Prove the sequence of partial products converges uniformly. Suppose $f:A\to\mathbb{C}$ and that $\forall n, \exists M_n$ such that $\sup_{x\in A}|f_n(x)| \leq M_n$. Also suppose $\sum_1^\infty M_n$ converges to some finite number. Define $$g_k(x) = \prod_1^k (1+f_n(x)).$$
Prove $g_k$ converges uniformly.
This looks analogous to the Weierstrass M-test, but for products, so I'm thinking that may be required in the proof. I haven't been able to get anywhere with it yet, though.
Suggestions?
 A: It is assumed that $f_n(x)\ne 1$, else the product is $0$.
Since $f_n\in \mathbb{C}$, then we can write $f_n(x)=|f_n(x)|e^{i\phi_n(x)}$, where $\phi_n(x) =\arg(f_n(x))$.  
We are given that $ |f_n(x)|\le M_n$ and $\sum_{n=1}^\infty M_n<\infty$.  Therefore, $\lim_{n\to \infty}|f_n(x)|= 0$.  So there exists a number $n_0$ such that $|f_n(x)|<1/2$ for $n\ge n_0$..  
Taking $k\ge n_0$, we note that 
$$\begin{align}
\log\left(\prod_{n=n_0}^k(1+f_n(x))\right)&=\sum_{n=n_0}^k\log(|1+f_n(x)|)+i\sum_{n=n_0}^k\arg(1+|f_n(x)|e^{i\phi_n(x)})\\\\
&=\frac12\sum_{n=n_0}^k\log(1+|f_n(x)|^2+2|f_n(x)|\cos(\phi_n(x)))\\\\
&+i \sum_{n=n_0}^k \arctan\left(\frac{|f_n(x)|\sin(\phi_n(x))}{1+|f_n(x)|\cos(\phi_n(x))}\right)
\end{align}$$

We have the estimates 
$$\begin{align}
\left|\sum_{n=n_0}^k\log(1+|f_n(x)|^2+2|f_n(x)|\cos(\phi_n(x)))\right|&\le 2\sum_{n=n_0}^k\log(1+|f_n(x)|)\\\\
&\le 2\sum_{n=n_0}^k |f_n(x)|\\\\
&\le 2\sum_{n=n_0}^k M_n
\end{align}$$
Therefore, the real part of $\prod_{n=1}^k(1+f_n(x))$ converges uniformly.

In addition, we have the estimates
$$\begin{align}
\left|\sum_{n=n_0}^k \arctan\left(\frac{|f_n(x)|\sin(\phi_n(x))}{1+|f_n(x)|\cos(\phi_n(x))}\right)\right|&\le \sum_{n=n_0}^k \frac{|f_n(x)|\,|\sin(\phi_n(x))|}{1+|f_n(x)|\cos(\phi_n(x))}\\\\
&\le 2 \sum_{n=n_0}^k |f_n(x)|
\end{align}$$
Therefore, the imaginary part of $\prod_{n=1}^k(1+f_n(x))$ converges uniformly.

Inasmuch as the real and imaginary parts of $\prod_{n=1}^k(1+f_n(x))$ converge uniformly, then $\prod_{n=1}^k(1+f_n(x))$ converges uniformly.
