= Solution
For the <binomial branching process>, the offspring <probability generating function> is $f(s)=(1-p+ps)^n$. The <Galton-Watson extinction fixed point> gives $1-\rho=f(1-\rho)=(1-p\rho)^n$.
\b[The printed finite-binomial upper bound is false without an additional asymptotic qualification.] For $n=2$ and $p=(1+\varepsilon)/2$, solving the fixed-point equation exactly gives
$$
\rho=\frac{4\varepsilon}{(1+\varepsilon)^2}.
$$
For instance $\varepsilon=1/20$ gives $\rho=80/441>1/10=2\varepsilon$. More generally, at every fixed $n>1$ the <Taylor expansion> gives $\rho=2n\varepsilon/(n-1)+O_n(\varepsilon^2)$, again contradicting that upper bound for sufficiently small positive $\varepsilon$.
Here is the corrected <binomial branching survival correction>. The <Poisson branching process> of mean $1+\varepsilon$ has <branching survival probability> $r$ satisfying $-\log(1-r)=(1+\varepsilon)r$. Expanding the <logarithm> gives
$$
\varepsilon=\sum_{j\geq1}\frac{r^j}{j+1},\qquad
\frac r2\leq\varepsilon\leq\frac r{2(1-r)}.
$$
Thus $2\varepsilon/(1+2\varepsilon)\leq r\leq2\varepsilon$, which implies the requested lower estimate $r\geq2\varepsilon-4\varepsilon^2$. Since $(1-p+ps)^n\leq e^{np(s-1)}$ on $[0,1]$, iteration of the two offspring <probability generating functions> gives $\rho\geq r$.
For the actual <binomial branching process>, expansion of its exact fixed-point equation gives
$$
\varepsilon=\sum_{j\geq1}\frac{1-np^{j+1}}{j+1}\rho^j.
$$
If $(1+\varepsilon)^2<n$, all coefficients are nonnegative. Keeping the first term proves
$$
\boxed{\frac{2\varepsilon}{1+2\varepsilon}\leq\rho\leq
\frac{2\varepsilon}{1-(1+\varepsilon)^2/n}.}
$$
In particular, for $n\to\infty$ and $\varepsilon=o(1)$ this gives $\rho=2\varepsilon+O(\varepsilon^2+\varepsilon/n)$. The printed bounds also hold for the <binomial branching process> in an explicit large-$n$ regime: $0<\varepsilon\leq1/4$ and $n\varepsilon\geq2$. Indeed these conditions give $np^2\leq\varepsilon$ and $np^3\leq1/4$. In the preceding nonnegative series, evaluation at $x=2\varepsilon$ gives
$$
\sum_{j\geq1}\frac{1-np^{j+1}}{j+1}x^j
\geq\varepsilon(1-np^2)+\frac43\varepsilon^2(1-np^3)
\geq\varepsilon-\varepsilon^2+\varepsilon^2=\varepsilon.
$$
The series is increasing, and its value at the actual <branching survival probability> $\rho$ is $\varepsilon$, so $\rho\leq2\varepsilon$. Together with the lower bound already proved, this recovers $2\varepsilon-4\varepsilon^2\leq\rho\leq2\varepsilon$ 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-9-branching-survival.png]
{title=Finite-binomial survival exceeds the printed upper bound, while Poisson survival lies between the corrected comparison bounds}
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