For $p \in (0,1)$ and $b \in [0,p]$, define a probability measure by $$P(Z = k) = b(1-p)^{k-1}$$ if $k \ge 1$, $$P(Z=k) = 1 - b/p$$ if $k = 0$. Show that $s = \frac{1- b/p}{1-p}$ is a fixed point of the probability generating function of $Z$.
I simply don't understand why it doesn't work out the way I did it. By definiton:
$$g_Z(s) = \sum_{k = 0}^\infty P(Z = k)s^k = (1-b/p)\sum_{k = 1}^\infty b(1-p)^{k-1}s^k.$$ Multplying both sides by $(1-p)$ yields $$g_Z(s)(1-p) = b(1-b/p) \sum_{k=1}^\infty ((1-p)s)^k = b(1-b/p) \sum_{k=1}^\infty (1 - b/p)^k.$$ Starting with $k = 0$, this becomes the geometric series, so we shift the index and receive $$b(1-b/p)(1-b/p)\sum_{k=0}^\infty (1-b/p)^k = b(1-b/p)^2 \frac{1}{b/p} = (1- b/p)^2 p.$$
But $$(1- b/p)^2 p \neq 1 - b/p,$$ so even after dividing by $(1-p)$, this doesn't become by desired fixed point.
Your probability generating function is incorrect. It should be $$ g_{Z}(s)=\sum_{k=0}^\infty P(Z=k)s^k=1-b/p+\sum_{k=1}^\infty b(1-p)^{k-1}s^k=1-b/p+\frac{bs}{1-(1-p)s} $$ and solve for $g_{Z}(s)=s$.