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.

Here's a solution that exploits a comment of Oscar Lanzi under my other answer (using an observation that I learned from a note of Noam Elkies [pdf]). In particular, it avoids both the identity $\sum_{n \in \Bbb N} \frac{1}{n^2} = \frac{\pi^2}{6}$ and using integration.
Let $\Bbb P$ denote the set of prime numbers and $X$ the union of $\{2\}$ and the set of odd integers $> 1$; in particular $\Bbb P \subset X$, so where $E$ denotes the set of positive, even integers:
$$\sum_{p \in \Bbb P} \frac{1}{p^2} \leq \sum_{n \in X} \frac{1}{n^2} = \color{#00af00}{\sum_{n \in \Bbb N \setminus E} \frac{1}{n^2}} - \frac{1}{1^2} + \frac{1}{2^2}.$$
Now, $$\sum_{n \in \Bbb N} \frac{1}{n^2} < 1 + \sum_{n \in \Bbb N \setminus \{1\}} \frac{1}{n^2 - \frac{1}{4}} = 1 + \sum_{n \in \Bbb N \setminus \{1\}} \left(\frac{1}{n - \frac{1}{2}} - \frac{1}{n + \frac{1}{2}} \right) = 1 + \frac{2}{3} = \frac{5}{3};$$ the second-to-last equality follows from the telescoping of the sum in the third expression.
The sum over just the even terms satisfies
$$\sum_{m \in E} \frac{1}{m^2} = \sum_{n \in \Bbb N} \frac{1}{(2 n)^2} = \frac{1}{4} \sum_{n \in \Bbb N} \frac{1}{n^2} ,$$ and thus
$$\color{#00af00}{\sum_{n \in \Bbb N \setminus E} \frac{1}{n^2} = \left(1 - \frac{1}{4}\right) \sum_{n \in \Bbb N} \frac{1}{n^2} < \frac{3}{4} \cdot \frac{5}{3} = \frac{5}{4}}.$$
Substituting in the first display equation above yields $$\sum_{p \in \Bbb P} \frac{1}{p^2} \leq \sum_{n \in X} \frac {1}{n^2} < \color{#00af00}{\frac{5}{4}} - 1 + \frac{1}{4} = \frac{1}{2} .$$
