Absolute Value of Convergent Infinite Product

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Theorem

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

Let the infinite product $\displaystyle \prod_{n \mathop = 1}^\infty a_n$ converge to $a\in\mathbb K$.


Then $\displaystyle \prod_{n \mathop = 1}^\infty \norm{a_n}$ converges to $\norm{a}$.


Proof

By Absoltue Value of Limit of Sequence, $\displaystyle \prod_{n \mathop = 1}^\infty \norm{a_n} = \norm{a}$.

It remains to show that the product converges.

By the convergence, there exists $n_0\in\N$ such that $a_n\neq0$ for $n\geq n_0$.

Then $\norm{a_n} \neq 0$ for $n\geq n_0$.

Let $P_n$ denote the $n$th partial product of $\displaystyle \prod_{n \mathop = n_0}^\infty a_n$.

Then $\norm{P_n}$ is the $n$th partial product of $\displaystyle \prod_{n \mathop = n_0}^\infty \norm{a_n}$.

Let $P_n$ converge to $b\in\mathbb K\setminus\{0\}$.

By Convergence of Absolute Value of Sequence:

$\norm{P_n} \to \norm{b} \in\mathbb K\setminus\{0\}$.

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

$\blacksquare$


Also see