Prove that $e^{\sum 1/p_k^2} > \pi/2$

Question

The $k$th prime is $p_k.$ Prove (if inclined, and without a machine) that

$$\exp\left({\sum_{k=1}^{\infty}\frac1{p_k^2}}\right) > \frac{\pi}{2} $$

Answer 1

I thought I saw an answer here using the prime zeta function, which looked correct, so I didn't develop my answer. But as there are requests in the comments, here it is:

We start with Aaron's suggestion in the comments:

$$\exp\left( \sum_p \frac1{p^2}\right) = \prod_p \exp\left( \frac1{p^2}\right) > \prod_p \left( 1 + \frac1{p^2} \right) = \prod_p \left( \left(1 - \frac1{p^4}\right)\sum_n \frac1{p^{2n}}\right) = \left( \prod_p \sum_n \frac1{p^{2n}} \right) \prod_p \left(1 - \frac1{p^4} \right) = \frac{\pi^2}6 \prod_p \left( \sum_n \frac1{p^{4n}} \right)^{-1} = \frac{\pi^2}6 \frac{90}{\pi^4} = \frac{15}{\pi^2} $$

Throughout $p$ ranges over primes and $n$ ranges over nonnegative integers. I used the usual sum-to-product formula $\zeta(s) = \sum_n \frac1{n^s} = \prod_p \left( \sum_n \frac1{p^{ns}}\right)$, which follows from (and is equivalent to) unique factorization (and I think is due to Euler). I also used known zeta values $\zeta(2) = \frac{\pi^2}6$ and $\zeta(4) = \frac{\pi^4}{90}$. (I don't know a quick proof of the second of these; the first follows from two expansions of $\iint_0^1 \frac{\mathrm d x\, \mathrm d y}{1 + xy}$, if my memory is correct, and is in Proofs from the Book regardless.)

Unfortunately, estimating $\frac{15}{\pi^2}$ gives an answer that's a little less than $\frac{\pi}2$, but we'd win if we could make up a factor of $(\frac\pi 2 - \frac{15}{\pi^2}) / (\frac{15}{\pi^2}) < 3.36\%$. So let's see if we can squeeze a little bit more out. In the comments asatzhh suggests looking at explicit values for small $p$. We have: $$ \frac{\exp(\frac14)}{1 + \frac14} > 1.0272,\ \frac{\exp(\frac19)}{1 + \frac19} > 1.0057, \ \frac{\exp(\frac1{25})}{1 + \frac1{25}} > 1.0007 $$ which you can get, e.g. from Taylor expansion.

Thus our estimate above is an underestimate by at least $2.72\% + .57\% + .07\% = 3.38\%$. This completes the proof.

I should admit, I've been getting the decimal expansions from a calculator, and rounding in whichever direction makes the estimate worse. So I don't know how far out you need to look in Taylor expansion to get them by hand. But the Taylor expansion of $\exp(x)$ converges very quickly, and it's easy to estimate its error. And $\pi$ has many famous series expansions; see http://en.wikipedia.org/wiki/Approximations_of_%CF%80.

Answer 2

This answer is quite similar to TenaliRaman's and contains the details of my comments to the question. The basic idea is that the "prime zeta function" is comparable to the logarithm of the actual zeta function, via the Euler product:

$$\zeta(s) = \sum_{n > 0} \frac{1}{n^s} = \prod_{\text{$p$ prime}} \frac{1}{1 - p^{-s}}.$$

If we take the logarithm, we get (with some necessary analysis to prove that the homomorphism property applies to the infinite product in this case):

$$\log \zeta(s) = \sum_p \log \frac{1}{1 - p^{-s}} = \sum_p \sum_{n > 0} \frac{(p^{-s})^n}{n}.$$

The last equality is just a well-known power series for $\log (1 - x)$:

$$\log (1 - x) = -\sum_{n > 0} \frac{x^n}{n}.$$

Anyway, taking just the first term of the $n$ sum in the previous equation for each $p$, we get

$$\log \zeta(s) > \sum_p p^{-s}.$$

Taking $s = 2$ gives the statement of the problem, and using the known value $\zeta(2) = \pi^2/6$, we have

$$\sum_p p^{-2} < \log \frac{\pi^2}{6} \iff \exp \sum_p \frac{1}{p^2} < \frac{\pi^2}{6}.$$

The right-hand side is greater than $\pi/2$, so this inequality is consistent with the problem, but without a proof that the difference between the two sides of the inequality above is less than $\pi^2/6 - \pi/2$, it doesn't actually finish it.

Answer 3

An early attempt which is clearly wrong. Making this into a community wiki, in case someone wants to take a stab at it and see if something can be salvaged here.

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The exponent is called as the Prime Zeta Function. From Mathworld [1], $$\sum_{k = 1}^{\infty} \frac{1}{p_k^2} = \sum_{k = 1}^{\infty} \frac{\mu(k)}{k} \ln(\zeta(2k))$$ where $\mu(n)$ is the Möbius function and $\zeta(n)$ is the Riemann zeta function. Therefore, $$\sum_{k = 1}^{\infty} \frac{1}{p_k^2} > \frac{\mu(1)}{1} \ln(\zeta(2)) = \ln\left( \frac{\pi^2}{6} \right)$$ $$\exp\left( \sum_{k = 1}^{\infty} \frac{1}{p_k^2} \right) > \frac{\pi^2}{6} > \frac{3\pi}{6} = \frac{\pi}{2}$$

[1] http://mathworld.wolfram.com/PrimeZetaFunction.html