Do all values of $\zeta (s)$ follow from the Dirichlet series definition of $\zeta (s)$?

Asked Feb 16 '20 at 15:44

Active Feb 16 '20 at 15:59

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If we define $\zeta (s)$ by $$\zeta (s)\overset{\text{def}}{=}\sum_{n\gt 0}n^{-s},\, \operatorname{Re}s\gt 1,$$ does it follow from the definition that e.g. $\zeta (-1)=-\frac{1}{12}$?

Or is it necessary to use $$\zeta (s) \overset{\text{def}}{=} \begin{cases} \sum_{n\gt 0}n^{-s} &\mbox{if } \operatorname{Re}s\gt 1 \\ 2^s\pi ^{s-1}\sin\frac{\pi s}{2}\,\Gamma (1-s)\sum_{n\gt 0}n^{s-1}& \mbox{if }\operatorname{Re}s\le 0\\ \frac{1}{1-2^{1-s}}\sum_{n\gt 0}(-1)^{n+1}n^{-s}& \mbox{otherwise}\end{cases}$$ and then say that $\zeta (-1)=-\frac{1}{12}$?

edited Feb 16 '20 at 15:59

asked Feb 16 '20 at 15:44

Derazav Gunstig

3

The usual definition is that $\zeta(s)$ is the analytic continuation of the Dirichlet series, and that is enough to get $\zeta(-1).$ Note your second definition is not enough - is does include the cases where the real part of $s$ is in $[0,1].$ – Thomas Andrews Feb 16 '20 at 15:51
Oh, I see where's the problem now. – Derazav Gunstig Feb 16 '20 at 15:56
https://math.stackexchange.com/q/3159749/8530 – Mats Granvik Feb 18 '20 at 14:29
$$\lim_{k\to \infty } , \left(\sum _{n=1}^k \frac{1}{n^s}+\frac{k^{1-s}}{s-1}-\frac{k^{-s}}{2}+\frac{s}{12 k^{s+1}}\right)=-\frac{1}{12}$$ when $s=-1$ – Mats Granvik Feb 18 '20 at 15:05

Do all values of $\zeta (s)$ follow from the Dirichlet series definition of $\zeta (s)$?

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