= Solution
Put $q=1-p$. The count has the zero-based <geometric distribution>, whose <probability generating function> is $G_N(z)=p/(1-qz)$. Hence the aggregate has a <compound geometric distribution>, with
$$
\boxed{\kappa_S(t)=\log p-\log\bigl(1-qM_{X_1}(t)\bigr).}
$$
This transform requires $qM_{X_1}(t)<1$; it holds for every $t\le0$. To differentiate explicitly, let $g(t)=M_{X_1}(t)$. Then
$$
\kappa_S'(t)=\frac{qg'}{1-qg},\qquad \kappa_S''(t)=\frac{qg''}{1-qg}+\frac{q^2(g')^2}{(1-qg)^2},
$$
and
$$
\kappa_S'''(t)=\frac{qg'''}{1-qg}+\frac{3q^2g'g''}{(1-qg)^2}+\frac{2q^3(g')^3}{(1-qg)^3}.
$$
At zero, $g=1$ and $g^{(j)}=m_j$, giving
$$
\boxed{\kappa_{S,3}=\frac qp m_3+\frac{3q^2}{p^2}m_1m_2+\frac{2q^3}{p^3}m_1^3.}
$$
Every term is positive for $0<p<1$ and strictly positive claim sizes. Therefore \b[the unstandardized skewness is positive], even if the claim-size third <central moment> itself is negative. The zero-based convention matters: there is no factor $M_{X_1}(t)$ in the numerator of this aggregate transform.
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