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can anyone check if this formula is plausible ??

$ \frac{1}{\zeta (s)} = \sum_{n=0}^{\infty}\frac{ (-\pi)^{n}(s-1)s}{2n!(s+2n)(s+2n+1)} $

according to the authors this formula would be valid only $ 0 < \Re(s) <1/2 $

the paper may be found at http://arxiv.org/pdf/0709.1389.pdf

The authors claim

"Our representation of $\zeta^{−1}(s)$ for $\Re(s) ∈ (0, 1/2)$ - which seems to an anonymous referee "much too simple" to be true is analogical to the following - rather simple - series representations of $\zeta(s)$

$\zeta(s) = \frac{1}{1-2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s} \text{ for } \Re(s) >0, s\neq 1$

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    Related http://mathoverflow.net/questions/59770/are-the-non-trivial-zeros-of-zeta-simple/59774#597742012-07-03

3 Answers 3

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I think you can evaluate that sum. Multiply by $(-\pi)^{(s+1)/2}$, differentiate (this will cancel the $s+2n+1$ in the denominator); multiply by $(-\pi)^{1/2}$, differentiate (this will cancel the $s+2n$ in the denominator); what's left is essentially $e^{-\pi}$ times some simple function of $s$.

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    Oops, that should be ${}_2F_2$, not ${}_2F_1$.2012-07-04
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You could also try numerical evaluation at some particular $s$, say $s=1/3$. It's not at all close: $1/\zeta(1/3) \approx -1.027368851$, his sum $\approx -.2205319901$.

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    @PeterTamaroff Check my link in the comment to the other answer, that probably explains a lot...2012-07-04
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I might be missing something dumb, but I think this can't be right. The given sum converges uniformly on compact subsets of $Re(s)>0$. (Because the summand is $O(\pi^n/(n! n^2))$.) So it gives an analytic function on $Re(s)>0$. Meanwhile, $\zeta(s)^{-1}$ does not have an analytic extension to $Re(s)>0$, since there are zeroes on the critical line.

I can't make head or tail of the paper you link, so I don't know where this claim coms from.

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    http://arxiv.org/find/math/1/au:+Madrecki_A/0/1/0/all/0/12012-07-03