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I think that I can say with a fair amount of assurance that $$\sum_{n\in \mathcal P} \frac{1}{n^2}$$ has no closed form (assuming that $\mathcal P$ represents the full set of primes)

I currently know the following:

a) The sum converges at $ 0.45224742\dots$

b) The definition of a closed from expression is that you can express the constant with the following:

  • addition, multiplication, subtraction, and division
  • raising a number to a power that exists (Includes reciprocals, roots, powers, and powers to fractional exponents)
  • known constants such as $\pi,e, \gamma, \phi \cdots$ and no others
  • trigonometric functions and their inverses ($\sin,\text{arccsc}, \cot, \text{arcsec}\cdots\text{\etc.}$)
  • hyperbolic trig functions and their inverses ($\text{arccosh},\text{tanh},\text{arccoth}$)
  • You may not use anything that involves limits. A few off limits :) might be limits, derivatives, integrals, infinite sums, infinite products, and so on.

c) I know, at least, that the number is irrational. The irrationality of this concept [in part] inspired this question.

The reason I think that there is no closed form is due to the randomness of primes and that often times even random sums have no closed form. But really, I am not looking for whether or not this has a closed form, but a proof that there exists no closed form. Go as complicated as necessary to solve. Thanks for any help.

I realize that the a proof of the closed form is often extremely difficult, and even unknown for even the simplest of constants ($\gamma$ and $\zeta(3)$). I did know of the difficulty of a proof like this before asking this question. The reason I thought it might be easier was due to how random the distribution of primes were. (Remember we are trying to prove that there is no closed form not the opposite.)

PS: As a high school student, could you please namedrop the names of theorems please in your proof so I could learn more?

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    $\begingroup$ If you want a mathematical proof, you'll need to specify a mathematical definition. $\endgroup$ – pre-kidney Nov 8 '15 at 1:23
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    $\begingroup$ You are probably right. But wouldn't your thoughts expressed there imply that $$\prod_{n \in P}\left(1-\frac{1}{n^2}\right)$$ also has no closed form? $\endgroup$ – GEdgar Nov 8 '15 at 1:24
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    $\begingroup$ @pre-kidney Its kinda rude to dissect the small stuff and ignore the question. Which you are obviously doing. $\endgroup$ – user253055 Nov 8 '15 at 1:34
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    $\begingroup$ Proving that there is no closed form for some expression can be extremely difficult. No one knows, for example, whether there is a closed form for $\zeta(3)=\sum n^{-3}$, or for Euler's $\gamma$. $\endgroup$ – Gerry Myerson Nov 8 '15 at 1:44
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    $\begingroup$ Now posted to MO (with no link from either site to the other), mathoverflow.net/questions/223064/… $\endgroup$ – Gerry Myerson Nov 9 '15 at 4:47
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Your sum is equal to $P(2)$, where $P(n)$ denotes the prime zeta function defined as

$$ P(n)=\sum_{p\in\mathbb{P}}\frac{1}{p^n}. $$

No closed-form expression is known for any values of the prime zeta function.

c) I know, at least, that the number is irrational.

The irrationality of the prime zeta function at any positive integer $n\geq2$ is unknown. A proof of your statement would immediately gain you considerable fame.

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