Q:

Panjer-Distribution and limit

Consider a discrete random variable $X \geq 0 $. Then there are $a,b \in \mathbb{R}: a+b>0 $, such that $$p_k = \left(a+ \frac{b}{k}\right) p_{k-1} $$ with $p_k=P(X=k)$
Now consider the case $a<0 \Rightarrow b >0$ The sequence then $X_n:= a+ \frac{b}{n}$ is decreasing with limit $a<0$. So you can conclude, that there is a $m \in \mathbb{N}$, such that $X_m >0$ and $ X_{m+1} \leq 0$ 
How can I conclude now, that $p_k>0  \ \forall k \leq m$ and $p_k=0 \ \forall k \geq m+1$

A:

If $p_0 = 0$, then the recurrence relation would imply $p_k = 0$ for all $k\geqslant 0$.
This obviously cannot be , since $\sum_{k\geqslant 0} p_k$ must equal $1$.
It follows that $p_0>0$.
Now we use induction.

Claim: For each $k\leqslant m$, $p_k>0$.

Proof: The claim is true for $k=0$ as we've shown above.
Now, suppose the claim holds for some $0\leqslant k < m$.
We show it holds for $k+1$.
Indeed, because $k<m$, we have that $k+1\leqslant m$ and hence $a + \frac b{k+1} > 0$.
By the induction hypothesis, $p_k > 0$ and so
$$p_{k+1} = \left(a + \frac b{k+1}\right)p_k > 0$$
as the product of two positive numbers, which proves the claim. $\square$
If we now use the recursion, we'd get
$$p_{m+1} = \underbrace{\left(a+\frac b{m+1}\right)}_{\leqslant 0} \underbrace{p_m}_{>0} \leqslant 0,$$
Since the $p_k$ cannot be negative, we must have $p_{m+1} = 0$.
At this point, inducion with the recurrence relation will take care of ensuring that al $p_j=0$ for all $j\geqslant m+1$, much like we did with the claim.

