Generating functions and the Riemann Zeta Function The generating function for the terms of the harmonic series:
$\frac{1}{n}$
is $-\ln(1 - x)$.
Does an ordinary generating function exist for the terms of the zeta function $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$
for any $ s > 1$?
That is, does there exist any function $f(x)$ that can be expressed in terms of elementary functions such that $f(x) = \sum_{n=1}^\infty \frac{1}{n^s}x^n$ for some $s > 1$?
I'm assuming that such a function in fact does not exist.  Can this be proven to be the case?
 A: The function you are looking for is the Li$_s(z)$, the polylogarithm.
A: This is an answer to the initial question (asking for a generating function of $\zeta(s)$) :

*

*$−\ln(1−x)$ is the generating function of $\ \frac 1n$.

*$-\frac{\ln(1−x)}{1-x}$ is the generating function of the harmonic number $\ H_n=\sum_{k=1}^n\frac 1k$
The generating function of the generalized harmonic number $\ H_{n,s}:=\sum_{k=1}^n\frac 1{k^s}\ $ is given by :
$$\frac{\operatorname{Li}_s(x)}{1-x}$$
with $\operatorname{Li}_s$ the polylogarithm.
Should you simply want $\ \displaystyle\sum_{n=1}^\infty \frac{x^n}{n^s}=\operatorname{Li}_s(x)\ $ then dot dot's answer is right of course !
Note that a generating function for $\zeta(n)$ is known as the digamma function :
$$\psi(1+x)=-\gamma-\sum_{n=1}^\infty \zeta(n+1)\;(-x)^n$$
while the reflection formula allows to get the even values of $\zeta$ directly as :
$$\pi\;x\;\cot(\pi\;x)=-2\sum_{n=0}^\infty \zeta(2n)\;x^{2n}$$
A generating function for the polylogarithm was obtained too :
$$z\,\Phi(z, 1, 1-x)=\sum_{n=0}^\infty\;\operatorname{Li}_{n+1}(z)\;x^n$$
using the Lerch zeta function $\displaystyle\;\Phi(z, s, \alpha) := \sum_{k=0}^\infty \frac { z^k} {(k+\alpha)^s}$.
