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I noticed that $(1+S)^{-1}=\sum_{n=0}^{\infty}S^n$, where S is a square matrix. Is there any theorem related this identity?

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    Perhaps you mean $(I - S)^{-1} = \sum_{n=0}^\infty S^n$. There are cases where the right hand side fails to converge, so it holds only conditionally. But it is a pretty important idea, cropping up in lots of contexts. – hardmath May 05 '16 at 20:47

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This is only true (i.e., converges), if each eigenvalue of $S$ has absolute value less than $1$. Is has been discussed often on MSE, e.g.,here, or here. If $S$ is nilpotent with $S^n=0$, then $$ (1+S)^{-1}=I-S+S^2\cdots \pm S^{n-1}. $$

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Dietrich Burde
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