If $\prod\limits_{n}(1+a_n)$ converges, does $\sum\limits_{n}\frac{a_n}{1+a_n}$ converge?

Question

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!

Answer 1

The infinite product $\prod_n (1+a_n)$ converges (to a nonzero value) if and only if $\sum_n \log(1+a_n)$ converges (for a branch of $\log$ that is analytic in a neighbourhood of $1$, with $\log(1) = 0$). This does not in general imply $\sum_n \dfrac{a_n}{1+a_n}$ converges. For example, take a positive sequence $b_k \to 0 + $ and let $c_k = 1/(1+b_k) - 1$. Thus $\log(1+c_k) = -\log(1+b_k)$. On the other hand, $$ \dfrac{c_k}{1+c_k} + \dfrac{b_k}{1+b_k} = \dfrac{-b_k^2}{1+ b_k} \ne 0$$ Consider a sequence $a_n$ that, for each $k$, repeats $b_k, c_k$ enough times for the partial sum of $a_n/(1+a_n)$ to change by $1$, before progressing to the next $k$. Then $\sum \log(1+a_n)$ converges to $0$ but $\sum a_n/(1+a_n)$ diverges.

Answer 2

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.