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I saw some place where someone wrote $\frac{1}{1-f(x)} = \sum_{n=0}^{\infty} f(x)^n, \forall x \in \mathbb{C}, $ where $f$ is some function s.t. $\vert f(x)\vert < 1, \forall x \in \mathbb{C}$.

Then he proved that the RHS $\sum_{n=0}^{\infty} f(x)^n$ absolutely converges $, \forall x \in \mathbb{C}$, in order that the RHS is well-defined and finite. I was wondering why we have to have the absolute convergence? Isn't that $\vert f(x)\vert < 1, \forall x \in \mathbb{C}$ can guarantee $\frac{1}{1-f(x)} = \sum_{n=0}^{\infty} f(x)^n, \forall x \in \mathbb{C}$?

I admit I am not familiar with complex analysis, and I am not sure if there is also a similar situation in Real analysis? I would appreciate if someone could point out what kinds of materials (such as Wikipedia articles or other internet links) will help me in this regard, besides explanation. Thanks for clarification!

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    @Jonas: On a piece of paper handwritten by some math student. I also asked another math student, who agreed with what is on the paper, but I can't understand what he said. He said something like analytical function, for the series to converge, etc.2011-03-08

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Yeah, it's not clear to me why he would need absolute convergence, given that he knowns |f(x)|<1. He might have wanted to show the RHS is continuous, but that seems obvious since it is equal to the LHS, which is continuous.

It is possible that, for some reason, the author originally needed this argument, but, due to a later simplification, he failed to drop this line of reasoning, not realizing it had become redundant.