Absolute Value of Divergent Infinite Product

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Theorem

Let $\struct {\mathbb K, \norm{\,\cdot\,}}$ be a valued field.

The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$ if and only if $\displaystyle \prod_{n \mathop = 1}^\infty \norm{a_n}$ diverges to $0$.


Proof

1 implies 2

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 $0$.

$\Box$

2 implies 1

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 $0$.

$\blacksquare$

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