= Solution
Let $P(z)=\sum_{n\ge0}p_nz^n$, $F(z)=\sum_{j\ge1}f_jz^j$, and $G(z)=\sum_{k\ge0}g_kz^k$ be the count, claim-size and aggregate <probability generating functions>. Independence in the <aggregate claims model> gives
$$
G(z)=\mathbb E\bigl(F(z)^N\bigr)=P(F(z)).
$$
The <Panjer claim-count class> recurrence can be written $np_n=(an+b)p_{n-1}$. Multiply by $z^{n-1}$ and sum to obtain
$$
P'(z)=a\bigl(zP'(z)+P(z)\bigr)+bP(z),
\qquad (1-az)P'(z)=(a+b)P(z).
$$
Using the <chain rule> in $G=P\circ F$ then yields
$$
(1-aF(z))G'(z)=(a+b)F'(z)G(z).
$$
The coefficient of $z^{k-1}$ on the left is $kg_k-a\sum_{j=1}^k f_j(k-j)g_{k-j}$, where the $j=k$ term is zero. On the right it is $(a+b)\sum_{j=1}^k jf_jg_{k-j}$. Rearranging gives the <Panjer recursion>
$$
\boxed{g_0=p_0,\qquad
g_k=\sum_{j=1}^k\left(a+\frac{bj}{k}\right)f_jg_{k-j},\quad k\ge1.}
$$
The initial value is $p_0$ because all claim sizes are positive: zero total loss occurs exactly when $N=0$. Only already computed aggregate probabilities occur on the right. The count normalization determines $p_0$ from $a,b$ as well: with $w_0=1$ and $w_n=\prod_{j=1}^n(a+b/j)$, we have $p_n=p_0w_n$ and $p_0=(\sum_{n\ge0}w_n)^{-1}$. The assumed valid count law makes this sum finite.
Finally $p_0$ must be positive. If it were zero, the recurrence would make every $p_n$ zero, contradicting normalization. At $n=1$ the recurrence therefore gives
$$
\boxed{a+b=\frac{p_1}{p_0}\ge0.}
$$
This proves the sign constraint without assuming any particular count family. The three parts below also provide explicit initial probabilities for the aggregate recursion.
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