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