TITLE: Looking for a reference to a proof of $(I - A)^{-1} = I + A + A^2 + A^3 + \ldots$
QUESTION [5 upvotes]: On some online forum, there is the claim:
Given some square matrix:
$$(I - A)^{-1} = I + A + A^2 + A^3 + \ldots$$

This is true if the right side converges, which is true if and only if
  all of the eigenvalues of A have absolute value smaller than $1$.

Reference https://www.physicsforums.com/threads/matrix-inverse-equals-power-series.423897/
I really like this result, because it relies on the more intuitive spectral radius rather than matrix norm which is defined as: 
\begin{align} \|A\| &= \sup\{\|Ax\| : x\in K^n \mbox{ with }\|x\|= 1\} \\ &= \sup\left\{\frac{\|Ax\|}{\|x\|} : x\in K^n \mbox{ with }x\ne 0\right\}. \end{align} 
Can someone provide a reference to the proof of this claim?

REPLY [2 votes]: You can also prove it by relating spectral radius with matrix norm.
For a linear operator $A$,
$$\operatorname{spr}(A) \le \|A\|$$
where
$$\begin{align}
\|A\| &= \underset{\|x\|=1}{\sup} \|Ax\| \\
\operatorname{spr}(A) &= \max\left\lbrace |\lambda|: \lambda \text{ is an eigenvalue of } A \right\rbrace
\end{align}$$
 The spectral radius $\operatorname{spr}(A)$ and $\|A\|$ are equal iff the operator $A$ is normal.
So $\|A\| < 1 \Rightarrow \operatorname{spr}(A) < 1 \Rightarrow $ all of the eigenvalues of $A$ have absolute value smaller than 1.
See Exercise I.2.6 of Rajendra Bhatia's Matrix Analysis where he calls it the Neumann Series.