This question originated from this problem. Can anyone help me simplifying the infinite series below:
$$\sum_{n=0}^{+\infty}\frac{1}{\Gamma(\beta_2-n)}\frac{(-e^{-t})^n}{n!}$$
The only idea I have is $\sum_{n=0}^{+\infty}\frac{(-e^{-t})^n}{n!}=e^{-e^{-t}}$ and I am not sure what to do with $\frac{1}{\Gamma(\beta_2-n)}$
Let us first simplify the expression by writing $x=-e^{-t}$ and $\beta_2=a+1$. Then we are left
with evaluating $~\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!~(a-n)!}$ , which, after employing Euler's reflection formula for
the $\Gamma$ function, in conjunction with that of the binomial series, becomes $-(1+x)^a~\cdot~\Gamma(-a)~\cdot$
$\cdot~\dfrac{\sin a\pi}\pi$.