Sum of (only certain) prime reciprocicals

Question

It is well known that

$$ \sum_{p\ is\ prime}\frac1{p}$$

diverges.

Is there a simple proof that

$$ \sum_{p\equiv 1\pmod 4}\frac1{p}$$

and

$$ \sum_{p\equiv 3\pmod 4}\frac1{p}$$

also diverge?

(p denotes prime numbers in both expressions above)

What I found so far are fairly complex considerations related to some more general issues, this must be simpler than that.

I think that maybe the formula $\pi_{n,a}(x)\sim \dfrac{1}{\varphi(n)}\text{Li}(x)$ will be a good place to start. For details see here. — , Sep 05 '15 at 12:22
(stronger forms of) Dirichlet's theorem on arithmetic progressions may help too (the primes are asymptotically evenly distributed between $1$ and $3$). — Raymond Manzoni, Sep 05 '15 at 12:40

score 3 · Accepted Answer · answered Sep 05 '15 at 12:20

There is an easy proof, see Proposition $5$ on page $4$ in the paper Prime reciprocals and primes in arithmetic progressions by Daniel Litt. The proof makes use of the ring of Gaussian integers $\mathbb{Z}[i]$ and its Euclidean norm, i.e., that $p=x^2+y^2$ for primes $p\equiv 1 \bmod 4$ The analytic part of the proof is quite simple. The idea can be also adapted to the case $p\equiv 3\bmod 4$, replacing the ring of Gaussian integers with the ring of Eisenstein integers, see remark $2$.

score 2 · Answer 2 · answered Sep 05 '15 at 12:28

You just need to prove that $$\sum_{p>2}\frac{\chi(p)}{p}\tag{1}$$ is a convergent series to prove that both your series are divergent. Here $\chi(n)$ equals zero if $n$ is even, $1$ if $n=4k+1$ and $-1$ if $n=4k-1$. $\chi$ is a multiplicative function, and you may use the following consequence of the Euler's product $$ \prod_{p>2}\left(1-\frac{\chi(p)}{p}\right)^{-1}=\sum_{n\equiv 1\!\!\pmod{2}}\frac{\chi(n)}{n}=\sum_{m\geq 0}\frac{(-1)^m}{2m+1}=\arctan(1)=\frac{\pi}{4} \tag{2}$$ to prove that $(1)$ is converging. Notice that $(1)$ is essentially the logarithm of the LHS of $(2)$.

Sum of (only certain) prime reciprocicals

2 Answers2