Series of reciprocals of primes squared

Question

I have to check the convergence of series $\sum_{p \in \mathbb{P}}^\infty \frac{1}{p^2}$ where $\mathbb{P}$ is the set of all primes . I was thinking that I could compare this with the series $\sum_{n=1}^\infty \frac{1}{n^2}$, which is bigger than our series since it contains more elements and by comparison test our series also converges. Is this a good enough argument?

Alternatively, this is a $p$-series (not to confuse $p$ with a prime) with $p=2$ and thus it converges. However I am still not sure if this argument is valid since we miss a lot of terms compared to $n=1 \to \infty$.

Answer 1

Yes, using $p_n \geq n$ where $p_n$ denotes the n-th prime number proves your claim if you use https://en.wikipedia.org/wiki/Direct_comparison_test.

Answer 2

This is $P(2)$, where $P(s)$ is the Prime Zeta Function. The converge of $P(2)$ is indeed trivial, and its values is approximately $$ P(2)=\sum_p \frac{1}{p^2}=0.4522474200410654985065, $$ see this OEIS entry for references. The other series is $\zeta(2)=\sum_n \frac{1}{n^2}=\frac{\pi^2}{6}$.