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I would like to sum the following series:

$$\sum_{n=0}^\infty \frac{u^n}{n!}\sum_{k=1}^{n+1}(-1)^{k-1}(k-1)! S_{n+1,k}p^k,$$

where $S_{n,k}$ are Stirling numbers of the second kind, $0

I tried to attack this with generating functions, and came up with something that seems certainly wrong, but it seems like a clean closed form solution may exist.

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    Could you show your attempt? – 2017-01-17
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    Hi dimpol. If Marko's answer below is correct, it looks like I got the generating function for S_{n+1,k} wrong. I thought the term in front of the z^{n+1} was (n+1), not (n+1)!. Still don't understand why it's (n+1)!, actually. – 2017-01-18

1 Answers 1

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We seek to simplify

$$\sum_{n\ge 0} \frac{u^n}{n!} \sum_{k=1}^{n+1} (-1)^{k-1} (k-1)! {n+1\brace k} p^k.$$

Using formal power series we have

$${n+1\brace k} = (n+1)! [z^{n+1}] \frac{(\exp(z)-1)^k}{k!}.$$

We obtain for the sum

$$\sum_{n\ge 0} \frac{u^n}{n!} (n+1)! [z^{n+1}] \sum_{k=1}^{n+1} (-1)^{k-1} \frac{1}{k} (\exp(z)-1)^k p^k \\ = \sum_{n\ge 0} u^n (n+1) [z^{n+1}] \log\left(1+p(\exp(z)-1)\right) \\ = \left.\log\left(1+p(\exp(z)-1)\right)'\right|_{z=u}.$$

This yields $$\bbox[5px,border:2px solid #00A000]{ \frac{p\exp(u)}{1+p(\exp(u)-1)}.}$$