How to prove $\sum_p p^{-2} < \frac{1}{2}$? I am trying to prove $\sum_p p^{-2} < \frac{1}{2}$, where $p$ ranges over all primes. I think this should be doable by elementary methods but a proof evades me. 
Questions already asked here (eg. What is the value of $\sum_{p\le x} 1/p^2$? and Rate of convergence of series of squared prime reciprocals) deal with the exact value of the above sum, and so require some non-elementary math.

We can deduce this quickly, and without knowing the numerical value of $\pi$, from the fact that
$$\sum_{n \in \Bbb N} \frac{1}{n^2} = \frac{\pi^2}{6},$$ for which there are numerous proofs available.
Let $E$ denote the set of even numbers; the sum of the squares of all such numbers is
$$\sum_{n \in E} \frac{1}{n^2} = \sum_{k \in \Bbb N} \frac{1}{(2 k)^2} = \frac{1}{4} \sum_{k \in \Bbb N} \frac{1}{k^2} = \frac{1}{4} \cdot \frac{\pi^2}{6} = \frac{\pi^2}{24}.$$
Now, let $X$ denote the union of $\{2\}$ and all positive odd integers $> 1$. In particular, $X$ contains the set $\Bbb P$ of all prime numbers as a subset, and so
\begin{align}
\sum_{p \in \Bbb P} \frac{1}{p^2}
    &\leq \sum_{n \in X} \frac{1}{n^2} \\
    &= \sum_{n \in \Bbb N} \frac{1}{n^2} - \sum_{n \in E} \frac{1}{n^2} - \frac{1}{1^2} + \frac{1}{2^2} \\
    &= \frac{\pi^2}{6} - \frac{\pi^2}{24} - 1 + \frac{1}{4} \\
    &= \frac{\pi^2}{8} - \frac{3}{4} .
\end{align}
So, it suffices to show that
$$\frac{\pi^2}{8} - \frac{3}{4} < \frac{1}{2},$$
but rearranging shows that this is equivalent to $\pi^2 < 10$, and $\pi < \frac{22}{7}$ implies
$$\pi^2 < \left(\frac{22}{7}\right)^2 = \frac{484}{49} < \frac{490}{49} = 10.
$$
