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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 $1-\log n$.

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    $\begingroup$ Break it up into two sums, and notice that the first one is $0$. $\endgroup$
    – Lucian
    Commented Sep 14, 2014 at 17:51

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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?

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