Absolute Value of Divergent Infinite Product

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

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

Case 2
Let $n_0 \in \N$ such that $a_n \ne 0$ for $n \ge n_0$.

Then $\norm {a_n} \ne 0$ for $n \ge n_0$.

By definition of divergent product:
 * $\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = n_0}^N a_n = 0$

By Absolute Value of Limit of Sequence:
 * $\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = n_0}^N \norm {a_n} = 0$

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

Case 2
Let $n_0 \in \N$ such that $\norm {a_n} \ne 0$ for $n \ge n_0$.

Then $a_n \ne 0$ for $n \ge n_0$.

By definition of divergent product:
 * $\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = n_0}^N \norm {a_n} = 0$

By Absolute Value of Limit of Sequence:
 * $\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = n_0}^N a_n = 0$

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

Also see

 * Absolute Value of Infinite Product, for related results