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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.

edited Mar 11 2012 at 15:06

asked Mar 11 2012 at 15:01

tmy1018
55●5

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). – suVRit Mar 11 2012 at 18:04

Igor Rivin · Accepted Answer · 2012-03-11 19:22:15Z UTC

3

See Steven Landsburg's note.

answered Mar 11 2012 at 19:22

Igor Rivin
31.2k●3●37●71

The relationship between Stirling number of the second kind and the polylogarithm

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