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Does anyone know of any papers or resources dealing with the following question: For which values of $s=\sigma+it$ does the following sum of Stieltjes constants hold, $$\sum_{n=0}^\infty\frac{(-1)^n}{n!}\gamma_n(s-1)^n=0,$$ where $\gamma_n$ is the $n$-th Stieltjes constant. See here for information on the Stieltjes constants. I'm investigating when we have $\zeta(s)=1/(s-1)$ for $s\neq 1$ and thought this might help me given that $$\zeta(s) = \frac{1}{s-1}+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\gamma_n(s-1)^n.$$

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2 Answers 2

up vote 1 down vote accepted

Building on the answer of Manzoni, the Mathematica command

s[theta_, eps_, k_] := ZetaZero[k] + eps Exp[I theta]

Plot[{Abs[1/(s[theta, .1, 1] - 1)], Abs[Zeta[s[theta, .1, 1]]]}, {theta, 0, 2 Pi}, PlotStyle -> {Directive[Thick, Red], Directive[Thick, Green]}]

Generates this plot

plot

This shows that $$ \frac{1}{|s-1|}<|\zeta(s)| $$ for $s$ on the circle of radius $.1$ around the lowest nontrivial Riemann zero $\rho=1/2+i14.13\ldots$.

By Rouche's Theorem, $\zeta(s)$ and $\zeta(s)-1/(s-1)$ have the same number of zeros inside the circle, namely $1$.

One could presumably do this for higher zeros as well.

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I don't know papers about your question but some numerical investigations return zeros near the Riemann zeros (because $\frac 1{s-1}$ is small for Riemann's zeros and of the smooth character of $\zeta'$ near $\zeta$ zeros) :

\begin{array} {llll} &0.5+14.1347251417i&\mapsto &0.47940300622352\cdots&+14.04751638282159\cdots i\\ &0.5+21.0220396387 i&\mapsto &0.507180820934280248966\cdots&+20.980648953392265333\cdots i\\ &0.5+25.0108575801i&\mapsto &0.4893602291391196345\cdots&+24.9838122121001409936\cdots i\\ \end{array}

To find the approximative value of these zeros let's consider the first order expansion of $\zeta$ near one of its zeros $s_i$ : $$\zeta(s_i+\epsilon_i)=\zeta(s_i)+\epsilon_i\zeta'(s_i)+O(\epsilon_i^2)$$ Since $\zeta(s_i)=0$ and since you want $\ \zeta(s_i+\epsilon_i)=\frac 1{s_i+\epsilon_i-1}\ $ you will have : $$\frac 1{s_i+\epsilon_i-1}\approx \epsilon_i\zeta'(s_i)$$ and, to first order, the zeros will be nearly : $$s_i+\epsilon_i \approx s_i+\frac 1{\zeta'(s_i)(s_i-1)}$$ With the values of $\zeta$ and $\zeta'$ for the three first zeros we get the approximate table :

\begin{array} {ll} &0.482882 &+ 14.0472i\\ &0.508179 &+ 20.9810i\\ &0.489891 &+ 24.9835i\\ \end{array}

Not sure it will help much...

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