TITLE: Prove that $e^{\sum 1/p_k^2} > \pi/2$
QUESTION [13 upvotes]: The $k$th prime is $p_k.$ Prove (if inclined, and without a machine) that
$$\exp\left({\sum_{k=1}^{\infty}\frac1{p_k^2}}\right) > \frac{\pi}{2} $$

REPLY [2 votes]: 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.