This question is a duplicate of an existing MO question, but that other MO question has an accepted answer that does not actually answer the question, and I'm not sure how to fix that other than by re-asking the question.

On page 76 of Serre's book A Course in Arithmetic, he writes:

[T]here exist sets having an analytic density but no natural density. It is the case, for example, of the set $P^1$ of prime numbers whose first digit (in the decimal system, say) is equal to 1. One sees easily, using the prime number theorem, that $P^1$ does not have a natural density and on the other hand Bombieri has shown me a proof that the analytic density of $P^1$ exists (it is equal to $\log_{10}2 = 0.301029995\ldots$).

There is a slight misstatement here because literally speaking, $P^1$ has natural density zero, but clearly the intent is to speak of the relative density of $P^1$ inside the set $P$ of all primes. In other words, Bombieri's result is a kind of "Benford's law for primes": $$\lim_{s\to1} {\sum_{m\in P_1} m^{-s} \over \sum_{p\in P} p^{-s}} = \log_{10}2.$$

My question is, how does one prove that the above limit (which goes by various names—relative analytic density, relative Dirichlet density, relative zeta density) exists? Serre does not say anything about this.

The accepted answer to the duplicate MO question cites two papers, one by Cohen and Katz, and one by Raimi. But the paper by Cohen and Katz simply restates what Serre says without giving a proof. The paper by Raimi cites a paper by R. E. Whitney (Initial digits for the sequence of primes, Amer. Math. Monthly 79 (1972), 150–152) but Whitney's paper considers logarithmic density rather than Dirichlet density: $$\lim_{N\to\infty} {\sum_{m\in P_1, m\le N} 1/m \over \sum_{p\in P, p\le N} 1/p}.$$ It's not clear to me that this implies Bombieri's result.

In a comment to a now-deleted MO question, KConrad suggested looking in the book Prime Numbers by Ellison and Ellison, but I did not find the answer there either.

  • 1
    $\begingroup$ It takes a certain amount of chutzpah to accuse Serre of making a misstatement. In number theory, it is extremely common to talk about the densities of various sets of primes relative to the set of all primes; this is the most common use of the word. The fact that the quote previously mentions the concepts of "analytic" density (which only makes sense for sets of prime numbers) makes it even clearer that this is the context in which Serre is working. You may as well say that "even numbers have density 1/2" is a misstatement because the set of even integers inside the reals has density zero. $\endgroup$ – user113030 Aug 9 '17 at 0:07
  • $\begingroup$ @user113030 : You're right, Serre does actually define "natural density" in the preceding paragraph as relative density. My bad. $\endgroup$ – Timothy Chow Aug 9 '17 at 0:33

If a set of primes has a logarithmic density, then it also has a Dirichlet density, and the two agree --- and vice versa! For the direction you want (logarithmic -> Dirichlet), this follows by partial summation. Suppose $P$ is a set of primes with logarithmic density $\delta$, and let $S(t) = \sum_{p \le t,~p \in P} 1/p$. So by hypothesis, $S(t) = (\delta+o(1))\log\log{t}$, as $t\to\infty$. By partial summation, for any $s>1$, we have$$ \sum_{p \in P} 1/p^s = (s-1) \int_{1}^{\infty} S(t) t^{-s}\, dt. $$

The key to the rest of the proof is that if we replace $S(t)$ with $\log\log t$ on the right-hand side, then the right-hand side is exactly $\log(\frac{1}{s-1}) - \gamma$. Starting from this, a little fooling around will show that our assumption on $S(t)$ implies that our right-hand side is $(\delta + o(1))\log\frac{1}{s-1}$ as $s\downarrow 1$. Since $\sum_{p} \frac{1}{p^s} = (1+o(1)) \log \frac{1}{s-1}$ as $s\downarrow 1$, we get the desired conclusion.

The other direction isn't relevant to this question, so I'll only quickly sketch it. If $P$ has Dirichlet density $\delta$, then letting $s=1+1/\log{x}$, one deduces that as $x\to\infty$, one has $\sum_{p\in P} \frac{1}{p^{1+1/\log{x}}} = (\delta+o(1)) \log\log x$. Using elementary estimates on the distribution of primes, one argues that $$ \sum_{p \le x,~p\in P} \left(\frac{1}{p} - \frac{1}{p^{1+1/\log x}}\right) = o(\log \log x) $$ and that $$ \sum_{p > x,~p \in P} \frac{1}{p^{1+1/\log x}} = o(\log\log x), $$ so that $\sum_{p \le x,~p \in P} 1/p = (\delta+o(1)) \log \log x$, as $x\to\infty$. (In fact, both of the displayed sums are $O(1)$, not merely $o(\log \log x)$.)


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