Reciprocal of Absolutely Convergent Product is Absolutely Convergent

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

Let $(1+a_n)$ 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 $a\in\mathbb K\setminus\{0\}$.

Then $\displaystyle \prod_{n \mathop = 1}^\infty \frac1{1 + a_n}$ converges absolutely to $1/a$.

Proof
By continuity of $x\mapsto 1/x$, $\lim_{N\to\infty}\displaystyle \prod_{n \mathop = 1}^N\frac1{1 + a_n}=\frac1a$.

It remains to prove the absolute convergence.

Because $\displaystyle \prod_{n \mathop = 1}^\infty (1 + a_n)$ 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, $\displaystyle \sum_{n \mathop = 1}^\infty\left(\frac1{a_n+1}-1\right)$ converges absolutely.

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