Calculate the coefficients of the series $\frac{1}{1+\frac{p}{1-p}(1-z)^\lambda}$ I wish to calculate the coefficients of the series
$$\frac 1 {1+\frac p {1-p} (1-z)^\lambda},\quad p,\lambda\in(0,1).$$
The $k$-th coefficient is easily found to be
$$ (-1)^k\sum_{n=0}^\infty \left(\frac p {p-1}\right)^n\binom {\lambda n} k. $$
Now I am looking for some nice representation of this quantity, maybe in terms of some known special functions or sequence of constants.
I think that these constants form a probability distribution (a generalization of the geometric distribution with parameter $p$) and numerical computations say indeed that the first 15 coefficients are positive. A proof that they are all positive would be appreciated as well.
Thank you for your references and/or comments.
UPDATE: Changed formulation of the question.
 A: Hints: 
(1) Use the expansion $\frac1{1-at}=\sum\limits_{n\geqslant0}a^nt^n$ for $a=\frac{p}{p-1}$ and $t=(1-z)^\lambda$. 
(2) Use the expansion $(1-z)^\mu=\sum\limits_{k\geqslant0}(-1)^k{\mu\choose k}z^k$ for $\mu=n\lambda$, for each $n\geqslant0$.
(3) Exchange the order of the summations over $n$ and $k$... et voilà !
Edit: To show that every coefficient is positive, one can start from the expansion $(1-z)^\lambda=1-\sum\limits_{n\geqslant1}c_n^\lambda z^n$ where $c_n^\lambda=(-1)^{n+1}{\lambda\choose n}$ is positive, for every $n\geqslant1$, because $\lambda$ is in $(0,1)$. Hence
$$
\frac1{1+\frac{p}{1-p}(1-z)^\lambda}=\frac{1-p}{1-p\sum\limits_{n\geqslant1}c_n^\lambda z^n}=\sum_{i\geqslant0}(1-p)p^i\left(\sum\limits_{n\geqslant1}c_n^\lambda z^n\right)^i.
$$
Thus, the coefficient of $z^0$ is $1-p$ and, for every $k\geqslant1$, the coefficient of $z^k$ is
$$
\sum_{i\geqslant1}(1-p)p^i\cdot\sum_{n_1+\cdots+n_i=k}c_{n_1}^\lambda\cdots c_{n_i}^\lambda.
$$
In particular, the series has positive coefficients.
A realization of this series is as follows. Let $N$ denote a random variable with geometric distribution of parameter $p$, that is, $\mathbb P(N=i)=(1-p)p^i$ for every $i\geqslant0$. Let $(X_k)_{k\geqslant1}$ denote an i.i.d. sequence with distribution $\mathbb P(X_k=n)=c_n^\lambda$, for every $n\geqslant1$ (this exists since the coefficients $c_n^\lambda$ are positive and sum to $1$). Assume that $(X_k)_{k\geqslant1}$ is independent of $N$. Then,
$$
\frac1{1+\frac{p}{1-p}(1-z)^\lambda}=\mathbb E(z^Y),\qquad Y=\sum_{k=1}^NX_k.
$$
