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.

$\blacksquare$


Also see