Where does Laurent series for zeta function converge?

120 Views Asked by At

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?