The ordered bell numbers can be defined by the power coefficients

$\displaystyle \frac{1}{2-e^z} = \sum_{n=0}^{\infty} \frac{b(n)}{n!} z^n$ which is convergent in a neighbourhood of z=0.

Determine the first 6 terms of b(n). DONE

$\displaystyle b(0) = 1, b(1) = 1, b(2) = 3, b(3) = 13, b(4) = 75, b(5) = 541$.

Now show that $\displaystyle b(n)$~$\displaystyle \frac{n!}{2 (log(2))^{n+1}}$

Now the idea here is to find the value of z such that we have a pole in the $\displaystyle \frac{1}{(2-e^z)}$ part.

So doing this gives us $\displaystyle z = \ln(2)$ (but I think there should be some sort of $\displaystyle + n2\pi$ part here as well maybe?!?)

Then we find the residue at this part (called the principal part) and use that somehow to find the series... But I cannot figure out how this is being done!

EDIT: Should probably say I found the residue (by using L'Hopitals) to be...

$\displaystyle \lim_{z \to ln 2} \frac{1}{2-e^z} = \lim_{z \to z_0} \frac{z - z_0}{2-e^z} = \lim_{z \to z_0} \frac{1}{-e^z} = -\frac{1}{2}$ so I seem to be looking at doing something with...

$\displaystyle \frac{-1}{2(z-\ln(2))}$ which is the principal part at $\displaystyle z=z_0$

It seems to be you compute the residue of the closest pole (or in the case in the example, two equally close poles) and then sum then and SOMEHOW get the series. I have no idea how to get the series...

Below is an example of another question done. (I didn't do it)...