Consider a branching Markov chain $(X_n)_{n \geq 0}$, where $X_0 = 1$ and the offspring distribution is given by $$p(0) = \frac 1 6, p(1) = \frac 1 2, p(2) = \frac 1 3.$$ The number of offspring produced by each individual is independent of the number of offspring produced by other individuals.
$(a)\quad$ Find the probability of ultimate extinction.
Now, suppose that the offspring distribution remains unchanged, but that $X_0$ has a geometric distribution with PMF $$\mathbb{P}(X_0 = m) = (1 - \pi)\pi^{m - 1}\ \forall\ m \in \mathbb{Z}^+.$$
$(b)\quad$ Find the probability of ultimate extinction.
My working
$(a)\quad$ We can clearly see that the mean number of offspring produced by each individual is more than one, so the probability of ultimate extinction is less than one. Let this probability be $s$ and we have $$\frac 1 6 + \frac 1 2 s + \frac 1 3 s^2 = s.$$ Now, solving this equation will give $$s = \frac 1 2.$$
Edit
Following the discussion in the comments, my working for part $(b)$ is as follows:
The mean number of offspring produced by each individual is still more than one, so the probability of ultimate extinction, $s$, is still less than one.
$$\begin{aligned} \because \rho_{10} & = \frac 1 2\\[1 mm] & = (\rho_{20})^{\frac 1 2}\\[1 mm] & = (\rho_{30})^{\frac 1 3}\\[1 mm] & = \dots,\\[1 mm] \therefore \rho_{n0} & = 2^{-n}\ \forall\ n \in \mathbb{Z}^+.\\[1 mm] \implies s & = \sum^{\infty}_{n = 1} \mathbb{P}(X_0 = n) \rho_{n0}\\[1 mm] & = (1 - \pi) \sum^{\infty}_{n = 1} \pi^{n - 1}\left(2^{-n}\right)\\[1 mm] \end{aligned}$$
However, this is where I am stuck. We can observe that the sum is the sum to infinity of a geometric series with first term $\frac 1 2$ and common ratio $\frac 1 2 \pi$. Since the common ratio is not less than one, this sum is not finite i.e. $s = \infty$, but how can we have infinite probability? Moreover, this contradicts my initial statement that the ultimate probability of extinction is less than one, but I am unable to figure out where I have gone wrong, so any intuitive explanations will be greatly appreciated :)