[Vassilev-Missana - A note on prime zeta function and Riemann zeta function](http://nntdm.net/volume-22-2016/number-4/12-15)¹ claims the following remarkable identity:
$$
P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-...}}}},
$$
for integer $s>1$, where $\zeta(s)$ is the Riemann zeta function defined as $\sum\_{n=1}^\infty \frac{1}{n^s}$ and $P(s)$ denotes the prime zeta function $\sum\_{p\in\mathbb{P}}^\infty \frac{1}{p^s}$, and $\mathbb{P}$ is the set of prime numbers.


I am not a number theoretician, but the paper looks like it might contain some errors. I have already raised the question [here](https://math.stackexchange.com/questions/2570344/proving-an-identity-involing-the-riemann-zeta-function), and it looks like I may be right. However, after performing some calculations using Wolfram Alpha, it looks like the identity might have some truth to it.


If we denote by $\epsilon(s)$ the error term $P(s)-\left(1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-...}}\right)$, we get the following results for successive finite approximations to the nested radical, in which the $k^{\rm th}$ approximation is $\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{...-\sqrt{\frac{2}{\zeta(2^{k-1}s)}}}}}$, and in this case by taking $s=2$:


\begin{array}{|c|c|}
\hline
k & \epsilon(2)\\ \hline
3 & 0.19732\ldots \\ \hline
4 & -0.13198\ldots \\ \hline
6 & 0.04839\ldots \\ \hline
5 & -0.035665\ldots \\ \hline
7 & 0.007512\ldots \\ \hline
8 & -0.013753\ldots \\ \hline
9 & -0.00304\ldots \\ \hline
\end{array}


so indeed it looks like $\epsilon(2)$ is converging to $0$ (note the expression "*looks like*", as I have not run any calculations for *really* large $k$).


Additionally, by taking $s=3$ the identity above would remarkably imply that:


$$
\zeta(3) = 2\left((1-P(3))^2+\sqrt{\frac{2}{\zeta(6)}-\sqrt{\frac{2}{\zeta(12)}-\sqrt{\frac{2}{\zeta(24)}-...}}}\right)^{-1},
$$


and indeed taking the finite nested radical approximation for $k=9$, the error term is already $\epsilon=-0.002516\ldots$ and seems to approach $0$ as $k\to\infty$ (note again the expression "*seems to*").


Is it possible that the identity still holds, nonwithstanding the issues in the paper, and if so, how may one prove it? Or do the approximations for large $k$ actually **not** produce $\epsilon\to0$? If so, how can the apparent convergence be explained?




---


¹ Note that the journal in question is a reputed one, with **Paul Erdős** being at some point the editor (between '95 and '96), not a second-rate journal.



$P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-...}}}}$
This has been answered in the [comments](https://mathoverflow.net/questions/288847/ps-1-sqrt-frac2-zetas-sqrt-frac2-zeta2s-sqrt-frac2-zeta#comment715366_288847) by [Lucia](https://mathoverflow.net/users/38624/lucia). The identity $$P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-\cdots}}}}$$ is false. By subtracting $1$ from both sides and squaring, we have that $$(1-P(s))^{2}+1-\frac{2}{\zeta(s)}=1-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-\cdots}}},$$ which by the identity implies that $$(1-P(s))^{2}+1-\frac{2}{\zeta(s)}=P(2s).$$ This equation, which appears as theorem $1$ in the paper, is not true.


