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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I added "on a Banach space" to make the statement correct. Please adjust if it was an incorrect assumption. –  Jonas Meyer Jan 9 '13 at 22:23

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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