Let $P(s)$ be the [Prime zeta function](http://mathworld.wolfram.com/PrimeZetaFunction.html).


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](https://oeis.org/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 ?
> 
> 
> 



Conjectured relation between alternating Prime zeta series and Riemann zeta
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 $n=3$.


