= Solution
Expand the analytic factor around zero by the <binomial series>:
$$
(14-5z)^{-1/2}=\frac1{\sqrt{14}}\sum_{j=0}^\infty
\frac{\binom{2j}{j}}{4^j}\left(\frac5{14}\right)^jz^j.
$$
Every coefficient is nonnegative. Adding $2/3$ to the constant coefficient gives probabilities $p_j$ with sum $g(1)=2/3+1/3=1$. Define $\Pr(T=j)=p_j$ for $j\geq0$. This is a nonnegative integer-valued <random variable> and its <probability generating function> is exactly $g$. The power series has radius $14/5>1$, so the evaluations and derivatives used below are finite.
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