= Solution
For the <binomial distribution>, $G_N(z)=(1-p+pz)^n$. Thus the <compound binomial distribution> has
$$
\boxed{\kappa_S(t)=n\log\bigl(1-p+pM_{X_1}(t)\bigr).}
$$
Set $h(t)=1-p+pM_{X_1}(t)$. Since $h(0)=1$ and $h^{(j)}(0)=pm_j$, differentiating $n\log h$ three times gives
$$
\boxed{\kappa_{S,3}=n\bigl(pm_3-3p^2m_1m_2+2p^3m_1^3\bigr).}
$$
The first two <cumulants>, for comparison, are $npm_1$ and $n(pm_2-p^2m_1^2)$.
Choose deterministic claim sizes $X_i=d>0$. Then $S=dN$, so $m_j=d^j$ and
$$
\kappa_{S,3}=nd^3p(1-p)(1-2p).
$$
This is strictly negative for $p>1/2$ and $n\ge1$. A concrete \b[negative-skewness example] is $n=1$, $p=3/4$, $d=1$, for which
$$
\boxed{\kappa_{S,3}=-3/32<0.}
$$
There is no contradiction with positive claims: bounded Bernoulli-type totals can have more mass near their upper value and a longer left tail.
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