# Double sequence convergence

Let be $$S(m,k)$$ number of partitions of a $k$ element set into $m$ nonempty parts investigating with generating functions I get this very interesting equation$$\sum_{k=0}^ {\infty}\sum_{m=0}^ {\infty}S(m,k)\frac{1}{k!}=e^{e-1}$$ Can someone tell me if I am right.

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How did you arrive at this conclusion? Off hand, I know only one way to do it: the one that's currently in the Wikipedia article on Dobinski's formula. I actually suspect I've read another derivation of it at some point, but I don't remember any more than that. – Michael Hardy Sep 8 '11 at 12:41

The inner sum actually has a finite number of non-zero members: $$\sum_{m=0}^ {\infty}S(m,k)= \sum_{m=0}^ {k}S(m,k)=B_k$$ where $B_k$ are the Bell numbers. The exponential generating function for them is $$\sum_{k=0}^\infty \frac{B_k x^k}{k!}=e^{e^x-1}.$$ Putting $x=1$ gives the required equality.

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+1. You could have mentioned something about Stirling subset numbers though... :) – J. M. Sep 8 '11 at 11:45
@J. M. They're mentioned in the reference, so I decided not to expand :) – Andrew Sep 8 '11 at 12:18

http://en.wikipedia.org/wiki/Dobinski%27s_formula

and (2) the conventional notation for Stirling numbers is that $S(m,k)$ is the number of partitions of a set of size $m$ into $k$ subsets, not the number of partitions of a set of size $k$ into $m$ subsets. http://en.wikipedia.org/wiki/Stirling_number

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