region of the complex plane and $\prod(1+a_{n}(x))$ converges uniformly in that region, then the infinite product represents an analytic function in that region.
They are taken from Special Functions page 596-597.
I want to prove Theorem A.1.5 and Corollary A.1.6.
Proof: There exists a convergent series $\sum_{n=1}^{\infty}M_{n}$ such that $|a_{n}(x)|\leq M_{n}$, for all $x\in A\subseteq \mathbb{C}$. Note that 
$|1+a_{m}(x)|\leq 1+|a_{m}(x)|<2M_{m}$, so $$|p_{n}(x)|<2\prod_{m=k}^{n}M_{m}\leq 2\prod_{m=1}^{n}M_{m}=:2C_{n}.$$ We know that $M_{m}$ is bounded for all $1\leq m\leq n$, so the partial products $C_{n}$ must be bounded too. Therefore $2\sum_{n=1}^{\infty}C_{n}<\infty$, and this shows that the infinite product is uniformly convergent on $A$. 
Is this correct? Is there another way to prove it? I am not sure how to provethe corollary. Any hints?

No, that's totally wrong.  For the series to be convergent, you certainly need $M_n \to 0$.  But  $1 + |a_n| > 1$, which is certainly not bounded by $2 M_n$.
Hint: What you need to do is use logarithms:
$$\prod_{n=1}^N (1 + a_n(x)) = \exp\left(\sum_{n=1}^N \log(1+a_n(x))\right)$$
As for the corollary, that's just the fact that a uniform limit of analytic functions in a region is analytic there.
