I'm working on the following exercise:
Show that $G(s)=1-\alpha(1-s)^{\beta}$ is the probability generating function of a nonnegative integer valued random variable when $\alpha, \beta\in(0,1)$.
I tried the following:
The probability generating function of a discrete random variable $X$ is defined by $G_X(s)=\mathbb{E}(s^X)=\sum_{k=0}^{\infty}s^k\cdot\mathbb{P}(X=k)$, and thus $G_X^{(k)}(0)=k!\cdot\mathbb{P}(X=k)$, where $G_X^{(k)}(s)$ denotes the $k$'th derivative with respect to $s$. Thus $\mathbb{P}(X=k)=\frac{G_X^{(k)}(0)}{k!}$. Working this out I find:
$$\begin{align}\mathbb{P}(X=0)&=\frac{G_X^{(0)}(0)}{0!}=G_X^{(0)}(0)=G_X(0)=1-\alpha\\ \mathbb{P}(X=1)&=\frac{G_X^{(1)}(0)}{1!}=\alpha\beta\\ \mathbb{P}(X=2)&=\frac{G_X^{(2)}(0)}{2!}=\frac{\alpha\beta(1-\beta)}{2!}\\ & \ \ \vdots\\ \mathbb{P}(X=n)&=\frac{G_X^{(n)}(0)}{n!}=\frac{\alpha\beta(1-\beta)(2-\beta)\cdots(n-\beta)}{n!}\\ \end{align}$$
If $\alpha, \beta\in(0,1)$ it follows that all probabilities $\mathbb{P}(X=k)$, for $k\in\mathbb{N}\cup\{0\}$ are in $(0,1)$. For this nonnegative integer valued random variable I will assume that $\mathbb{P}(X=x)=0$ for $x\not\in\mathbb{N}\cup\{0\}$.
To show that this is indeed the probability generating function of some nonnegative integer valued random variable I have to show that $$\sum_{k=0}^{\infty}\mathbb{P}(X=k)=1-\alpha+\sum_{k=1}^{\infty}\frac{\alpha\beta(1-\beta)(2-\beta)\cdots(k-\beta)}{k!}\\=1-\alpha+\alpha\sum_{k=1}^{\infty}\frac{\beta(1-\beta)(2-\beta)\cdots(k-\beta)}{k!}$$ equals $1$. I didn't succeed to show this, but I have the feeling I have to use the Binomial Theorem somehow. Any ideas? Thanks in advance!
For $\alpha, \beta \in (0,1)$, your $G$ satisfies
So $G$ is a probability generating function.
Edit: You are trying to use that $G(s)=\sum_{n=0}^{\infty}\frac{G^{(n)}(0)}{n!}s^n$, but you do not have to calculate the power series on the RHS and plug in $s=1$. You already have its closed form (the LHS) and you can do this directly.