Given that pgf is $G(H(\xi))=\frac{1+\xi}{3-\xi}$. Where $H(\xi)=\frac{1}{2}(1+\xi)$ and $G(\xi) = \frac{\xi}{2-\xi}$. And that $G(H(\xi))$ is the pgf of some random variable $Y$. How does one get the value of $P(Y=r)$, I realize that I have to differentiate $r$ times and then set $\xi$ to $0$. But in practice..
$$\frac{d^r}{d\xi^r} G(H(\xi)) \Bigr|_{\substack{\xi=0}}$$
I was thinking of making a substitution $H(\xi)=u$, but $\frac{d^r}{d\xi^r} \neq \frac{d^r}{du^r} \frac{d^r u}{d\xi^r} $
$$ \frac{1+\xi}{3-\xi}=\frac13\frac{1+\xi}{1-\frac\xi3}=\frac13(1+\xi)\sum_{k=0}^\infty\left(\frac\xi3\right)^k\;. $$