Absolute Value of Divergent Infinite Product

Theorem
Let $\mathbb K$ be a field with absolute value $\left\vert{\, \cdot \,}\right\vert$. The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$  $\displaystyle \prod_{n \mathop = 1}^\infty \left\vert{a_n}\right\vert$ diverges to $0$.

Case 1
Let there be infinitely many $n\in\N$ such that $a_n=0$.

Then there are infinitely many $n\in\N$ such that $|a_n|=0$.

Thus $\displaystyle \prod_{n \mathop = 1}^\infty |a_n|$ diverges to $0$.

Case 2
Let $n_0\in\N$ such that $a_n\neq0$ for $n\geq n_0$.

Then $|a_n|\neq0$ for $n\geq n_0$.

By the divergence, $\displaystyle\lim_{N\to\infty}\prod_{n=n_0}^Na_n=0$.

By Absolute Value of Limit of Sequence, $\displaystyle\lim_{N\to\infty}\prod_{n=n_0}^N|a_n|=0$.

Thus $\displaystyle \prod_{n \mathop = 1}^\infty |a_n|$ diverges to $0$.

Case 1
Let there be infinitely many $n\in\N$ such that $|a_n|=0$.

Then there are infinitely many $n\in\N$ such that $a_n=0$.

Thus $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$.

Case 2
Let $n_0\in\N$ such that $|a_n|\neq0$ for $n\geq n_0$.

Then $a_n\neq0$ for $n\geq n_0$.

By the divergence, $\displaystyle\lim_{N\to\infty}\prod_{n=n_0}^N|a_n|=0$.

By Absolute Value of Limit of Sequence, $\displaystyle\lim_{N\to\infty}\prod_{n=n_0}^Na_n=0$.

Thus $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$.

Also see

 * Absolute Value of Infinite Product, for related results