Originally Posted by
JoachimAgrell Is it possible to prove this using the fact that $\displaystyle \sum_{p\in P}\frac{1}{p}$ diverges (where $\displaystyle P=\{n\in\mathbb{N}: n \text{ is prime}\}$)?
No. Dirichlet's theorem on arithmetic progressions is much stronger than this statement. In fact, since $\displaystyle (123456789,10^{10})=1$, Dirichlet's theorem on arithemtic progressions assures us that
$\displaystyle \sum_{p \equiv 123456789 \mod 10^{10}}\frac{1}{p}$
diverges. That $\displaystyle \sum_{p}\frac 1 p$ diverges is not enough!