Road testing Ansus' continuum product formula

[Gottfried]

If we have a base 1<b<=exp(exp(-1)) the progression of consecutive f°k(0) approaches zero, because the values approach a limit, so the inner sums should be convergent.

Not sure whether I understand you. Where does the iteration of f occur in the inner sum?

... here:
\(
\sum_{n=1}^{\infty} \frac{f^{(n-1)}(0)}{n!} {n \choose k} B_{n - k}
\)
there is the f°(n-1)(0) in the inner sum, where n is the running index

--- well, upps, or was possibly the n-1'th derivative meant??

Gottfried