# No closed form for $\sum_{n\in P} \frac{1}{n^2}$

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.)

• If you want a mathematical proof, you'll need to specify a mathematical definition. Nov 8, 2015 at 1:23
• 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? Nov 8, 2015 at 1:24
• @pre-kidney Its kinda rude to dissect the small stuff and ignore the question. Which you are obviously doing.
– user253055
Nov 8, 2015 at 1:34
• 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 $\gamma$. Nov 8, 2015 at 1:44
• Now posted to MO (with no link from either site to the other), mathoverflow.net/questions/223064/… Nov 9, 2015 at 4:47

Your sum is equal to $$P(2)$$, where $$P(n)$$ denotes the prime zeta function defined as
$$P(n)=\sum_{p\in\mathbb{P}}\frac{1}{p^n}.$$
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.