# Can the inverse of the Riemann zeta function in $s > 1$ be expressed as a series?

In this post, we are interested in the Rimenann zeta function $$\zeta(s)$$ in $$s > 1$$ only where it is strictly decreasing rather than $$s$$ in the entire complex plane. We have the Stieltjes series expansion of the Riemann Zeta function. I inverted the first few terms of this series using series reversion and showed that if $$s > 1$$ and $$\zeta(s) = a$$, then,

$$s = \zeta^{-1}(a) = 1 + \frac{1}{a - \gamma_0} - \frac{\gamma_1}{1!(a - \gamma_0)^2} + \frac{\gamma_2}{2!(a - \gamma_0)^3} - \frac{\gamma_3 - 12\gamma_1}{3!(a - \gamma_0)^4} + \mathcal O(a^{-5})$$

It seems that $$\zeta^{-1}(a)$$ can be expressed in the form

$$1 + \sum_{n=0}^{\infty} (-1)^n\frac{f(\gamma_1, \gamma_2, \ldots, \gamma_n)}{n!(a - \gamma_0)^{n+1}}$$

where $$f(\gamma_1, \gamma_2, \ldots, \gamma_n)$$ is some polynomial function of the Stieltjes constants $$\gamma_n$$.

Question: I am looking for a closed or a recurrence formula for $$f$$? Also any reference in this series in literature?

• Can you please provide a handful more coefficients, if not explicite then at least numerical approximations might be helpful for me to detect a pattern. (I think I've tried something like this years ago and might find some old notes when I see the pattern or numerical values) – Gottfried Helms Aug 10 at 10:57
• @GottfriedHelms Actually, I have computed these terms and that too way back in $2005$ so I afraid I no longer have the detailed notes. However, I have one more term of for a related series. – NiloS Aug 10 at 12:04
• For every real $x \ge 1$ there exists a positive real $c_x$ such that ${\displaystyle{ \zeta\Big(1+\frac{1}{x-1+c_x}\Big) = x. }}$. The first few terms of the asymptotic expansion of $c_x$ in terms of $n$ and the Stieltjes constants $\gamma_i$ are $$c_x = 1-\gamma_0 + \frac{\gamma_1}{x-1} + \frac{\gamma_2 + \gamma_1 - \gamma_0 \gamma_1}{(x-1)^2} + \frac{\gamma_2 +2\gamma_2 - 2\gamma_2 \gamma_0 + \gamma_1 - 2\gamma_1 \gamma_0 + \gamma_1 \gamma_0^2 - \gamma_1^2}{(x-1)^3} + O\Big(\frac{1}{x^4}\Big)$$ – NiloS Aug 10 at 12:04