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?
I've just found an answer in the Kadison and Ringose book on "Fundamentals of the theory of operator algebras".
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 $A$.