A connection (discovered by me) between prime numbers and the Hermite Constant in 3D

Question

Let $P(n)$ be a characteristic function of prime numbers, such that

$$ P(n)= \begin{cases} 1, & \small \text{if } n \text{ is prime} \\ 0, & \small \text{otherwise} \end{cases} $$

For instance: $P(n) = \pi(n)-\pi(n-1)$, where $\pi(x)$ is the prime-counting function. Knowing all that, we can evaluate the following identity (conjeture?) discovered by me

$$ \sum_{i=1}^\infty \cfrac{(-1)^{P(i)}}{i^2}= \sum_{i=1}^\infty \cfrac{(-1)^{\pi(i)-\pi(i-1)}}{i^2}=\frac{\pi}{3\sqrt{2}} $$

and it seems to approach (converge to) the Hermite Constant in 3 dimensions. So, obviously the question is: can you prove that sum is true?

Thanks

Answer 1

Expanding on Gerry Myerson's comment; $\zeta(2)=\dfrac{\pi^2}{6}$ and the first $104$ digits of $P(2)=\sum_{p}p^{-2}$ can be found at: http://oeis.org/A085548. We now see that $$\zeta(2)-2P(2)=0.7404392267660955...$$, while $$\dfrac{\pi}{3\sqrt{2}}=0.7404804896930609...$$ So no, you can't prove it because, sadly, it's false.

Answer 2

Let $\chi _{{{\mathbb {P}}}}(n)$ be a characteristic function of prime numbers. Then, it's easy to see that: $$ (-1)^{\chi _{{{\mathbb {P}}}}(i)}= 1- 2 \chi _{{{\mathbb {P}}}}(i) $$ therefore, $$ \sum_{i=1}^\infty \cfrac{(-1)^{\chi _{{{\mathbb {P}}}}(i)}}{i^2}= \sum_{i=1}^\infty \cfrac{1- 2 \chi _{{{\mathbb {P}}}}(i)}{i^2} = \sum_{i=1}^\infty \cfrac{1}{i^2}- 2 \sum_{i=1}^\infty \cfrac{\chi _{{{\mathbb {P}}}}(i)}{i^2} = \\ \\ =\zeta(2)-2 \sum_p \cfrac{1}{p^2} = \zeta(2)-2 P(2)= \\ \\ \\ = 0.7404392267660954394593\ldots $$ where P(2) means Prime zeta function of 2. The conclusion is that

$$ \frac{\pi}{3\sqrt{2}}\neq \zeta(2)-2 P(2) $$ because $$ \frac{\pi ^2 -\pi \sqrt{2}}{12}<P(2) $$ so, I was wrong about my conjeture, but Gerry Myerson and Mastrem are right.

Regards