The series I'm working with is
$$\sum_{k=0}^\infty \binom{z}{k}(-1)^k ( 1-\frac{\Gamma(k,-\log n)}{\Gamma(k)} )$$
with $z$ a complex variable and $\Gamma(k, -\log n)$ the upper incomplete gamma function.
Can this be expressed in a closed form (possibly involving special functions)?
As a bit of extra information, I already have a proof of the special case
$$\lim_{z \rightarrow 0}z^{-1}(-1+\sum_{k=0}^\infty \binom{z}{k}(-1)^k ( 1-\frac{\Gamma(k,-\log n))}{\Gamma(k)} ) = -\Gamma(0,-\log n)-\pi i -\log\log n - \gamma$$
and it seems empirically that for $z=-1$, the sum reduces to $1-\log n$.
Alright, an answer, though without a rigorous proof. If anyone can fill in the details here, I'd be happy to accept their answer.
With a bit of manipulation in Mathematica, I ended up with the very tidy:
$$\sum_{k=0}^\infty \binom{z}{k}(-1)^k( 1 - \frac{\Gamma(k,-\log n)}{\Gamma(k)}) = L_{-z}(\log n),$$
where $L_z(n)$ is the Laguerre Polynomials (Mathworld link).
To end up here, I used an identity J.M. had shown in an answer to a previous question of mine (this one):
$$(-1)^k( 1 - \frac{\Gamma(k,-\log n)}{\Gamma(k)}) = \frac{(-1)^{k+1}}{\Gamma(k)}\int_{-\log n}^0 t^{k-1}e^{-t} dt$$
and then I swapped the order of the sum and the integral, and Mathematica immediately recognized the resulting terms as the Laguerre Polynomials.
Anyway, I know now from a bunch of tests that my sum does indeed seem to equal $L_{-z}(\log n).$ Can this be shown more tidily?