Equivalence of Definitions of Absolute Convergence of Product of Complex Numbers

Theorem
Let $\left\langle{a_n}\right\rangle$ be a sequence of complex numbers.

Let $\log$ denote the complex logarithm.

Then the following definitions of absolute convergence of a product are equivalent:

1 iff 2
Follows directly from Equivalence of Definitions of Absolute Convergence of Product.

2 implies 3
By Terms in Convergent Series Converge to Zero, there exists $n_0\in\N$ such that $|a_n|\leq \frac12$ for $n>n_0$.

Thus $a_n\neq-1$ for $n>n_0$.

By Bounds for Complex Logarithm, $|\log (1+a_n)| \leq \frac32|a_n|$ for $n>n_0$.

By the Comparison Test, $\displaystyle \sum_{n \mathop = n_0+1}^\infty \log(1+a_n)$ is absolutely convergent.

3 implies 2
By Terms in Convergent Series Converge to Zero, $\log(1+a_n)\to 0$.

By Complex Exponential is Continuous, $1+a_n\to1$.

That is, $a_n\to0$.

Let $n_1\in\N$ be such that $|a_n|\leq\frac12$ for $n>n_1$.

By Bounds for Complex Logarithm, $\frac12|a_n| \leq |\log (1+a_n)|$ for $n>\max(n_0,n_1)$.

By the Comparison Test, $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent.

Also see

 * Logarithm of Infinite Product of Complex Numbers for a refined relation between the series $\displaystyle \sum_{n \mathop = 1}^\infty \log(1+a_n)$ and the product $\displaystyle \prod_{n \mathop = 1}^\infty (1+a_n)$