Let $X$ be a discrete random variable with probability generating function $G(t)=\frac1 {2-t}$, what distribution has $X$?
I know that to get the $G(t)$ I need to apply the definition of probability generating function which is:
\begin{align} G(t)=E(t^X)=\sum_{x=0}^{\infty} t^x P(X=x) \end{align}
And in this case $P(X=x)$ is just $X$, my question is: is there an opposite function to apply to $G(t)$ so I can get the probability function and see what kind of distribution appears to be? My guess is that $X$ is continuous uniform distribution but I don't know how to prove it.
Plug in $t=0$ to get $\mathbb P(X=0)=\frac{1}{2}$. So then $$\frac{1}{2-t}=G(t)=\frac{1}{2}+\sum_{k=1}^\infty\mathbb P(X=k)\cdot t^k.$$ Subtract $\frac{1}{2}$ from each side, divide by $t$ and plug in $t=0$ (strictly speaking we should be taking the limit $t\to0$) again to get $\mathbb P(X=1)=\frac{1}{4}$. So then $$\frac{1}{4-2t}=\frac{G(t)-1/2}{t}=\frac{1}{4}+\sum_{k=2}^\infty\mathbb P(X=k)\cdot t^{k-1}.$$ You can extract the whole probability mass function by continuing like this.
A trick for this specific PGF: you might know that $\frac{1}{1-t}=1+t+t^2+\dots$ (e.g. from the geometric series formula, or Taylor expanding), so then $$G(t)=\frac{1}{2-t}=\frac{1}{2}\left(\frac{1}{1-t/2}\right)=\frac{1}{2}+\frac{t}{4}+\frac{t^2}{8}+\dots$$ and we can instantly read off $\mathbb P(X=k)=(1/2)^{k+1}$.
You might recognise this as the geometric distribution -- in this case, the number of tails you flip before you flip your first head, for a fair coin.