How do I show that the infinite product converges. Let $(a_n)_n$ be a sequence in $\mathbb{C}$, $a_n \neq -1$.
Prove that $\displaystyle\sum_{n=1}^{\infty}|a_n| < \infty,$ then $\displaystyle\prod_{n=1}^{\infty}(1+a_n)$ converges to a non-zero number.
Here is what I have so far
$\displaystyle\prod_{n=1}^{\infty}|1+a_n|
\leq \displaystyle\prod_{n=1}^{\infty}(1+|a_n|)
\leq \displaystyle\prod_{n=1}^{\infty}e^{|a_n|} = \displaystyle\exp(\sum_{n=1}^{\infty}|a_n|) < \infty$
So I only showed that the infinite product is bounded above, but how do I show the following


*

*$\displaystyle\prod_{n=1}^{\infty}(1+a_n)$ converges. (By Monotone Convergence Theorem, we know that $\displaystyle\prod_{n=1}^{\infty}(1+|a_n|)$ converges, but I am not sure where to go from there

*How to show that $\displaystyle\prod_{n=1}^{\infty}|1+a_n| >0$,
In the real case, this follows from $e^{-|a_n|} \leq 1+|a_n| \leq e^{|a_n|}$
But I am not sure how to generalize that to the complex norm
 A: Taking for granted that $\exp$ is continuous, we just need to show that $\sum_{k=1}^\infty \log (1+a_k)$ converges absolutely if $\sum_{k=1}^\infty a_k$ does. Note that the assumption implies that $a_k \to 0$ so that the (principal) logarithm is well defined for large enough $k$; without loss $k=1$ is large enough.
For this, note the (imprecise) inequality from e.g. Taylor expansion that
$$ |\log(1+a_k)| ≤ 10|a_k|$$
so that $$\sum_{k=1}^K|\log(1+a_k)| ≤ 10\sum_{k=1}^K |a_k| \to 10 \sum_1^\infty |a_k| < \infty$$
This implies that $\sum_{k=1}^\infty \log (1+a_k)$ converges absolutely, and therefore the product
$$ \prod_{k=1}^\infty (1+a_k) := \lim_{K\to \infty}\exp \left(\sum_{k=1}^K  \log(1+a_k)\right) = \exp \left(\sum_{k=1}^\infty  \log(1+a_k)\right)$$
converges. 
It is a positive number because $\exp$ restricted to the reals has positive range, and $∏_{k=1}^\infty |1+a_k| = \exp (\sum_{k=1}^\infty\log|1+a_k|)$ (with the same definition of infinite product above) is the exponential of a real number.
