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For $n\ge 1$, substitute $s=e^{t/n}$ in the probability generating function. The moment-generating function of $X_n/n$ isThe 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:Its conditional moment-generating function is $e^{t/n}/[n+1-n{e}^{t/n}]$, 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 $x\ge 0$, the strict upper tail is exactlySince $\log r_n=-1/n+O(n^{-2})$ and $\lfloor nx\rfloor /n\to x$,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 $n$.

For a nonnegative integer-valued random variable, the probability generating function iswith $0\le s\le 1$ always allowed. Its derivative at $1$ from below gives the expectation when finite.

The waiting time $N$ has the positive-integer geometric distribution $P(N=j)=p{q}^{j-1}$. For the capped geometric waiting time, $X=j<k$ exactly when $N=j$, while $X=k$ occurs whenever $N\ge k$. Thus the terminal mass is $q^{k-1}$, givingThe finite sum is a polynomial; any apparent singularity of the rational expression is removable. This includes $k=1$, where $G_X(s)=s$.

Differentiating the rational form at $s=1$ yieldsAs a second interpretation, the expectation is the sum of the tail probabilities $P(X\ge j)=q^{j-1}$ for $1\le j\le k$. Their finite geometric sum gives the same answer.

Condition on $N$. Since the amounts are independent and identically distributed random variables and are independent of $N$, their moment-generating functions multiply:Summing against the positive-support geometric distribution gives the geometric-sum moment-generating functionThe finite-transform domain is $M_X(t)<\infty$ and $(1-p)M_X(t)<1$. The terms are nonnegative for real $t$, so if the latter condition fails the sum diverges. For $t\le 0$ both conditions always hold, since the claims are positive; thus the same calculation always provides the Laplace transform of a nonnegative random variable. Positivity alone does not guarantee a finite moment-generating function on a positive neighbourhood of zero.

There is at least one strictly positive amount in the portfolio. Hence $\mathbb{P}(S=0)=0$; zero carries no atom of a measure to add to the continuous laws in the next two parts.