Does the sum of reciprocals of primes congruent to $1 \mod{4}$ diverge?

Question

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

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

Answer 1

For any coprime $k,m$ the sum of the reciprocals of the primes $\equiv k\pmod m$ diverges.

Answer 2

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. $\diamondsuit$