Is it possible to express the following sum in terms of the hypergeometric function $_2F_1$:
$$ f(x) = \sum_{n=0}^\infty\frac{(-ax)^n}{n!~\Gamma(b-n)} $$
with $a$ and $b$ constant values ($x>0$ may also be treated as constant). If not, what other methods could be used?
For $|ax|<1$ the series converges to
$$\frac{(1-ax)^{b-1}}{\Gamma(b)}$$
We know it converges for $|ax|<1$ due to the ratio test:
$$ \lim_{n\to\infty}\Bigg|\left(\frac{(-ax)^{n+1}}{(n+1)!\Gamma(b-n-1)}\right)\left(\frac{n!\Gamma(b-n)}{(-ax)^n}\right)\Bigg|=|ax| $$ where I've used the fact that $\Gamma(b-n)/\Gamma(b-n-1)\sim -(n+1)$ as $n\to\infty$.
One way to evaluate the sum is to rearrange
\begin{align} \sum_{n=0}^\infty \frac{(-ax)^n}{\Gamma(b-n)n!}&=\frac{1}{\Gamma(b)}\sum_{n=0}^\infty \frac{\Gamma(b)}{\Gamma(b-n)}\frac{(-ax)^n}{n!}\\ &=\frac{1}{\Gamma(b)}\sum_{n=0}^\infty (b-1)_n\frac{(-ax)^n}{n!} \end{align} where $$ (x)_n = x(x-1)(x-2)...(x-n+1) $$ is the the falling factorial, and then use the exponential generating function $$ \sum_{n=0}^\infty (x)_n\frac{t^n}{n!}=(1+t)^x $$