# Absolutely convergent series of complex numbers

If $\sum_{n=0}^{\infty} a_n$ is an absolutely convergent series of complex numbers and $a_n \ne -1$ $\forall$ $n$, prove that the series

$\sum_{n=0}^{\infty} \frac{a_n}{1+a_n}$ is absolutely convergent.

I'm not sure about this series in $\Bbb C$... I haven't worked with complex series before so I'm not quite sure how to prove this. I'm trying to start with a proof in $\Bbb R$, but I'm not getting very far. Placing a series in a series is tough to wrap my mind around when the only thing I know is that $a_n \ne -1$. Which is an obvious fact since then the fraction would be undefined.

Since $a_n\xrightarrow{n\to\infty}0$ so for sufficiently large $n$ we have
$$0\le \vert a_n\vert\le\frac12$$ hence
$$\left\vert\frac{a_n}{1+a_n}\right\vert\le \frac{\vert a_n\vert}{1-\vert a_n\vert}\le 2\vert a_n\vert$$ and then we conclude by comparison.
So we have $|a_{n}|<1/2$ for $n\geq N$, then $\left|\dfrac{a_{n}}{1+a_{n}}\right|\leq\dfrac{|a_{n}|}{1-|a_{n}|}\leq 2|a_{n}|$ for all such $n$, so it is absolutely convergent.
Observe that $$\frac{a_n}{1+a_n}=\frac{1}{1+\frac{1}{a_n}}$$ whence $$\left|1+\frac{1}{a_n}\right|\ge\frac{1}{|a_n|}.$$ You should have enough tools to conclude the result via comparison.