factorial moments from generating functions by taylor expansion at z=1 Let $X$ be some positive, integer valued random variable with probability generating function $G(z):=\mathbb E(z^X)$.
I computed this for Poisson and geometric distribution. Then, i did a taylor expansion at $z=1$ and got a series of the form
$$G(z)=\sum_{k\geq 0}\frac{\mathbb E[(X)_k]}{k!}(z-1)^k$$
where $\mathbb E[(X)_k]$ is the $k$-th factorial moment of $X$. I was asking myself, whether this is a general property.
The answer seems just one binomial away (as usual)...
$$G(z)=\sum_{k\geq 0}(z-1+1)^k\,\mathbb P(X=k)=\sum_{k= 0}^\infty\sum_{i=0}^k\binom{k}{i}(z-1)^i\mathbb P(X=k)\\
\sum_{k= 0}^\infty\sum_{i=0}^k\frac{k!}{(k-i)!}\frac{(z-1)^i}{i!}\mathbb P(X=k)
$$
Now it has the factorials, but the summations is not right... How can i fix that?

Just upon inverting the two sums
$$G(z)=\sum_{k\geq 0}(z-1+1)^k\,\mathbb P(X=k)=\sum_{k= 0}^\infty\sum_{i=0}^k\binom{k}{i}(z-1)^i\mathbb P(X=k)\\
= \sum_{i=0}^\infty \frac{(z-1)^i}{i!} \sum_{k=i}^\infty(k)_i\mathbb P(X=k)
$$
Then because $k<i \implies (k)_i = 0$
$$= \sum_{i=0}^\infty \frac{(z-1)^i}{i!} \sum_{k=0}^\infty(k)_i\mathbb P(X=k) = \sum_{i=0}^\infty \frac{(z-1)^i}{i!} \Bbb E[(X)_k]
$$
