I am stopped by the following algebra.
Let $A(z) = 1-az-...-a_pz^p$ and $A(z)=0$ has no roots on $|z| \leq1$, so that $B(z) = 1/A(z) $ is well-defined. Now, let $B(z)=\sum_{i=0}^{\infty}b_iz^i$. We already know that $\sum_{i=0}^{\infty}|b_i|<\infty$ Then, under what condition do we have $\sum_{i=0}^{\infty}i|b_i|<\infty$?
The paper I am now reading expresses this as an exponential decay, and proceeds taking this exponential decay as holding true with no specific reference.
No additional condition needed. Let $r=\min\{|z|:A(z)=0\}$. Then $r>1$, and $B(z)$ converges absolutely in $|z|<r$. Then so does $B'(z)=\sum_{n=1}^{\infty}nb_n z^{n-1}$; in particular, $B'(1)$ is absolutely convergent.