Absolute Value of Divergent Infinite Product

Theorem
Let $\struct {\mathbb K, \norm{\,\cdot\,}}$ be a valued field. The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$  $\displaystyle \prod_{n \mathop = 1}^\infty \norm{a_n}$ 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 $\displaystyle \prod_{n \mathop = 1}^\infty \norm{a_n}$ diverges to $0$.

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

Then $\norm{a_n}\neq0$ 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, $\displaystyle\lim_{N\to\infty}\prod_{n=n_0}^N \norm{a_n} = 0$.

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

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

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 $\norm{a_n}\neq0$ for $n\geq n_0$.

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

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

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 $0$.

Also see

 * Absolute Value of Infinite Product, for related results