I am stuck on exercise 11.2 From Grimmett's probability on graphs. Here is a link to the pdf on his website.
Consider a branching process whose family-sizes have the binomial distribution bin$(n, \frac{\lambda}{n})$. Show that the extinction probability converges to $\eta(\lambda)$ as $n \rightarrow \infty$, where $\eta(\lambda)$ is the extinction probability of a branching process with family-sizes distributed as $\text{Po}(\lambda)$.
To solve this exercise, we use a theorem from his other book, which states that if you have a branching process whose family sizes are $X$ distributed then the extinction probability of this branching process is the smallest non-negative fixed point of the equation $s = G(s)$ where $G$ is the probability generating function $G(s) = \sum_{k = 0}^\infty \mathbb{P}(X = k)s^k$.
My next step was computing these probability generating functions for $\text{bin}(n,\lambda)$, these are $G_n(s) = (1+\frac{\lambda( s -1)}{n})^n$. When the family sizes are Poisson distributed with parameter $\lambda$ we get $G(s) = e^{\lambda(s-1)}$. and so clearly $G_n$ converges pointwise to $G$.
This however is not enough, as we want to show that the extinction probabilities also converge, so we have $s_n$ extinction probabilities of the $n$-th process, i.e. $s_n$ is the smallest non-negative solution to $s = G_n(s)$, and we want to show that this converges the smallest non-negative solution of $s = G(s)$. I hoped I could use some theorem which states that if you have pointwise convergence then the fixed points also converge, but this is false, see this counterexample.
During writing this I thought of an attempt to solve this using Hurwitz theorem, namely, show that the smallest non-negative fixed point of $G(s)$ is smaller than 1, then show that the function $G'$ is complex differentiable with non-zero derrivative, use inverse function theorem, get an open neighbourhood of our fixed point, find sequence of fixed points which converge to the smallest non-negative fixed point $\eta(\lambda)$ of $G$. Here I am stuck trying to show that $s_n$ are the smallest non-negative fixed points of $G_n$.
How to proceed?
For each $n$ large enough, let $q_n(\lambda)$ denote the extinction probability of the binomial branching process with parameters $(n,\lambda/n)$. Let $q(\lambda)$ denote the extinction probability of the Poisson branching process with parameter $\lambda$ (this is your $\eta(\lambda)$).
For every fixed $s$ in $[0,1)$, the sequence $(G_n(s))$ increases to $G(s)$, each $q_n(\lambda)$ is the smallest solution in $[0,1]$ of the equation $s=G_n(s)$, and $q(\lambda)$ is the smallest solution in $[0,1]$ of the equation $s=G(s)$, hence the sequence $(q_n(\lambda))_n$ is increasing and $q_n(\lambda)<q(\lambda)$ for every $n$ (to prove this, use that every $G_n$ is increasing on $[0,1]$).
Let $r(\lambda)$ denote the limit of the sequence $(q_n(\lambda))_n$. Then $q_n(\lambda)<r(\lambda)\leqslant q(\lambda)$ for every $n$, in particular, $G_n(r(\lambda))<r(\lambda)$, which implies that $G(r(\lambda))=\lim G_n(r(\lambda))\leqslant r(\lambda)$. Now, $G(r(\lambda))\leqslant r(\lambda)$ is equivalent to $r(\lambda)\geqslant q(\lambda)$, hence all this proves that $\lim q_n(\lambda)=r(\lambda)=q(\lambda)$, as desired.