EDIT: I've rephrased the question
I'm interested in ways to formally sum the series $$s=\sum_{n=0}^\infty (-1)^n\left(\alpha n\right)!$$
When $\alpha$ < 1, this series can be calculated using Borel summation.
Using Borel summation we get $s = \int_{0}^{\infty}e^{-t}\left(\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}\left(\alpha n\right)!\left(t\right)^{n}}{n!}\right)dt$. To have this value actually converge, we need to be careful of how fast each of the infinities grow. We need to ensure in the expression $\int_{0}^{N_0}e^{-t}\left(\sum_{n=0}^{N}\frac{\left(-1\right)^{n}\left(\alpha n\right)!\left(t\right)^{n}}{n!}\right)dt$ that $N_0$ is much smaller than $N$.
But for $\alpha$ >1, Borel summation doesn't work, since $\lim_{n \to \infty}\frac{\left(\alpha n\right)!}{\left(n\right)!}t^n = \infty$.
Using another method, I have managed to get some results for ways to sum $\sum_{n=0}^\infty (-1)^n \left(\alpha n\right)! x^n$, but I'm not sure if these results are valid. I'm hoping to see a solution to these sums with another method to check my answers. I'm thinking it's potentially possible to find the value of this sum using a differential equation approach, a stronger form of Borel summation, or possibly using Ramanujan summation, but I'm unfamiliar with these approaches.
Any help would be appreciated!
Best Answer
The following Borel-like resummation works: $$ \sum\limits_{n = 0}^\infty {( - 1)^n (\alpha n)!} = \sum\limits_{n = 0}^\infty {( - 1)^n \int_0^{ + \infty } {e^{ - t} t^{\alpha n} dt} } = \int_0^{ + \infty } {e^{ - t} \sum\limits_{n = 0}^\infty {( - 1)^n t^{\alpha n} } dt} = \int_0^{ + \infty } {\frac{{e^{ - t} }}{{1 + t^\alpha }}dt} . $$