How to find the probability generating function?

Question

Suppose for $d>0$ $$ p_d=\frac{1}{-\log(1-e^{-\beta})}\left(\frac{e^{-\beta d}}{d}\right) $$

How do I find the generating function

Answer 1

Since you have defined $p_0 = 0$ you can start the generating function at the index $d=1$. Hence, for all $0< x < e^\beta$ we have:

$$\begin{equation} \begin{aligned} g(x) = \sum_{d=1}^\infty p_d x^d &= \frac{1}{| \log (1- e^{-\beta}) |} \sum_{d=1}^\infty e^{- \beta d} \frac{x^d}{d} \\[8pt] &= \frac{1}{| \log (1- e^{-\beta}) |} \sum_{d=1}^\infty e^{- \beta d} \int \limits_0^x r^{d-1} dr \\[8pt] &= \frac{1}{| \log (1- e^{-\beta}) |} \int \limits_0^x \Bigg( \sum_{d=1}^\infty e^{- \beta d} r^{d-1} \Bigg) dr \\[8pt] &= \frac{1}{| \log (1- e^{-\beta}) |} \int \limits_0^x \Bigg( e^{- \beta} \sum_{d=0}^\infty (r e^{- \beta})^{d} \Bigg) dr \\[8pt] &= \frac{1}{| \log (1- e^{-\beta}) |} \int \limits_0^x \frac{e^{- \beta}}{1 - r e^{-\beta}} dr \\[8pt] &= \frac{1}{| \log (1- e^{-\beta}) |} \Big[ - \log (1-r e^{-\beta}) \Big]_{r=0}^{r=x} \\[8pt] &= \frac{| \log (1-x e^{-\beta}) |}{| \log (1- e^{-\beta}) |} \\[8pt] &= \frac{ \log (1-x e^{-\beta}) }{ \log (1- e^{-\beta}) }. \\[8pt] \end{aligned} \end{equation}$$

(For $x \geqslant e^\beta$ the summation in this expression diverges to infinity and so the generating function diverges. Hence, the radius of convergence for this generating function is $e^\beta$.)