Does Artin's conjecture imply that the reciprocal sum of primes with a given primitive root would diverge? Artin conjectured that every non-square integer $a\ne -1$ is a primitive root for infinitely many primes. Here it is on Wikipedia: Artin's conjecture on primitive roots. The conjecture also includes a possible asymptotic density for the primes having $a$ as a primitive root, under certain conditions. It's not clear to me whether the existence of that density, and the fact that it is positive, would imply that the reciprocals of all such primes form a divergent series.
I know that the sum of reciprocals of all primes diverges, and that the sum of reciprocals of primes in an arithmetic sequence diverges (at least I think I remember reading that somewhere), and that leads me to think that a positive asymptotic density is sufficient, but I don't know how to make that entirely clear.
Any insights are appreciated.
 A: If $A$ is a set of primes that has a positive asymptotic density in the set $\mathbb{P}$ of all primes, then the sum of the reciprocals of the elements of $A$ diverges. For the existence of the asymptotic density implies the existence of the logarithmic density, and these two densities are equal. The logarithmic density, if it exists, is defined as
$$\lambda := \lim_{x\to \infty} \frac{\sum_{\substack{p\in A \\ p \leqslant x}} \frac{1}{p}}{\sum_{\substack{p \in \mathbb{P} \\ p \leqslant x}} \frac{1}{p}} = \lim_{x\to\infty} \frac{1}{\log \log x} \sum_{\substack{p\in A \\ p \leqslant x}} \frac{1}{p}\,.$$
If $\lambda > 0$, then clearly
$$\sum_{\substack{p \in A \\ p \leqslant x}} \frac{1}{p} \sim \lambda\log \log x$$
is unbounded.
A positive asymptotic density is more than needed to have the sum of the reciprocals diverge. It is evidently sufficient that the set has a positive upper logarithmic density, i.e.
$$\lambda^{\ast} := \limsup_{x\to\infty} \frac{1}{\log \log x} \sum_{\substack{p\in A \\ p \leqslant x}} \frac{1}{p} > 0\,.$$
It is however not sufficient that $A$ has a positive upper asymptotic density in $\mathbb{P}$. If $(a_k)_{k \in \mathbb{N}}$ is a sufficiently fast growing sequence, e.g. $a_k = 2^{k^2}$, then $A = \{ p \in \mathbb{P} : (\exists k)(a_k < p \leqslant ka_k)\}$ has positive upper asymptotic density in $\mathbb{P}$ but the sum of reciprocals converges. If $\pi_A(x) = \operatorname{card}\:\{ p \in A : p \leqslant x\}$, then
$$\frac{\pi_A(ka_k)}{\pi(ka_k)} \geqslant \frac{\pi(ka_k) - \pi(a_k)}{\pi(ka_k)} = 1 - \frac{\pi(a_k)}{\pi(ka_k)} = 1 - \frac{a_k(\log k + \log a_k)}{ka_k\log a_k}(1+o(1)) \to 1,$$
so the upper asymptotic density of $A$ is $1$. But from the prime number theorem with (not the strongest known) error bounds we obtain
$$\sum_{a_k < p \leqslant ka_k} \frac{1}{p} = \log \frac{\log (ka_k)}{\log a_k} + O\biggl(\frac{1}{(\log a_k)^2}\biggr) = \frac{\log k}{\log a_k} + O\biggl(\biggl(\frac{\log k}{\log a_k}\biggr)^2\biggr)\,,$$
so
$$\sum_{p \in A} \frac{1}{p} < +\infty \iff \sum_k \frac{\log k}{\log a_k} < +\infty\,.$$
Of course having a positive lower asymptotic density in $\mathbb{P}$ suffices, since that implies a positive lower logarithmic density and hence a positive upper logarithmic density. But a positive lower asymptotic density is not necessary, as $\mathbb{P}\setminus A$ for the above $A$ shows.
