Double sequence convergence

Question

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.

Answer 1

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.

Answer 2

Two comments: (1) There is a wikipedia article about this:

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