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.

Solution 1:

All primes but 2 are odd numbers so $$\sum_p p^{-2} < 1/4 + \sum_{k=1}^\infty \frac{1}{(2k+1)^2}$$ Using the fact that $1/x^2$ is convex the sum is bounded by $$ \sum_{k=1}^\infty \int_{k-1/2}^{k+1/2}\frac{1}{(2x+1)^2}dx = \int_{1/2}^\infty \frac{1}{(2x+1)^2}dx = 1/4$$

Solution 2:

If you know $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ then you could simply say $$ \displaystyle \sum_{p \text{ prime}} \frac{1}{p^2} $$ $$\lt \frac{\pi^2}{6} - \frac{1}{1^2}- \frac{1}{4^2}- \frac{1}{6^2}- \frac{1}{8^2}- \frac{1}{9^2}- \frac{1}{10^2}- \frac{1}{12^2}- \frac{1}{14^2}- \frac{1}{15^2}- \frac{1}{16^2} $$ $$ \approx 0.49629 $$ $$ \lt \frac12.$$

Alternatively if you do not know that, instead use $\displaystyle \frac{1}{k^2} \le \int_{x=k-1}^k \frac{1}{x^2}\, dx = \frac{1}{k-1} - \frac{1}{k}$ so $\displaystyle \sum_{n=k}^\infty \frac{1}{n^2} \le \int_{x=k-1}^\infty \frac{1}{x^2}\, dx = \frac{1}{k-1}$ and you can say: $$ \displaystyle \sum_{p \text{ prime}} \frac{1}{p^2} \lt \frac{1}{2^2}+ \frac{1}{3^2}+ \frac{1}{5^2}+ \frac{1}{7^2}+ \frac{1}{11^2}+ \frac{1}{13^2}+ \frac{1}{17-1} \approx 0.4982 \lt \frac12.$$