Asymptotic expansion of $\sum_{k=0}^{\infty} k^{1 - \lambda}(1 - \epsilon)^{k-1}$

Question

I'm seeing a physics paper about percolation (http://arxiv.org/abs/cond-mat/0202259). In the paper the following asymptotic relation is used without derivation. $$ \sum_{k=0}^{\infty} k P(k) (1 - \epsilon)^{k-1} \sim \left<k\right> - \left<k(k - 1)\right> \epsilon + \cdots + c \Gamma(2 - \lambda) \epsilon^{\lambda - 2}, $$ where $P(k) = c k^{-\lambda}$ with $\lambda > 2$ is a power-law probability mass function of $k$ and the bracket means average. It's the equation number (12) of the paper. I have no idea how to get the relation. Especially, from where does the $\epsilon^{\lambda - 2}$ term come?

Answer 1

I have not been able to see how is coming the $\epsilon^{\lambda - 2}$ term. However, and, may be, this could be a track $$ \sum_{k=0}^{\infty} k P(k) (1 - \epsilon)^{k-1}=\frac{c \Phi (1-\epsilon,\lambda -1,0)}{1-\epsilon}$$ when $P(k) = c k^{-\lambda}$ ($\Phi$ being the the Hurwitz-Lerch transcendent $\Phi(z,s,a)$ function). May be some asymptotic developments ..?

Answer 2

$$ \sum_{k=0}^{\infty} k P(k) (1 - \epsilon)^{k-1} = \sum_{k=0}^\infty k P(k) \left(\sum_{j=0}^{k-1} {k-1 \choose j}(-\epsilon)^j \right)\\ = \sum_{k=0}^\infty k P(k) { k-1 \choose 0} - \sum_{k=0}^\infty k P(k) { k-1 \choose 1} \epsilon + \ldots\\ = \langle k \rangle - \langle k(k-1) \rangle\epsilon + \ldots $$