# Limit of a convergent sequence of invertible elements in a Banach algebra

Question: Let $$\{A_{n}\}$$ be a sequence of invertible elements in a Banach algebra $$\mathfrak{A}$$ and suppose $$\{A_{n}\}$$ has a limit A that commutes with each $$A_{n}$$. Let $$r(A_{n}^{-1})$$ denote the spectral radius of $$A_{n}^{-1}$$. Show that $$A$$ is invertible if $$r(A_{n}^{-1})$$ is bounded.

My ideas: I can estimate:

$$r(1-A_{n}^{-1} A) = r(A_{n}^{-1} (A_{n} - A)) \leq r(A_{n}^{-1}) r(A_{n} - A) \to 0.$$

This shows that $$(1-A_{n}^{-1} A)$$ is a generalized nilpotent in $$\mathfrak{A}$$. But how does this imply $$\| 1-A_{n}^{-1} A \| \to 0$$. Or am I on the wrong track?

From the estimate that I've done, it follows that for large $$n$$ the spectrum of $$(A_{n}^{-1} A)$$ lies in a small disk with center 1. Hence, $$A_{n}^{-1} A$$ is invertable and so is $$A$$.