For $0<z<1, B(z)=\sum_{n=1}^{\infty}(A_\epsilon(z))^n,$ where
$A_\epsilon(z)=\sum a_\epsilon(k) z^k;$
$B(z)=\sum b(k)z^k;$
$b(k)=\sum_{n=1}^{\infty}a_\epsilon^{*n}(k);$
$a_\epsilon^{*n}$ denotes the $n-$th convolution power of function $a_\epsilon(k)=(1-\epsilon) \epsilon^k, k \in Z^+$
Notes: This is an intermediate step in Markov Chains and Stochastic Stability by Meyn and Tweedie p186 Lemma 8.2.3. The author merely stated that it is true. I'm not very familiar with computation with convolution. Any exaplanation is welcomed.