Absolute Value of Divergent Infinite Product

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Theorem

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

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

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 $0$.

$\Box$


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

$\blacksquare$


Also see

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