We know that the existence of natural density implies that of the logarithmic density over the natural numbers. Now, I can analogously define the relative density of a set $P$ of prime numbers to be the limit $\lim_{x \rightarrow \infty} \frac{P(x)}{\pi(x)}$ when it exists and I can analogously define the logarithmic density of the set $P$ as the limit
$$\lim_{x \rightarrow \infty} \frac{\sum_{p \in P, p \leq a} \frac{1}{a}}{\sum_{p \leq x} \frac{1}{p}} = \lim_{x \rightarrow \infty} \frac{1}{\log \log x} \sum_{p \in P, p \leq a} \frac{1}{a}$$
also subject to existence. I read here (Does Artin's conjecture imply that the reciprocal sum of primes with a given primitive root would diverge?) that the corresponding analogue of the result I stated in the beginning about the equality of the natural and logarithmic densities also holds for the above two densities. However, I have not found any reference containing the proof, nor have I been able to prove the same (I tried using Abel Summation). I would be really grateful for any references/ hints/proofs.
Aside 1: We also know that the upper and lower logarithmic densities over the natural numbers are always sandwiched between the upper and lower natural densities. Does the corresponding analogue hold for the above two densities defined above?
Aside 2: Also in the link provided above, in order to show that the sum of reciprocals of a set of primes having positive relative density $\alpha>0$ is divergent, the accepted answer goes through logarithmic density, however I feel that there should be more elementary arguments for the same. Does anyone know of any? Thanks.
Best Answer
That the existence of a natural (relative) density implies the existence of a (relative) logarithmic density — these two are then of course equal — holds more generally.
For $A \subset \mathbb{N}$ (using the convention $0 \notin \mathbb{N}$ here), define \begin{align} N_A(x) &:= \# \{ a \in A : a \leqslant x\}\,,\\ L_A(x) &:= \sum_{\substack{a \in A \\ a \leqslant x}} \frac{1}{a}\,. \end{align} Then if we have a substantial $S \subset \mathbb{N}$, that is, a set with $\lim_{x \to \infty} L_S(x) = \infty$, for all $B \subset S$ the inequalities $$\liminf_{x \to \infty} \frac{N_B(x)}{N_S(x)} \leqslant \liminf_{x \to \infty} \frac{L_B(x)}{L_S(x)} \leqslant \limsup_{x \to \infty} \frac{L_B(x)}{L_S(x)} \leqslant \limsup_{x \to \infty} \frac{N_B(x)}{N_S(x)} \tag{$\ast$}$$ hold. This answers your "aside 1" in the affirmative, and yields the implication stated above, since the left and right terms of $(\ast)$ are equal if $B$ has a (relative) natural density in $S$. We can prove $(\ast)$ via Abel summation:
Let $c$ and $C$ be the left and right hand side of $(\ast)$ respectively. If $c = 0$ there's nothing to prove for the left inequality, otherwise for every $0 < \gamma < c$ there is an $x_{\gamma}$ with $N_B(x) \geqslant d\cdot N_S(x)$ for all $x \geqslant x_{\gamma}$. Then, for $x > x_{\gamma}$ we have \begin{align} L_B(x) &= \frac{N_B(x)}{x} + \int_1^x \frac{N_B(t)}{t^2}\,dt \\ &\geqslant \gamma\cdot \Biggl(\frac{N_S(x)}{x} + \int_1^x \frac{N_S(t)}{t^2}\,dt\Biggr) - \int_1^{x_{\gamma}} \frac{\lvert \gamma N_S(t) - N_B(t)\rvert}{t^2}\,dt \\ &= \gamma\cdot L_S(x) - \int_1^{x_{\gamma}} \frac{\lvert \gamma N_S(t) - N_B(t)\rvert}{t^2}\,dt \\ &\geqslant \gamma\cdot L_S(x) - \log x_{\gamma} \end{align} and consequently $$\liminf_{x \to \infty} \frac{L_B(x)}{L_S(x)} \geqslant \gamma\,.$$ (This uses $\lvert\gamma N_S(t) - N_B(t)\rvert \leqslant t$, which follows from $0 \leqslant N_B(t) \leqslant t$ and $0 \leqslant \gamma N_S(t) \leqslant \gamma t \leqslant t$ [clearly $c \leqslant 1$, hence $\gamma < 1$] and $\frac{\log x_{\gamma}}{L_S(x)} \to 0$.) Since $\gamma < c$ was arbitrary the left inequality of $(\ast)$ follows. The middle inequality follows from the definitions of $\liminf$ and $\limsup$, and the right inequality of $(\ast)$ is proved similar to the left. Let $\Gamma > C$. Then there is an $x_{\Gamma}$ such that $N_B(x) \leqslant \Gamma\cdot N_S(x)$ for $x \geqslant x_{\Gamma}$. Essentially the same computation, just with the sense of the inequalities reversed, $\gamma$ replaced with $\Gamma$, the last integral added instead of subtracted, yields $$\limsup_{x \to \infty} \frac{L_B(x)}{L_S(x)} \leqslant \Gamma\,.$$ Again, since this holds for all $\Gamma > C$, the right inequality of $(\ast)$ follows.
Concerning your second "aside",
this is a bit delicate. I sketched an example of a set $A$ of primes having upper (relative) natural density $1$ in the set of all primes such that the series of the reciprocals of the primes in $A$ converges. Thus you should not expect a very simple argument. On the other hand, as I also mentioned over there, we don't need a (relative) natural density, a positive lower (relative) natural density suffices. The proof is above: Abel summation yields $$L_B(x) \geqslant \gamma L_S(x) - \log x_{\gamma}\,,$$ and the right hand side tends to $\infty$ for $x \to \infty$. The argument is fairly elementary, I'd say, but of course not totally trivial. I can't think of a more elementary argument.