1
$\begingroup$

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 $1-\log n$.

$\endgroup$
  • $\begingroup$ Break it up into two sums, and notice that the first one is $0$. $\endgroup$ – Lucian Sep 14 '14 at 17:51
1
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.