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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.

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    $\begingroup$ Surely the conjectured density refers to the analytic density, in which case...yes. The sum of the reciprocals of the relevant primes must diverge. If it converged then the numerator would be finite while the denominator tends to $\infty$, which would give a density of $0$. $\endgroup$ – lulu Oct 17 '17 at 15:38
  • $\begingroup$ If you'd care to write this up formally as answer, I'd accept it. $\endgroup$ – G Tony Jacobs Oct 17 '17 at 15:46
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    $\begingroup$ I'll defer to others who can speak with greater authority. My sense is that people tend to default to the analytic density because it is A. more tractable and B. when other densities (like the natural density) exist then the analytic density exists (and they coincide) but the converse is not true. But I'm sure some of the users here are expert on the issue and can supply more detail. $\endgroup$ – lulu Oct 17 '17 at 15:51
  • $\begingroup$ Honestly, your link to the analytic density article has basically answered my question. Thanks for that. $\endgroup$ – G Tony Jacobs Oct 17 '17 at 15:54
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    $\begingroup$ The wikipedia article on Artin's conjecture speaks of asymptotic density. That is the natural density. If that exists, also the logarithmic density and the Dirichlet density exist, and they all coincide. In particular, a positive asymptotic density (in the set of primes) implies that the sum of reciprocals is infinite. $\endgroup$ – Daniel Fischer Oct 17 '17 at 17:50
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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.

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  • $\begingroup$ The naive model, supporting the conjecture, would be that the order of $a \bmod p$ as $p$ varies is quite uniformly distributed in $1 \ldots,p-1$, so we are looking at something close to $\sum_p \frac{1}{p}\frac{\phi(p-1)}{p-1}$. Using the random model again it is $\approx \sum_n \frac{1}{n \log n} \frac{\phi(n-1)}{n-1}\approx \sum_n \frac{1}{n \log n} \frac{\phi(n)}{n}$ which diverges since $\sum_{n=1}^\infty \phi(n)n^{-s}= \frac{\zeta(s-1)}{\zeta(s)}$ has a pole at $s=2$ thus $\sum_{n=2}^\infty \phi(n)\frac{n^{-s}}{\log n}= C+\int_3^s (\frac{\zeta(z-1)}{\zeta(z)}-1)dz$ is unbounded at $s=2$. $\endgroup$ – reuns Dec 29 '17 at 17:38
  • $\begingroup$ (I meant the order of $a \bmod p$ would be distributed in the divisors of $p-1$ as if $a$ was chosen randomly) $\endgroup$ – reuns Dec 29 '17 at 17:47

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