Sum of (only certain) prime reciprocicals 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.
 A: 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 $2$.
A: 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 $(2)$.
