Sums of reciprocals of prime numbers: $p \equiv a \!\! \mod m$ vs. $p \equiv b \!\! \mod m$ 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
$c_{a(m),b(m)} := \lim_{n \rightarrow \infty} f_{a(m),b(m)}(n)$?
 A: There is a detailed survey
Comparative prime number theory: A survey
Greg Martin, Justin Scarfy.
It contains numerous results on this "comparative prime number theory", 
which goes back to Chebyshev, Knapowski and Turan, and others.
In this area there are several results which indicate that primes in "quadratic residue classes" occur (in some precise technical sense) less often than primes in quadratic non-residue classes.
For example, for your question modulo 3, Theorem 5.6 is of interest,
which answers your question for a different weight.
$\lim_{x \rightarrow \infty} \sum_p \chi_3(p) \frac{\log p}{p^{1/2}} 
\exp(-(\log^2 p)/x)=-\infty$. (Here $\chi_3(p)=1$, if $p=1 \bmod 3$, and $\chi_3(p)=-1$ if $p=2 \bmod 3$,
A: 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).
