I am trying to find the PGF with the following definition: $$ F(z) = <z^n> = E[z^n] = \sum_{n=0}^{\infty}P_nz^n $$ Considering a discrete probability distribution given by $P_n = p(1-p)^n$ for $(n=0,1,2,...)$.
My attempt: $$ F(z) = <z^n> = E[z^n] = \sum_{n=0}^{\infty}p(1-p)^nz^n $$ $$ = \sum_{n=0}^{\infty}p[(1-p)z]^n $$ which is just a geometric series so you can simplify as $$ F(z) =\frac{p}{1-qz} $$ where $q=1-p$. I don't know where I went wrong but just looking around online I think the PGF for a Bernoulli Trial type distribution should be: $$ PGF = q+pz $$ Any help would be much appreciated!