given the sum
$$ \sum_{n=0}^{\infty}n^k \exp(-n\epsilon), $$ for given $ k >0 $
how can it be Abel regularizable?
According to this paper the regularized value agrees up to some pole term to zeta function regularization that is
$$ \sum_{n=0}^\infty n^k \exp(-n\epsilon)= \zeta(-k)+ \frac{(-1)^k \Gamma (k+1)}{\epsilon ^{k+1}} $$
which is a consequence of the expansion $$ \frac{\epsilon}{e^{\epsilon}-1}=1+\sum_{n=1}^\infty \frac{B_n \epsilon^n}{n!}$$
however i get it wrong ::( and do not get the number $ \zeta (-k) $ which is supposed to be
also how could i prove that $$ \sum_{n=0}^\infty n^{-1} \exp(-n\epsilon)=\gamma + \frac{1}{\epsilon}+O(\epsilon)$$
from equation $$ \sum_{n=0}^\infty n^k \exp(-n\epsilon)=-\log\frac{1}{1-e^{-\epsilon}}$$