Product of Absolutely Convergent Products is Absolutely Convergent

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

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

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

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

Proof
We have:
 * $a_n b_n - 1 = \left({a_n - 1}\right) \left({b_n - 1}\right) + \left({a_n - 1}\right) + \left({b_n - 1}\right)$

By the Triangle Inequality:
 * $\left\lvert{a_n b_n - 1}\right\rvert \le \left\lvert{a_n - 1}\right\rvert \left\lvert{b_n - 1}\right\rvert + \left\lvert{a_n - 1}\right\rvert + \left\lvert{b_n -1}\right\rvert$

By the absolute convergence, $\displaystyle \sum_{n \mathop = 1}^\infty \left\lvert{a_n - 1}\right\rvert$ and $\displaystyle \sum_{n \mathop = 1}^\infty \left\lvert{b_n -1}\right\rvert$ converge.

By Inner Product of Absolutely Convergent Series, $\displaystyle \sum_{n \mathop = 1}^\infty \left\lvert{a_n - 1}\right\rvert \left\lvert{b_n -1}\right\rvert$ converges.

By the Comparison Test, $\displaystyle \sum_{n \mathop = 1}^\infty \left\lvert{a_n b_n - 1}\right\rvert$ converges.

Also see

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