Suppose we start with $k$ primes $p_1,p_2,\ldots ,p_k$ (not necessarily consecutive) and a residue class for each prime $r_1,r_2,\ldots ,r_k$.

We denote the least integer not covered by the arithmetic progressions $r_i+m\cdot p_i$ as $r_{k+1}$ which is going to be the new residue class for a (random) prime $p_{k+1}$.

We proceed in this way "covering" the natural numbers (without changing the $r_i$'s.)

**Question**:Is it true that $\sum\limits_{n=1}^\infty\frac{1}{r_n}=+\infty$?

**Motivation**:

This would directly imply Dirichlet's theorem on primes in arithmetic progressions if we make use of this

**Lemma**:

Let $a_n$ be a sequence of natural numbers, strictly increasing with $\gcd(a_i,a_j)=1$.

Then if $\sum\limits_{n=1}^\infty\frac{1}{a_n}=+\infty$ then the sequence contains infinitely many prime numbers.

I tried to modify some proofs which show the divergence of $\sum\frac{1}{p}$ but without much success.

Thank you very much in advance!

**EDIT:** The primes are distinct and the residue classes are not reduced modulo $p$.Suppose we start with the primes $2,3,7$ and their residue classes are $(1)_2 , (2)_3 , (4)_7$ which means $r_1=1$, $r_2=2$ and $r_3=4$. The least number not covered by the progressions $1+2k , 2+3k , 4+7k$ is $6$.We define then $r_4=6$ and $6$ is going to be a new residue class for a new prime (random choice) let's say $p_4=17$.Then $r_5=10$ and we choose a new prime (Let's say $p_5=5$) and continue in this direction.

The $r_i$'s could be much greater than the $p_i$'s

12more comments