 Solution

Write $g(s)=G_T(s)$. For $0<s<1$, differentiate the power-series identity from part (ii):Because $g(s)\le 1$ and $\lambda <1$, its denominator is positive. Using $g(s)=s{e}^{\lambda (g(s)-1)}$ givesAs $s\uparrow 1$, almost-sure finiteness and bounded convergence give $g(s)\to 1$. On the other hand, monotone convergence in $g^{\prime }(s)={\sum }_{k\ge 1}k\mathbb{P}(T=k)s^{k-1}$ gives $g^{\prime }(s)\to \mathbb{E}T$, initially allowing infinity. The displayed formula makes this limit finite and provesThis establishes finiteness rather than assuming it when rearranging an expectation equation. It is the subcritical branching-process total-progeny mean.