5

Let $\zeta^{(n)}(2)$ be the $n$-th derivative of the Riemann zeta function, evaluated at $2$. Numerical experiments seem to suggest that $|\zeta^{(n)}(2)|\sim n!$, in the sense that $|\zeta^{(n)}(2)|/n!\rightarrow 1 $ as $n\rightarrow \infty$. Is this a theorem, or is it just another of those numerical illusions that often happen when experimenting with the zeta function?

Valerio
  • 340

1 Answers1

3

$$\sum_{k\ge 0} (\frac{\zeta^{(k)}(2)}{k!}-(-1)^k) (s-2)^k =\zeta(s)-\frac1{s-1}$$ It extends analytically to the whole complex plane thus the Taylor series converges for all $s$ and for $r$ arbitrary large, $$\lim_{k\to \infty} (\frac{\zeta^{(k)}(2)}{k!}-(-1)^k)r^k =0$$

reuns
  • 77,999