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?

Solution 1:

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 $\displaystyle\;\Phi(z, s, \alpha) := \sum_{k=0}^\infty \frac { z^k} {(k+\alpha)^s}$.