# Invertibility of $I-A$ if the spectral radius of the operator $A$ is less than $1$

I want an explication of the following fact:

If the spectral radius of a bounded operator $A$ on a Banach space is less than one, then $I - A$ is invertible.

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If $\rho(A)<1$, then the operator $$\sum_{n=0}^\infty A^n$$ is defined (why?) and is in fact $(I-A)^{-1}$.

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This is trivial from the definition of spectrum. If the spectral radius is less than one, then in particular $1$ is not in the spectrum, which means $I-A$ is invertible.

Cameron Buie has answered a more interesting question.

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You can find hint at Neumann series wikipedia article. Furthermore the question is maybe a duplicate, I think you can find the answer at math.se.

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