Series expansion of $\zeta(s)$ using the derivatives of $\zeta(0)$

Question

When playing with series expansions, I stumbled upon the following relation:

$$\zeta(s) := \sum \limits_{n=0}^{\infty} \left( \zeta^{(n)}(0) \frac{s^n}{\Gamma(n+1)} \right)$$

that seems to hold for all $|s|<1,s \in \mathbb{C}.$ Note that $\zeta^{(n)}(0)$ is the n-th derivative of $\displaystyle \lim_{s \to 0} \zeta(s)$.

I have the strong feeling this must be related to the Laurent series expansion of $\zeta(s)$ (see Laurent series), however I can't get it reconciled.

Does anybody see the link? And if so, could it be analytically continued?

EDIT 1:

The question has been answered below, however despite the constraint of $|s<1|$ the following difference still seems to converge for $s=i$:

$$\displaystyle \sum_{n \ge 0} \frac{\zeta^{(n)}(0)}{n!} i^n -\zeta(i) =\frac12 +\frac12 i$$

EDIT 2:

Based on the comments below, I now know that for a power series you could have either convergence or divergence exactly at the radius of convergence (which in this case is the unit circle). All values on the unit circle, except for $s=1$, indeed do seem to converge, but the infinite sum then no longer equates $\zeta(s)$. However, the following difference does yield interesting results again, but I haven't managed to explain them:

$$D(z):= \displaystyle \sum_{n \ge 0} \frac{\zeta^{(n)}(0)}{n!} z^n -\zeta(z)$$

with $z \in \mathbb{C}$ and of the type: $e^{ti}=\cos(t)+i \sin(t)$.

Some values are:

• $D(-1) = \frac12$
• $D(i) = \frac12+\frac12 i$
• $D(-i) = \frac12-\frac12 i$
• $D(\frac12 \sqrt{2}+\frac12 \sqrt{2} i) = -\frac12-\frac12 \sqrt{2} i$
• $\dots$

How could I reconcile these numbers from the known formulas?

Answer 1

That is just Maclaurin's series: $\Gamma(n + 1) = n!$, so: $$ \zeta(s) = \sum_{n \ge 0} \frac{\zeta^{(n)}(0)}{n!} s^n $$

Answer 2

Hmm, I don't [update] didn't [/update] know what you mean, because if I do this I get a meaningful result.
[update] Using the imaginary unit $i$ for $x$ the first zeta-formula (that with the coefficients near $-1$ ) gets outside the range of convergence: the partial sums do not converge. If you take the second version of the zeta-series, where the ${1\over1-x}$ part is removed you get the correct result to any digits even if you use only few (say 15 or 20) coefficients.
I leave the initial comments untouched because they may still be instructive [/update]

I'm getting the coefficients for the powerseries as

$ \qquad \displaystyle \small \zeta(x)= -0.500000000000 - 0.918938533205 x - 1.00317822795 x^2 - 1.00078519448 x^3 \\ \small \qquad \qquad - 0.999879299501 x^4 - 1.00000194090 x^5 - 1.00000130115 x^6 \\ \small \qquad \qquad - 0.999999831384 x^7 - 1.00000000576 x^8 - 1.00000000091 x^9 \\ \small \qquad \qquad - 0.999999999850 x^{10} - 1.00000000001 x^{11}+ O(x^{12}) $

and if I substitute $0.5$ for $x$ in this I get using the first 64 terms

subst(Pol(psz),x,0.5) 
 %833 = -1.46035450881

while the original call to zeta(0.5) gives

zeta(0.5)
 %832 = -1.46035450881

So something else must be wrong. Perhaps some error becomes more obvious if we remove the systematic $-1.0$ from the coefficients and separate the series for ${1 \over 1-x}$ from it. We get then a series with reduced coefficients (which should also be entire) as

$ \qquad \displaystyle \small \zeta(x)+{1 \over 1-x} =0.500000000000 + 0.0810614667953 x - 0.00317822795429 x^2 - 0.000785194477042 x^3 \\ \small \qquad \qquad \qquad + 0.000120700499429 x^4 - 0.00000194089632046 x^5 - 0.00000130114601396 x^6 \\ \small \qquad \qquad \qquad + 0.000000168615826389 x^7 - 0.00000000576467597995 x^8 - 0.000000000911016489231 x^9 \\ \small \qquad \qquad \qquad + 1.49700759419 E-10 x^{10} - 9.40689566567 E-12 x^{11}+ O(x^{12}) $