TITLE: If $\prod\limits_{n}(1+a_n)$ converges, does $\sum\limits_{n}\frac{a_n}{1+a_n}$ converge?
QUESTION [5 upvotes]: I have a sequence of complex numbers $a_1,a_2,...$ such that $a_i \neq -1$. I then have the infinite product $\prod\limits_{n=1}^{\infty}(1+a_n)$ which I know converges to a non zero complex number. I was wondering if this is enough to guarantee the convergence of $\sum\limits_{n=1}^\infty\frac{a_n}{1+a_n}$. 
I couldn't come up with an obvious counterexample and my first idea for proving it was assuming $\prod\limits_{n=1}^{\infty}(1+a_n)=S$, so $\sum\limits_{n=1}^{\infty}\ln(1+a_n)=\ln S$ converges but have no idea how to deal with this since $1+a_i$ is complex. 
If this diverges in general, I was wondering if there are any necessary and sufficient conditions for the sum to converge such as $\sum\limits_n|a_n|<\infty$ or that the $a_n$ are positive real numbers.
Any help would be awesome,and thanks for taking a look!

REPLY [2 votes]: It's OK for absolute convergence.  But not in general.
Take $\log(1+a_n) = (-1)^n/\sqrt{n}$ for $n \ge 1$, so
$\sum \log(1+a_n) = \sum (-1)^n/\sqrt{n}$ converges (conditionally).  But
$$
\frac{a_n}{1+a_n} = \frac{(-1)^n}{\sqrt{n}} - \frac{1}{2n} +O\left(\frac{1}{n^{3/2}}\right)
$$
and thus $\sum a_n/(1+a_n)$ diverges.