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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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    I understand absolute convergence--your series converges absolutely for $|f(x)|<1$. What you mean by continuous convergence?2011-03-07
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    @Fabian: Thanks! Corrected.2011-03-07
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    $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$ holds formally and is valid, as functions, for $|x|<1$. replacing $x$ with $f(x)$ this still holds for any $x$ such that $|f(x)|<1. you don't _need_ absolute convergence, but it will converge absolutely (just review why a geometric series converges).2011-03-07
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    Absolute convergence is not much more than convergences. Inside the convergence radius the series always converges absolutely. Just on the boundary it can be different... In you case the series converges absolutely for $|f(z)|<1$.2011-03-07
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    "I saw some place where someone wrote": Any chance you remember where or who?2011-03-08
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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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