Riemann Zeta formula

Question

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 $\zeta(s)$

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

Answer 1

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 $s$.

Answer 2

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.

Answer 3

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 $\approx -.2205319901$.