I am finding some trouble on calculating the following double summation:
$ \sum_{k=1}^\infty \frac{b^k}{k!k}\sum_{n=0}^{k-1}\frac{(b*x)^n}{(n-1)!} $
Note that the inside sum gives: $\sum_{n=0}^{k-1}\frac{(b*x)^n}{(n-1)!}=\frac{b*x*e^{bx}Γ(k-1,bx)}{Γ(k-1)}$
But I cannot simplify the second sum as well. Any ideas?