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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?

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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 $A$.

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