Suppose someone claims that for $\alpha\in(0,1)$, the function
$f(u):=1-(1-u)^\alpha$
is a probability generating function of a nonnegative, nonconstant integer values random variable N. What properties of f must you check? Is the hypothesis $\alpha>0$ used? What happens in the cases $\alpha=0$, $\alpha=1$ and $\alpha>1$?
Check:
(i) That $f(1)=1,\;$ since $f(1) = \sum_n {P(N=n)}$.
(ii) For a pgf, $P(N=n) = f^{(n)}(0)/n!,\;$ where $^{(n)}$ indicates the $n^{th}$ derivative. So $f$ must be infinitely differentiable and $f^{(n)}(0)/n!$ must be non-negative.
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For $f(u) = 1-(1-u)^\alpha,\;$ we have,
(i) $f(1) = 1$ holds iff $\alpha \neq 0$.
(ii)
\begin{eqnarray*} f^{'}(u) &=& \alpha(1-u)^{\alpha-1} \\ f^{''}(u) &=& -\alpha(\alpha-1)(1-u)^{\alpha-2} \\ \text{and generally,}\qquad\qquad && \\ f^{(n)}(u) &=& (-1)^{n+1}\alpha(\alpha-1)\cdots(\alpha-n+1)(1-u)^{\alpha-n} \\ \text{so}\qquad\qquad\qquad\qquad && \\ f^{(n)}(0)/n! &=& (-1)^{n+1}\alpha(\alpha-1)\cdots(\alpha-n+1)/n!. \\ \end{eqnarray*}
This is non-negative $\forall\; n \geq 0 $ iff $ \alpha\in(0,1)$.
So $f$ is a probability generating function iff $ \alpha\in(0,1)$.