Reciprocal of Absolutely Convergent Product is Absolutely Convergent

Theorem
Let $\mathbb K$ be a field with absolute value $\left\vert{\, \cdot \,}\right\vert$.

Let $\left\langle{1 + a_n}\right\rangle$ 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, $|a_n|\leq\frac12$ for $n$ sufficiently large.

Thus $\left\vert\frac1{a_n+1}-1\right\vert = \left\vert\frac{a_n}{a_n+1}\right\vert \leq 2|a_n|$ 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.