Let $Q$ be a M/G/1-queue. We denote by $B$ the busy period of the queue, that is defined as follows: $$B:=\inf \{ t>0: Q(t+T_1)=0 \},$$ where $T_1$ is the arriving time of the first costumer.
Theorem: $$\Pr(B<\infty) \begin{cases} =1 & \text{ if } \rho \leq 1, \\ <1 & \text{ if } \rho >1. \end{cases}$$
How can I explicitly compute $\Pr(B<\infty)$ in the case when $\rho >1$?
$Z_n$ denotes the number of customers in the $n$th generation, where $Z_0=1$. Then we know that $$\Pr(B<\infty)=\Pr(Z_n=0 \text{ for some } n).$$ Now I have seen, that $$\lim_{n \rightarrow \infty}G_n(0)=\Pr(Z_n=0 \text{ for some } n),$$ where $G_n(s):=\mathbb{E}[s^{Z_n}]$ is the generating function of $Z_n$.
But is it possible to compute $G_n(s)$?
Once the $n$th generation of customers has been suitably defined (which is not a trivial task), the branching process representation mentioned in the question shows that the extinction probability $P(B\lt\infty)$ solves $$\theta=E(\mathrm e^{-\lambda(1-\theta)X}),\tag{$\ast$}$$ where $\lambda$ denotes the rate of arrivals and $X$ the (random) service time of any given customer. When $\rho=\lambda E(X)$ is such that $\rho\gt1$, then the root $\theta$ of $(\ast)$ which is in $(0,1)$ is the extinction probability.
There is usually no simple formula for the generating functions $G_n$.