I am really stuck on something. I need to show the following:

Let $U$ be a domain in $\mathbb{C}$. If $f_n: U \to \mathbb{D}$ are analytic functions satisfying that $\sum |f_n - 1|$ converges locally uniformly on $U$, then $$\lim_{n \to \infty} \prod f_n = f$$ converges absolutely and locally uniformly on $U$ to some analytic $f$. Additionally, $$\sum \frac{f_n'}{f_n} \to \frac{f'}{f}$$

I think the second part is just taking some logarithmic derivative and should be clear once the first part is established. We know the result for a product of complex numbers (namely that $\prod c_n$ converges absolutely iff $\sum |1-c_n|$ converges), and we know the Weierstrass factorization theorem, but that's pretty much it.

I assume the proof is to pick a point $c$ and then use some compactness argument. I think it starts out something like... pick a small simply connected precompact neighborhood $V$ of an arbitrary $c \in U$. Then we'll just look at the tails of the sum/product, choosing an $N_0$ big enough so that $f_n(V)$ is away from $0$ and thus has a log, which we can do by the convergence of the sum.

Now I want to do something like $|\frac{log(f(z)}{f(z) - 1} - 1| < \frac{1}{2}$ uniformly on $V$ (recall it's precompact in $U$), from which I can get the first part of the theorem. When dealing just with points, we can get this inequality by noting that $\lim_{z \to 1} \frac{log(z)}{z-1} = log(z)'$ at $1$, but I can't fix this method for the function version.

Can anyone see how to give a rigorous argument? Thanks a lot!

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    $\begingroup$ Since $\sum \lvert f_n(z) - 1 \rvert$ is locally uniformly convergent, you have an $N_0$ such that $\lvert f_n(z) - 1\rvert \leqslant \frac{1}{2}$ for all $n \geqslant N_0$ and all $z \in V$. Use the estimate $\lvert \log (1+w) - w\rvert \leqslant \frac{1}{2}\lvert w\rvert$ for $\lvert w\rvert \leqslant \frac{1}{2}$ (you set $w = f_n(z) - 1$) to deduce the uniform convergence of $\sum_{n \geqslant N_0} \log f_n(z)$ on $V$. $\endgroup$ – Daniel Fischer Feb 12 '16 at 12:05
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    $\begingroup$ @DanielFischer Or if, like me, one feels that the complex logarithm is evil, one can prove by induction and then use the fact that if $\sum_1^{n-1}|1-w_j|<1/2$ then $|1-\prod_1^nw_j|\le2\sum_1^n|1-w_j|$. $\endgroup$ – David C. Ullrich Feb 12 '16 at 14:09

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