Given $G(x)= c \ln(1-{x \over 2})$, $x \in \mathbb R$, for which $c \in \mathbb R$ is $G$ a probability generating function?
I am aware of the definition of a generating function but I'm having a hard time to figure out what it needs to suffice. Some hint or approach would be much appreciated.
An analytic function (or more generally a formal power series) $$A(z) = \sum_{n \geq 0}a_n z^n$$ is a probability generating function if and only if $a_n \geq 0$ for each $n$ and $A(1) = \sum a_n = 1$. In your example, we try to force $G(1) = 1$: $$c \ln(1/2) = 1 \implies c = - \frac{1}{\ln(2)}\,.$$ We then have to check that this choice of $c$ yields non-negative coefficients, which it does, therefore implying that $c = -1 / \ln(2)$ is the only answer.