3
$\begingroup$

It is shown here on Mathworld's page on Stirling number of the second kind that

$$ \sum_{k=1}^n S(n,k) (k-1)! z^k = (-1)^n \text{Li}_{1-n}(1+\frac{1}{z}) $$

where $S(n,k)$ is Stirling number of the second kind and $\text{Li}_{1-n}$ is the polylogarithm.

Can somebody provide me some reference on where this identity came from? It isn't shown on Mathworld's page.

$\endgroup$
  • $\begingroup$ Unless I am mistaken, this seems to simply the result of a Lagrange inversion of a series for the polylog function (see the Wikipedia page for the series for Li that is a candidate). $\endgroup$ – Suvrit Mar 11 '12 at 18:04
3
$\begingroup$

See Steven Landsburg's note.

$\endgroup$

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.