Absolute Value of Convergent Infinite Product

From ProofWiki
Jump to navigation Jump to search


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

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

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


By Absolute Value of Limit of Sequence:

$\ds \prod_{n \mathop = 1}^\infty \norm {a_n} = \norm a$

It remains to show that the product converges.

By the convergence, there exists $n_0 \in \N$ such that $a_n \ne 0$ for $n \ge n_0$.

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

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

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

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

By Convergence of Absolute Value of Sequence:

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

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


Also see