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Assume that $$G(s) = \sum_{n=0}^{\infty} (1-b)^ns^n = \frac1{1-(1-b)s},\quad0\le b \le 1,$$ where $G(s)$ is a probability generating function. The problem I see is that $(1-b)s$ can be greater than $1$ ($s = 2, b = 0.3$) and then the series will just diverge.

So how can the equivalence of the series to the fraction hold? This is taken from a solution to a question, and I don't understand how the equivalence is true with the problem I specified.

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You are of course right. What is the question? –  Jochen Wengenroth Apr 25 '12 at 11:03
    
In formal power series, the above equation holds regardless of convergence: Formally, $(1-x) \sum_{n=0}^\infty x^n = 1$, where $x$ is an indeterminate. –  Johannes Kloos Apr 25 '12 at 11:47

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What you need here is the notion of a formal power series. It’s a formal object, manipulated according to formal rules. For example, two formal power series are multiplied as one would multiply polynomials. Thus, $$\left(\sum_{n\ge 0}a_nx^n\right)\left(\sum_{n\ge 0}b_nx^n\right)=\sum_{n\ge 0}\left(\sum_{k=0}^na_kb_{n-k}\right)x^n\;,$$ and in particular $$\begin{align*} (1-x)\sum_{k\ge 0}x^k&=(1-x)(1+x+x^2+x^3+\dots)\\ &=1+(1-1)x+(1-1)x^2+\dots\\\\ &=1\;. \end{align*}$$

Thus, formally $1-x$ and $\sum_{k\ge 0}x^k$ are multiplicative inverses, and we write $$\sum_{k\ge 0}x^k=\frac1{1-x}\;.\tag{1}$$ Replace $x$ by $(1-b)s$, and you get your generating function.

Since these are simply formal objects in an indeterminate $x$, we aren’t evaluating them, and convergence simply isn’t an issue. It is true, however, that when the series on the lefthand side of $(1)$ converges, its value is given by the function on the righthand side: within its interval of convergence, this particular power series does represent the function to which it is formally equal. Typically, however, we’re interested in the coefficients of the formal power series, and the associated function is a way to perform certain useful manipulations on them easily.

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