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?


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

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


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?

  • $\begingroup$ concerning update 2: how do you sum that series at the given arguments? The absolute values of the terms approach 1 and the signs/args cycle on the complex unit-circle. If I use Euler-summation (one can as well use Cesaro summation) this "converges" then to the expected values such that D(argument)=0. Have you looked at your sums when the partial sums go to n,n+1,n+2 and n+3 terms? $\endgroup$ Mar 25, 2013 at 20:22
  • $\begingroup$ @Gottfried. You are right, subsequent partial sum results indeed cycle on the complex unit circle and change for n,n+1,n+2 and n+3 terms. So far I had only tested (as a strange habit) with odd partial sums of n=9,99,999... For $z=-1$ the outcome flips between $n$ and $n+1$ with $D_{oddN}(−1)=\frac12$ and $D_{evenN}(−1)=−\frac12$. What does this mean for the convergence when $n$ goes to infinite? Does it become zero? $\endgroup$
    – Agno
    Mar 25, 2013 at 21:33
  • $\begingroup$ :-) hm, it shall behave like the hands of a clock with an infinite battery... ;-) Only the introduction of a coherent concept of divergent summation can carry this to somewhere meaningful.. $\endgroup$ Mar 25, 2013 at 21:54
  • $\begingroup$ Thanks Gottfried. Point taken and have much appreciated your insights. In my infinite imagination I imagined imaginary hands powered by an imaginary battery... ;-) $\endgroup$
    – Agno
    Mar 25, 2013 at 22:19
  • $\begingroup$ Life is complex. It has real and imaginary aspects... :-) $\endgroup$ Mar 25, 2013 at 22:49

2 Answers 2


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

  • 2
    $\begingroup$ The singularity of $\zeta(s)$ nearest to $0$ is at $s=1$, so the formula is valid for $|s|<1$. $\endgroup$
    – GEdgar
    Mar 24, 2013 at 23:28
  • $\begingroup$ It's obvious when you see it... Thanks for the help. One last question: Does this also explain why $\displaystyle \sum_{n \ge 0} \frac{\zeta^{(n)}(0)}{n!} i^n -\zeta(i) =\frac12$ ? $\endgroup$
    – Agno
    Mar 24, 2013 at 23:32
  • $\begingroup$ Too late to edit the above comment, but I meant to say the difference between the real parts being equal to $\frac12$ $\endgroup$
    – Agno
    Mar 24, 2013 at 23:40
  • $\begingroup$ @user46149, do you even have series convergence for the infinite sum? $\endgroup$
    – user27126
    Mar 25, 2013 at 0:00
  • 2
    $\begingroup$ For a power series, you have convergence inside the radius of convergence, divergence outside, but exactly at the radius you could have either convergence or divergence. Note $(1+i)/2$ is inside, but $i$ is exactly on the radius of convergence. $\endgroup$
    – GEdgar
    Mar 25, 2013 at 13:46

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

 %833 = -1.46035450881

while the original call to zeta(0.5) gives

 %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}) $


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