# 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.

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

### Definition 1

The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + a_n}\right)$ is absolutely convergent if and only if $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + \left\vert{a_n}\right\vert}\right)$ is convergent.

### Definition 2

The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + a_n}\right)$ is absolutely convergent if and only if the series $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent.

### Definition 3

The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ 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 $\displaystyle \sum_{n \mathop = n_0 + 1}^\infty \log \paren {1 + a_n}$ 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 $|a_n|\leq \frac12$ 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.

$\blacksquare$

### 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.

$\blacksquare$