= Solution
For $n\geq1$, substitute $s=e^{t/n}$ in the <probability generating function>. The <moment-generating function> of $X_n/n$ is
$$
\boxed{M_{X_n/n}(t)=\frac{n-(n-1)e^{t/n}}{n+1-ne^{t/n}}},\qquad
\boxed{t<n\log(1+1/n)}.
$$
The domain follows from convergence of the generating series; at or above that positive threshold the expectation diverges, even though a rational continuation can be written.
Conditioning on survival, the <probability mass function> from part (ii) becomes a positive-integer <geometric distribution>:
$$
P(X_n=j\mid X_n>0)=\frac1{n+1}\left(\frac n{n+1}\right)^{j-1},\qquad j\geq1.
$$
Its conditional <moment-generating function> is $e^{t/n}/[n+1-ne^{t/n}]$, tending to $1/(1-t)$ for fixed $t<1$. We can prove the required limit directly, without invoking a continuity theorem. Set $r_n=n/(n+1)$. For every $x\geq0$, the strict upper tail is exactly
$$
P(X_n/n>x\mid X_n>0)=r_n^{\lfloor nx\rfloor}.
$$
Since $\log r_n=-1/n+O(n^{-2})$ and $\lfloor nx\rfloor/n\to x$,
$$
\boxed{P(X_n/n>x\mid X_n>0)\longrightarrow e^{-x}}.
$$
At $x=0$ both tails equal one. This is the <exponential limit of surviving geometric branching>. Conditioning is essential: the unconditional variable has <probability> $n/(n+1)$ of being zero, whereas the rare surviving population has size of order $n$.
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