Product of Absolutely Convergent Products is Absolutely Convergent

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

Let $\ds \prod_{n \mathop = 1}^\infty a_n$ converge absolutely.

Let $\ds \prod_{n \mathop = 1}^\infty b_n$ converge absolutely.

Then $\ds \prod_{n \mathop = 1}^\infty a_nb_n$ converges absolutely.

Proof
We have:
 * $a_n b_n - 1 = \paren {a_n - 1} \paren {b_n - 1} + \paren {a_n - 1} + \paren {b_n - 1}$

By the Triangle Inequality:
 * $\norm {a_n b_n - 1} \le \norm {a_n - 1} \norm {b_n - 1} + \norm {a_n - 1} + \norm {b_n -1}$

By the absolute convergence, $\ds \sum_{n \mathop = 1}^\infty \norm {a_n - 1}$ and $\ds \sum_{n \mathop = 1}^\infty \norm {b_n - 1}$ converge.

By Inner Product of Absolutely Convergent Series, $\ds \sum_{n \mathop = 1}^\infty \norm {a_n - 1} \norm{b_n -1}$ converges.

By the Comparison Test, $\ds \sum_{n \mathop = 1}^\infty \norm {a_n b_n - 1}$ converges.

Also see

 * Product of Convergent and Divergent Product is Divergent
 * Product of Convergent Products is Convergent