Reciprocal of Absolutely Convergent Product is Absolutely Convergent

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

Let $\sequence{1 + a_n}$ be a sequence of nonzero elements of $\mathbb K$.

Let the infinite product $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + a_n}\right)$ converge absolutely to $a \in \mathbb K \setminus \left\{ {0}\right\}$.

Then $\displaystyle \prod_{n \mathop = 1}^\infty \frac 1 {1 + a_n}$ converges absolutely to $1 / a$.

Proof
By continuity of $x \mapsto 1 / x$:
 * $\lim_{N \mathop \to \infty} \displaystyle \prod_{n \mathop = 1}^N \frac 1 {1 + a_n} = \frac 1 a$.

It remains to prove the absolute convergence.

Because $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + a_n}\right)$ converges absolutely, $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.

By Factors in Absolutely Convergent Product Converge to One, $\norm{a_n} \leq \frac 1 2 $ for $n$ sufficiently large.

Thus $\norm {\frac 1 {a_n+1} - 1} = \norm {\frac {a_n} {a_n+1}} \leq 2 \norm{a_n}$ for $n$ sufficiently large.

By the Comparison Test, $\displaystyle \sum_{n \mathop = 1}^\infty \left({\frac 1 {a_n + 1} - 1}\right)$ converges absolutely.

Thus $\displaystyle \prod_{n \mathop = 1}^\infty \frac1{1 + a_n}$ converges absolutely.