Given positive integers $a$, $m$ and $n$, let $s\_{a(m)}(n)$ denote the
sum of the reciprocals of the prime numbers less than or equal to $n$
which are congruent to $a$ modulo $m$.


Is there an integer $n$ such that $s\_{1(3)}(n) > s\_{2(3)}(n)$?


For small $n$, the function $s\_{2(3)}$ is clearly ahead --
for example it exceeds $1$ already at $n = 59$, while for
$s\_{1(3)}$ this takes until $n = 3560503$.


Even if the answer is *no*, are there still other
residue classes $a(m)$ and $b(m)$ such that the function
$$
 f\_{a(m),b(m)}: \mathbb{N} \rightarrow \mathbb{R}, \ \
 n \mapsto s\_{a(m)}(n) - s\_{b(m)}(n)
$$
has infinitely many sign changes?


Does $f\_{a(m),b(m)}(n)$ converge for $n \rightarrow \infty$,
and if so, is there anything known about the constants
$c\_{a(m),b(m)} := \lim\_{n \rightarrow \infty} f\_{a(m),b(m)}(n)$?



Sums of reciprocals of prime numbers: $p \equiv a \!\! \mod m$ vs. $p \equiv b \!\! \mod m$
Note that 
$$
s\_{1(3)}(n)-s\_{2(3)}(n) = \sum\_{p\le n} \frac{\chi\_{-3}(p)}{p} 
$$ 
where $\chi\_{-3}$ is the real Dirichlet character $\pmod 3$ (ie the Legendre symbol). 
This sum converges (as in the proof of Dirichlet's theorem). So if it starts out being negative for a long while, it will continue to be negative. The limiting value is 
$$ 
\log L(1,\chi\_{-3}) - \sum\_{k=2}^{\infty} \sum\_{p} \frac{\chi\_{-3}(p^k)}{kp^k}.
$$ 
Note that $L(1,\chi\_{-3}) = \pi/(3\sqrt{3})$ which is less than $1$ and the second term also makes a negative contribution.


Similarly for any other two residue classes. The function $f$ eventually will have constant sign (converging to a particular value indeed).


