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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

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… – Eugene Jul 3 '12 at 20:59
up vote 2 down vote accepted

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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I think what Gerry means is to replace $-\pi$ by a variable $t$. If $F(t) = \displaystyle \sum_{n=0}^\infty \dfrac{t^n (s-1)s}{2n! (s+2n)(s+2n+1)}$, then $$\dfrac{\partial}{\partial t} t^{1/2} \dfrac{\partial}{\partial t} t^{(s+1)/2} F(t) = \dfrac{s(s-1)}{8} t^{s/2-1} e^t$$ However, reversing the differentiations by integration leaves you with a not-very-pleasant integral. – Robert Israel Jul 4 '12 at 5:08
Maple, by the way, evaluates the sum as $$\dfrac{s-1}{2s+2} {}_2F_{1}(s/2,(s+1)/2;s/2+1,s/2+3/2; -\pi)$$ – Robert Israel Jul 4 '12 at 5:15
Oops, that should be ${}_2F_2$, not ${}_2F_1$. – Robert Israel Jul 4 '12 at 17:05

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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I'm curious why didn't they do that before publishing that paper, but maybe I should ask them. – Pedro Tamaroff Jul 4 '12 at 2:53
@PeterTamaroff Check my link in the comment to the other answer, that probably explains a lot... – N. S. Jul 4 '12 at 13:35

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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