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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

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    $\begingroup$ This is pretty much $\sum_Ci^{-2}-\sum_Pi^{-2}$, where the first sum is over the composites and the second over the primes, so it's pretty much $\zeta(2)-2\sum_Pi^{-2}$. The sum on the primes is sometimes called "the prime zeta function". $\pi/(3\sqrt2)$ seems highly unlikely to me. $\endgroup$ – Gerry Myerson Sep 12 '16 at 10:23
  • $\begingroup$ Indeed, little is known about that prime zeta thing. Related: mathoverflow.net/questions/53443/… $\endgroup$ – Ivan Neretin Sep 12 '16 at 10:25
  • $\begingroup$ what we know is $\sum_{p^k} \frac{1}{k p^{2k}} = \ln \zeta(2) = \ln \frac{\pi^2}{6}$, but the value of $\sum_{p} \frac{1}{p^{2}} = \sum_{k=1}^\infty \frac{\mu(k)}{k} \ln \zeta(2k)$ is unknown $\endgroup$ – reuns Sep 12 '16 at 15:39
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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.

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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

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  • $\begingroup$ however, it was interesting $\endgroup$ – user354674 Sep 12 '16 at 20:08

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