Convergence of Newton double series of reciprocal gamma function The reciprocal Gamma function has Newton series
$$\frac{1}{\Gamma\left(x+1\right)}=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x!}{\left(x-n\right)!}\sum_{k=0}^{n}\frac{\left(-1\right)^{k}}{k!^{2}\left(n-k\right)!}$$
See  this
I was wondering if you could provide a proof that this series converges. I tried both the ratio and the root test but in both cases the limit is 1 (i.e. the test is inconclusive) and I do not know where to proceed. I also thought about using Stirling's formula but wasn't sure how to do it as the series is alternating.
Thanks
 A: First, we note that
$$
\sum\limits_{k = 0}^n {\frac{{( - 1)^k }}{{k!^2 (n - k)!}}}  = \frac{{L_n (1)}}{{\Gamma (n + 1)}} = \sqrt {\frac{{\rm e}}{\pi }} \frac{1}{{n^{1/4} \Gamma (n + 1)}}\left( {\sin \left( {2\sqrt n  + \frac{\pi }{4}} \right) + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right),
$$
where $L_n$ is the $n$th Laguerre polynomial (see, e.g., here). By the reflection formula
$$
\frac{1}{{(x - n)!}} = ( - 1)^{n + 1} \frac{{\sin (\pi x)}}{\pi }\Gamma (n - x).
$$
We also have
$$
\frac{{\Gamma (n + a)}}{{\Gamma (n + b)}} = n^{a - b} \left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right)
$$
as $n\to +\infty$ with any fixed complex $a$ and $b$. See, for instance, here. Thus, the series in question converges iff the series
$$
\sum\limits_{n = 1}^\infty  {\frac{1}{{n^{x + 5/4} }}\left( {\sin \left( {2\sqrt n  + \frac{\pi }{4}} \right) + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right)} 
$$
converges. I note here that
$$
\sum\limits_{n = 1}^N {\sin \left( {2\sqrt n  + \frac{\pi }{4}} \right)}  =  - \sqrt N \cos \left( {2\sqrt N  + \frac{\pi }{4}} \right) + \mathcal{O}(1),
$$
i.e., the Dirichlet test is not applicable. We can certainly say that the series converges absolutely whenever $\operatorname{Re}(x)>-\frac{1}{4}$.
Addendum. Using summation by parts and the above asymptotics for the partial sum of the sine terms, it can be verified that
$$
\sum\limits_{n = 1}^N {\frac{1}{{n^{x + 5/4} }}\sin \left( {2\sqrt n  + \frac{\pi }{4}} \right)}  =  - \frac{1}{{N^{x + 3/4} }}\cos \left( {2\sqrt N  + \frac{\pi }{4}} \right) + \mathcal{O}\!\left( {\frac{1}{{N^{x + 5/4} }}} \right).
$$
Thus the original series converges absolutely precisely when $\operatorname{Re}(x)>-\frac{3}{4}$.
