# Is this a relation between the Riemann zeta function and the Prime zeta function?

I am partly repeating myself here again. Is this a correct relation between the Riemann zeta function and the Prime zeta function?

$$\zeta (s) = \sum\limits_{n=1}^{\infty}\frac{1}{n^{s}}$$

$\displaystyle \log(\zeta (s)) = \sum\limits_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$ where $\displaystyle a_{1} = 0\;$, $\displaystyle a_{n} = \frac{\Lambda(n)}{\log(n)}$ and $\Lambda(n)$ is the Mangoldt function defined by $\displaystyle \Lambda(n) = \log(p)$ if $n = p^{k}$ for $p$ a prime.

$\displaystyle ω_{1}(\log(\zeta (s))) = \sum\limits_{n=1}^{\infty}\frac{\log(a_{n})}{n^{s}},$ where $ω_{1}(\;)$ is an operation and $\displaystyle \log(a_n) = 0$ if $a_n = 0$ $$ω_{1}(\log(\zeta (s)))\zeta (s) = \sum\limits_{n=1}^{\infty}\frac{b_{n}}{n^{s}}$$

$\displaystyle ω_{2}(ω_{1}(\log(\zeta (s)))\zeta (s)) = \sum\limits_{n=1}^{\infty}\frac{e^{b_{n}}}{n^{s}}\;$ where $ω_{2}(\;)$ is another operation and $\displaystyle e^{b_{n}}=\exp(b_{n})$

$$\sum\limits_{p\;\text{prime}} \frac1{p^s} = \log(ω_2(ω_1(\log(\zeta (s)))\zeta (s)))$$

• Can you explain in more detail this "operation" $\omega_1$. Are $\omega_1$ and $\omega_2$ the same thing? What precisely is the definition. – Eric Naslund May 28 '11 at 21:59
• @Eric: I was puzzled by the same thing, but then I thought that this might be analytic number theorists' slang. I read it as follows: view the the Dirichlet series $\sum\limits_{n=1}^{\infty} \dfrac{a_n}{n^s} = D(s,(a_n))$ as a function of the sequence $(a_n)$. Then $\omega_1 D(s,(a_n))$ is the Dirichlet series $D(s,(\log{a_n}))$ and $\omega_2 D(s,(b_n)) = D(s, (e^{b_n}))$. I'm not sure whether this is well-defined if one thinks of $D(s,(a_n))$ as an actual function of $s$ without very strong hypotheses, but the "operation" seems to make sense this way. – t.b. May 28 '11 at 22:22
• mathworld.wolfram.com/PrimeZetaFunction.html gives a close-formed expression for ζ in terms of P and vice-versa. Both are simple but neither are Dirichlet series. This is far from an answer to Mats' question but maybe a useful hint to someone. – Dan Brumleve May 29 '11 at 5:14
• @Eric: I should have written unknown operations $ω_{1}$ and $ω_{2}$. – Mats Granvik May 29 '11 at 8:48

In the case $s=2$ we have $$\sum_{p\in\mathcal{P}}\frac{1}{p^2}\color{red}{\leq} \frac{1}{2}\sum_{p\in\mathcal{P}}\log\,\left(\frac{1+\frac{1}{p^2}}{1-\frac{1}{p^2}}\right)=\frac{1}{2}\log\frac{\zeta(2)^2}{\zeta(4)}=\log\sqrt{\frac{5}{2}}=0.458145365937\ldots$$ and that is quite a good approximation since $\frac{1}{2}\log\left(\frac{1+x}{1-x}\right) = x+\frac{x^3}{3}+\frac{x^5}{5}+\ldots$ in a neighbourhood of the origin. In general, by Moebius inversion formula

$$\sum_{p\in\mathcal{P}}\frac{1}{p^s}=\sum_{n\geq 1}\frac{\mu(n)}{n}\log\zeta(ns)\tag{1}$$ that is a series with a decent convergence speed, due to $\log\zeta(ns)\approx 2^{-ns}$ for any large $n$.

A variant of the relationship can be put like this in Mathematica:

s = 7;
nn = 1000;
N[Log[Sum[
1/(Exp[FullSimplify[
Sum[If[n/k == 1, Log[1],
If[Mod[n, k] == 0,
Log[Denominator[
FullSimplify[MangoldtLambda[n/k]/Log[n/k]]]], 0]], {k, 1,
n}]]]*n^s), {n, 1, nn}]], 12]
N[PrimeZetaP[s], 12]


which outputs two equal numbers:

0.00828383285613

0.00828383285613


The fake or the tweek is the use of Denominator[ which together with MangoldtLambda[n]/Log[n] should be possible to replace by something in terms of the PrimeZeta function and Riemann zeta function, which is a bit circular.

The same relationship as a product over the divisors instead of a sum:

s = 7;
nn = 1000;
N[Log[Sum[
1/(FullSimplify[
Product[If[n/k == 1, 1,
If[Mod[n, k] == 0,
Denominator[FullSimplify[MangoldtLambda[n/k]/Log[n/k]]],
1]], {k, 1, n}]]*n^s), {n, 1, nn}]], 12]
N[PrimeZetaP[s], 12]

0.00828383285613

0.00828383285613


The Dirichlet generating function is:

(*start*)
s = 7;
nn = 1000;
N[1 + Sum[
Denominator[FullSimplify[MangoldtLambda[n]/Log[n]]]/n^s, {n, 2,
nn}], 12]
N[Zeta[s] + Sum[n*PrimeZetaP[(n + 1)*s], {n, 1, 20}], 12]
(*end*)


$$1+\sum _{n=2}^{\infty } \frac{\text{Denominator}\left[\text{FullSimplify}\left[\frac{\Lambda (n)}{\log (n)}\right]\right]}{n^s}=\zeta (s)+\sum _{n=1}^{\infty } n P((n+1) s)$$