# 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.

The following definitions of the concept of Absolute Convergence of Product are equivalent:

### Definition 1

The infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent if and only if $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} }$ is convergent.

### Definition 2

The infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent if and only if the series $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent.

### Definition 3

The infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent if and only if there exists $n_0 \in \N$ such that:

$a_n \ne -1$ for $n > n_0$
The series $\ds \sum_{n \mathop = n_0 + 1}^\infty \log \paren {1 + a_n}$ is absolutely convergent

where $\log$ denotes the complex logarithm.

## Proof

### 1 iff 2

Follows directly from Equivalence of Definitions of Absolute Convergence of Product.

$\blacksquare$

### 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 $n > n_0$.

$\size {\map \log {1 + a_n} } \le \dfrac 3 2 \size {a_n}$

for $n > n_0$.

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

$\blacksquare$

### 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.

$\blacksquare$