= Solution
Write $g(s)=G_T(s)$. For $0<s<1$, differentiate the power-series identity from part (ii):
$$
g'(s)=e^{\lambda(g(s)-1)}+s\lambda e^{\lambda(g(s)-1)}g'(s).
$$
Because $g(s)\leq1$ and $\lambda<1$, its denominator is positive. Using $g(s)=s e^{\lambda(g(s)-1)}$ gives
$$
g'(s)=\frac{g(s)/s}{1-\lambda g(s)}.
$$
As $s\uparrow1$, almost-sure finiteness and bounded convergence give $g(s)\to1$. On the other hand, monotone convergence in $g'(s)=\sum_{k\geq1}k\mathbb P(T=k)s^{k-1}$ gives $g'(s)\to\mathbb ET$, initially allowing infinity. The displayed formula makes this limit finite and proves
$$
\boxed{\mathbb E[T]=\frac1{1-\lambda}.}
$$
This establishes finiteness rather than assuming it when rearranging an expectation equation. It is the subcritical branching-process total-progeny mean.
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