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#### chisigma

##### Well-known member

- Feb 13, 2012

- 1,704

http://mathhelpboards.com/challenge-questions-puzzles-28/simplifying-quotient-7235.html#post33008

... where has been obtained the expression...

$\displaystyle \zeta (s) = \frac{1}{1-2^{1 - s}}\ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}\ (1)$

... that allows the effective computation of $\zeta(*)$ in the half plane where $\text{Re} (s) > 0$. It is well known that $\zeta (0) = - \frac{1}{2}$ so that is...

$\displaystyle \lim_{s \rightarrow 0 +} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}} = \frac{1}{2}\ (2)$

What I'm interested about is how to demonstrate (2) independently from (1), i.e. whitout the preliminary knowledge that $\zeta (0) = - \frac{1}{2}$. I spent many hours in attempts but without success ...

Kind regards

$\chi$ $\sigma$