closed form solution to a sum of functions I need to find a closed form expression for the following: 
$$\sum_{n=0}^{\infty}\tfrac{a^n}{(n!)(c-bn)}e^{(c-bn)t} \text{ with } a,b,c<1$$ 
By closed form expression, I mean a formula that can be evaluated in a finite number of standard operations.
Thanks
 A: Let $f(t)$ be the sum of your series.
Differentiating term by term yields:
$$
\begin{split}
f^\prime (t) &= \sum_{n=0}^\infty \frac{a^n}{n!}\ e^{(c-bn)t} \\
&= e^{ct}\ \sum_{n=0}^\infty \frac{1}{n!}\ \left( \frac{a}{e^{bt}}\right)^n\\
&= e^{ct}\ \exp \left( ae^{-bt}\right)\\
&= \exp \left( ct+a\ e^{-bt}\right)
\end{split}$$
hence $f^\prime$ has a nice elementary expression.
Neverthless $f$ do not possess an elementary expression: in fact, as Raymond shows in his answer, $f$ can be expressed in terms of incomplete gamma functions which aren't elementary.
A: Let's define :
$$f(t)=\sum_{0}^{\infty}\tfrac{a^n}{(n!)(c-bn)}e^{(c-bn)t}$$
then
$$f'(t)=\sum_{0}^{\infty}\tfrac{a^n}{n!}e^{(c-bn)t}=e^{ct}\sum_{0}^{\infty}\tfrac{(a e^{-bt})^n}{n!}=e^{ct+ae^{-bt}}$$
EDIT (the $t$ was missing in $ct$ !)
for $u=ae^{-bt}$ that is $t=-\frac{\log(\frac ua)}b$ we have : 
$$\int e^{ct+ae^{-bt}} dt= -\frac 1b \int \frac{e^{-\frac{c\log(\frac ua)}b}e^{u}}u du=-\frac 1b \int \left(\frac ua\right)^{-\frac cb}\frac{e^{u}}u du$$
$$=-\frac {a^{\frac cb}}b \int u^{-1-\frac cb}e^{u} du=-\frac {(-a)^{\frac cb}}b\gamma\left(-\frac cb,-u\right)$$
with $\gamma$ the 'lower incomplete gamma function' getting : 
$$f(t)=C(a,b,c)-\frac {(-a)^{\frac cb}}b\gamma\left(-\frac cb,-ae^{-bt}\right)$$
where $C(a,b,c)=0$ I think.
This is a clearly non elementary result that we may rewrite as :
$$f(t)=\frac {(-a)^{\frac cb}}b\left[\Gamma\left(-\frac cb,-ae^{-bt}\right)-\Gamma\left(-\frac cb\right)\right]$$
The same result was obtained by Alpha ('Alternate form' assuming $a,b,c,t$ positive) : 

