Product of Absolutely Convergent Products is Absolutely Convergent
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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.
$\blacksquare$