I was looking for $\Pi_k$ of the queue M/M/1/-/K and I got stuck with this sum $\sum_{k=0}^K (\frac{a}{b})^k \cdot k!$ The only idea I have so far is to maybe use $\sum_{k=0}^n \binom{k}{n} \cdot a^kb^{n-k} =(a+b)^n $
I would use a hint on how to do it. Thank you in advance.
Just in case this could help you in any manner.
$$\sum_{k=0}^K x^k \, k!=-\frac{e^{-1/x}} x \left((-1)^K \,\Gamma (K+2)\, \Gamma \left(-(K+1),-\frac{1}{x}\right)+\Gamma \left(0,-\frac{1}{x}\right)\right)$$ where appear the complete and incomplete gamma functions.