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Let $P$ be the set of primes $p$ greater than $3$ such that $p\equiv1 \pmod{4}$. Does the following sum converge or diverge?

$$ \sum_{p\in P}\frac{1}{p} $$

Batominovski
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chuck
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2 Answers2

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This sum diverges (very, very, very slowly). In fact, so does the sum of the reciprocals of primes congruent to $3 \bmod 4$. Here's something amazing: asymptotically, these two sums diverge at the exact same rate.

We prove this using two pieces of information:

  1. Integration by parts (with respect to Riemann-Stieltjes integration), and
  2. An analytic statement of Dirichlet's theorem on primes in arithmetic progressions. In particular, if $\pi_{a,b}(x)$ denotes the number of primes up to $X$ that are congruent to $a \bmod b$, and if $\gcd(a,b) = 1$, then $$ \pi_{a,b}(X) = \frac{1}{\varphi(b)} \frac{X}{\log X} + O(X e^{-c \sqrt{\log X}})$$ for some absolute but unimportant constant $c > 0$.

Consider the sum $$\begin{align}\sum_{\substack{p \equiv a \bmod b \\ p \leq X}} \frac{1}{p} &= \int_1^X \frac{1}{t} d\lfloor \pi_{a,b}(t) \rfloor \\ &= \frac{\pi_{a,b}(t)}{t} \bigg|_1^X + \int_1^X \frac{\pi_{a,b}(t)}{t^2} dt \\ &= \frac{1}{\varphi(b)} \frac{1}{\log X} + \int_1^X \frac{1}{\varphi(b)}\frac{1}{t \log t} dt + O \left( \int_1^X \frac{e^{-c \sqrt{\log t}}}{t}dt \right) \\ &= \frac{1}{\varphi(b)} \log \log X + O(1), \end{align}$$ where I've used that the integral and first term converge as $X \to \infty$ to some constant.

So the reciprocals of primes do diverge, even when restricted to congruence classes. And, remarkably, they diverge at the same rate asymptotically in any permissible congruence class. $\diamondsuit$

davidlowryduda
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  • Nitpick: Fact #2 would more accurately be called the prime number theorem for arithmetic progressions. The analytic statement of Dirichlet's theorem is about the rate of divergence of $\sum_{p \equiv a \pmod b} p^{-s}$ as $s \to 1^+$, which also is enough to establish that the reciprocal sum diverges (and predates PNT). – Erick Wong Nov 06 '15 at 16:14
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For any coprime $k,m$ the sum of the reciprocals of the primes $\equiv k\pmod m$ diverges.

Charles
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    I dare you to prove that. – Race Bannon Aug 17 '15 at 18:09
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    @RaceBannon: It's easy given the Prime Number Theorem in arithmetic progressions -- or did you just mean that the PNT is hard to prove? – Charles Aug 17 '15 at 18:26
  • No, I was talking about the claim you posted. I would like to see a proof of your claim. – Race Bannon Aug 17 '15 at 18:28
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    @RaceBannon: As I said, it's pretty easy to prove given the PNT in APs. This even gives the correct asymptotic -- $$\sum_{p\le x,\ p\equiv k\pmod m}\frac1p\approx\frac{\log\log x}{\varphi(m)}$$. – Charles Aug 17 '15 at 18:33
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    @RaceBannon, what does "dare" mean in this context? What sort of game is it? Anyway, this is standard. – paul garrett Aug 17 '15 at 23:36