Sum of analytic functions converges uniformly but not the product We know that if $\sum_{n=1}^\infty |f_n(z)|$ converges uniformly on $S$ to a bounded function and $f_n(z)\neq -1$ on $S$ then $\prod_{n=1}^\infty (1+f_n(z))$ converges uniformly on $S$. 
I want a counterexample in above case if we remove the boundedness of the limit function, i.e. I want a set $S$ and a sequence of functions $\{f_n(z)\}$ defined on $S$ such that $\sum_{n=1}^\infty |f_n(z)|$ converges uniformly on $S$ to an unbounded function and $f_n(z)\neq -1$ on $S$ and $\prod_{n=1}^\infty (1+f_n(z))$ "does not" converge uniformly on $S$.
 A: One remark upfront. Such an example isn't very interesting, as the convergence one is interested in for analytic functions usually is locally uniform convergence, and the locally uniform convergence of $\sum \lvert f_n\rvert$ implies the locally uniform convergence of $\prod (1 + f_n)$.
To have $\sum \lvert f_n\rvert$ converge uniformly on $S$, all but a finite number of the $f_n$ must be bounded, and the tail of the series converges uniformly to a bounded function on $S$. So we can collapse the initial part containing the unbounded $f_n$ to one term, and assume that $f_0$ is unbounded, and $\sum\limits_{n=1}^\infty \lvert f_n\rvert$ converges uniformly to a bounded function.
The simplest way to achieve the desired result is then to choose $f_0$ with a pole in $z_0$ on the boundary of $S$, and the sequence $\bigl(f_n\bigr)_{n\geqslant 1}$ so that $\prod\limits_{n=1}^\infty (1 + f_n(z_0))$ converges to a nonzero complex number.
For example, let $S = \{ z : 0 < \lvert z\rvert < 1\}$ the punctured open unit disk (we need the puncture to avoid $f_0(z) = -1$, it is not important, isolated zeros of the product don't harm), and $f_0(z) = \frac{1}{z-1}$. For $n \geqslant 1$, let $f_n(z) = \frac{z^2}{n^2}$.
We then have
$$\prod_{n=0}^\infty \left(1 + f_n(z)\right) = \frac{z}{z-1}\prod_{n=1}^\infty \left(1 + \frac{z^2}{n^2}\right) = \frac{\sinh (\pi z)}{\pi(z-1)},$$
and since $\sinh \pi \neq 0$, the convergence of the product is not uniform on all of $S$. To show the latter, consider
$$\prod_{n=0}^N\left(1 + f_n(z)\right) - \prod_{n=0}^{N-1}\left(1+f_n(z)\right) = \frac{z^3}{N^2(z-1)}\prod_{n=1}^{N-1}\left(1 + \frac{z^2}{n^2}\right).$$
For $N \geqslant 2$, and $1 - \frac{1}{N^4} \leqslant z \leqslant 1$, the difference is larger in absolute value than
$$\frac{1/2^3}{N^2(N^{-4})} = \frac{N^2}{8}.$$
