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For the binomial branching process, the offspring probability generating function is $f(s)=(1-p+ps{)}^n$. The Galton-Watson extinction fixed point gives $1-\rho =f(1-\rho )=(1-p\rho {)}^n$.

The printed finite-binomial upper bound is false without an additional asymptotic qualification. For $n=2$ and $p=(1+\epsilon )/2$, solving the fixed-point equation exactly givesFor instance $\epsilon =1/20$ gives $\rho =80/441>1/10=2\epsilon$. More generally, at every fixed $n>1$ the Taylor expansion gives $\rho =2n\epsilon /(n-1)+O_n({\epsilon }^2)$, again contradicting that upper bound for sufficiently small positive $\epsilon$.

Here is the corrected binomial branching survival correction. The Poisson branching process of mean $1+\epsilon$ has branching survival probability $r$ satisfying $-\log (1-r)=(1+\epsilon )r$. Expanding the logarithm givesThus $2\epsilon /(1+2\epsilon )\le r\le 2\epsilon$, which implies the requested lower estimate $r\ge 2\epsilon -4{\epsilon }^2$. Since $(1-p+ps{)}^n\le e^{np(s-1)}$ on $[0,1]$, iteration of the two offspring probability generating functions gives $\rho \ge r$.

For the actual binomial branching process, expansion of its exact fixed-point equation givesIf $(1+\epsilon {)}^2<n$, all coefficients are nonnegative. Keeping the first term provesIn particular, for $n\to \infty$ and $\epsilon =o(1)$ this gives $\rho =2\epsilon +O({\epsilon }^2+\epsilon /n)$. The printed bounds also hold for the binomial branching process in an explicit large-$n$ regime: $0<\epsilon \le 1/4$ and $n\epsilon \ge 2$. Indeed these conditions give $n{p}^2\le \epsilon$ and $n{p}^3\le 1/4$. In the preceding nonnegative series, evaluation at $x=2\epsilon$ givesThe series is increasing, and its value at the actual branching survival probability $\rho$ is $\epsilon$, so $\rho \le 2\epsilon$. Together with the lower bound already proved, this recovers $2\epsilon -4{\epsilon }^2\le \rho \le 2\epsilon$ in that regime. In particular it applies eventually to part (ii). The counterexample shows why a regime condition is needed for a literal finite-$n$ statement.

Let $Z_n$ be the size of generation $n$. This is a Galton-Watson process with offspring meanThe expected generation size in a Galton-Watson process is thereforeIts probability generating function isBy the Galton-Watson extinction fixed point, the extinction probability $q$ is the smallest solution in $[0,1]$ of $G(q)=q$. Factoring givesso $q=(\sqrt{29}-5)/2$. The probability of production continuing forever is

Let $d_n$ be the probability of extinction by generation $n$. Conditioning on the initial individual's offspring givesThe events increase to eventual extinction, so $d_n\uparrow d$ and continuity gives $d=G_X(d)$. If $r\ge 0$ is any other fixed point, induction from $d_0=0\le r$ and monotonicity of $G_X$ gives $d_n\le r$ for every $n$. Hence $d\le r$, proving the Galton-Watson extinction fixed point characterization.