Road testing Ansus' continuum product formula

[bo198214]

\( b_k = \sum_{n=1}^{\infty} \frac{f^{(n-1)}(0)}{n!} {n \choose k} B_{n - k} \)

actually its not so difficult to see that the sum diverges for functions with finite convergence radius. Set \( f_n = \frac{f^{(n)}(0)}{n!} \) to be the coefficients of the corresponding powerseries. If the function has a finite convergence radius \( r \) then, by the root test \( r=\frac{1}{\limsup_{n\to\infty} \sqrt[n]{f_n} \), there is a subsequence that behaves asymptotically like \( 1/r^n \), roughly \( f_n \sim \frac{1}{r^n} \).

Now we use the asymptotic formula

\( |B_{2n}|\sim 4\sqrt{\pi n} \left(\frac{n}{\pi e}\right)^{2n} \)

A bit more sloppy written as
\( |B_{n}|\sim c\sqrt{ n} \left(\frac{n}{d}\right)^{n} \)
and put it into the summands in the beginning:
\( \frac{1}{n r^n} {n \choose k} c\sqrt{ n-k} \left(\frac{n-k}{d}\right)^{n-k} \)

but already
\( \left(\frac{n-k}{rd}\right)^{n-k} \)
goes strongly to infinity. (yes, ignored signs. this is only a very rough, though I think convincing, argumentation not a proper proof.)