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I am wondering in what region of the complex plane the Laurent series $$ \zeta(s)=\frac{1}{s-1} + \sum_{k=0}^{\infty} \frac{(-1)^k \gamma_k}{k!} (s-1)^k $$ converges. It is straight forward to derive this series formally from $\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}$ assuming Re($s)>1$, and so it should also hold in a punctured disk around $s=1$ by analytic continuation, but how far does this disk extend? There are some crude bounds on the Stieltjes constants $\gamma_k$ available such as $\frac{|\gamma_k|}{k!} \leq \frac{1}{2^{k+1}} $ which imply a radius of convergence of at least $\frac{1}{2}$ but can we do better? What are the best known bounds?

Dave
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    Since $\zeta(s)-\frac1{s-1}$ is an entire function, won't this series converge for all $s\ne1$? – Greg Martin Feb 23 '22 at 00:48
  • also per the wikipedia page one has $|\gamma_n| \le C e^{n \log \log n}$ which clearly shows that the radius of convergence is infinity by Stirling since $n! \ge e^{n \log n-n}$ – Conrad Feb 23 '22 at 00:53
  • @GregMartin Can you just conclude that? – Dave Feb 23 '22 at 00:53
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    Yes this is one of the many great theorems in complex analysis, analytic (locally equal to the Taylor series) on the whole complex plane is equivalent to having infinite radius of convergence for one (whence all) Taylor series. That $\zeta(s)-1/(s-1)$ is entire follows for example from the functional equation. – reuns Feb 23 '22 at 01:02
  • Ok I see. Thanks – Dave Feb 23 '22 at 01:11

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