Equivalence of Definitions of Absolute Convergence of Product of Complex Numbers

Theorem
Let $\sequence {a_n}$ be a sequence of complex numbers.

Let $\log$ denote the complex logarithm.

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 $\size {a_n} \le \dfrac 1 2$ for $n > n_0$.

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

By Bounds for Complex Logarithm:
 * $\size {\map \log {1 + a_n} } \le \dfrac 3 2 \size {a_n}$

for $n > n_0$.

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

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

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

That is, $a_n \to 0$.

Let $n_1 \in \N$ be such that $\size {a_n} \le \dfrac 1 2$ for $n > n_1$.

By Bounds for Complex Logarithm, $\dfrac 1 2 \size {a_n} \le \size {\map \log {1 + a_n} }$ for $n > \map \max {n_0, n_1}$.

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

Also see

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