# Conjectured relation between alternating Prime zeta series and Riemann zeta

Let $P(s)$ be the Prime zeta function.

Numerical evidence suggests these identities:

$$\sum_{k=1}^\infty \frac{(-1)^{k}P(3k)}{k}=\log{\bigg(\frac{1}{945}\frac{\pi^6}{\zeta(3)}\bigg)}\qquad\quad (1)$$

$$\sum_{k=1}^\infty \frac{(-1)^{k}P(nk)}{k}=\log{\bigg(\frac{1}{a(n)}\frac{\pi^{2n}}{\zeta(n)}\bigg)}\qquad (2)$$

for natural $n$, where $a(n)$ is OEIS A002432 Denominators of $~\dfrac{\zeta(2n)}{\pi^{2n}}$.

In $a(n)$ we have $\zeta(2n)$ and in (2) $\zeta(n)$.

Is (1) and/or (2) true ?

We have $$\sum_k\frac{(-1)^kP(nk)}{k}=\sum_{k,p}\frac{(-1)^k}{kp^{nk}}=-\sum_p\ln\left(1+\frac{1}{p^n}\right)=\sum_p\ln\left(\frac{1-\frac{1}{p^{n}}}{1-\frac{1}{p^{2n}}}\right)=\ln\frac{\zeta(2n)}{\zeta(n)}.$$

This computation shows that your guess is correct whenever the numerator of $\zeta(2n)/\pi^{2n}$ is equal to $1$ (more or less for $n$ up to $5$, if I understand correctly), in particular for $n=3$.

• I mean $(-1)^k$. Thank you. Indeed for n=8 I need additional numerator :-). – joro Sep 27 '14 at 14:08
• @joro: yes you are right, it's my mistake with signs, not yours :) – Vladimir Dotsenko Sep 27 '14 at 14:50
• In the second sum and later sums, $\sum_{k,p}$ do you assume $p$ is prime? – joro Sep 27 '14 at 16:13
• @joro: yes of course. The first equality is the definition of the prime zeta, the second is the Taylor series of logarithm, the third is basic algebra, the fourth is Euler product for zeta. – Vladimir Dotsenko Sep 27 '14 at 16:34
• Your suggestion to replace (-1)^k with (-1)^(k-1) leads to very similar result. – joro Sep 28 '14 at 11:08

There is a really nice formula which is valid for |q| < 1, and it involves the dirichlet convolution of arithmetic functions. First, let's denote $\frac{1}{n}$ to mean the arithmetic function that takes a natural number $k$ to $\frac{1}{k}$.

Suppose we have two arithmetic functions: $a,b$

If $a=\frac{1}{n}*b$, then we have the following formula which is valid for all $q<1$:

$$\frac{1}{(1-q)^{b(1)}(1-q^{2})^{b(2)}(1-q^{3})^{b(3)}...}=e^{((a(1)q+a(2)q^{2}+a(3)q^{3}+...))}$$

For your problem, we should consider the following arithmetic sequence for $a$:

$a(3k)= \frac{(-1)^{k}}{k}$ and $a(n)=0$ when $n$ is not divisible by $3$.

From this, we calculate $b(3)=-1$, $b(6)=1$, and $0$ everywhere else.

Given these arithmetic functions, let's set $q=\frac{1}{p}$ where $p$ is prime. Plugging all of this into our formula gives us

$$(1-\frac{1}{p^{3}})(\frac{1}{1-\frac{1}{p^{6}}})=e^{(\frac{-1}{p^{3}}+\frac{1}{2p^{6}}-\frac{1}{3p^{9}}+...))}$$

Each prime number gives us a local factor. Now we just need to multiply all of the local factors together and take the log of both sides and we should derive your first identity. You should be able to easily generalize this construction to obtain your second identity.

• Thank you. Can you convert the series to integral? Possibly for fixed $n$ or dropping (-1)^k? – joro Sep 29 '14 at 8:50
• I'll have to think about that a little more, but I think you can. You will have to come up with a more general formula than the one I gave above. In particular, you need to replace the infinite sum a(i)q^i with an integral. – six Sep 30 '14 at 19:08