# Strong form of Dirichlet prime number theorem and the 'smallest' large set

I have no background in number theory, but the statement of the Dirichlet's theorem from Wikipedia is easy enough to understand.

However, I'm confused about so called strong form (or 'stronger form') of this theorem:

Stronger forms of Dirichlet's theorem state that for any such arithmetic progression, the sum of the reciprocals of the prime numbers in the progression diverges and that different such arithmetic progressions with the same modulus have approximately the same proportions of primes.

• Is this 'strong form' proved? If so, please give me the link.

The strong form of Dirichlet's theorem implies that

$${\frac {1}{3}}+{\frac {1}{7}}+{\frac {1}{11}}+{\frac {1}{19}}+{\frac {1}{23}}+{\frac {1}{31}}+{\frac {1}{43}}+{\frac {1}{47}}+{\frac {1}{59}}+{\frac {1}{67}}+\cdots$$ is a divergent series.

This is the sum of reciprocals of the primes in the form $$4n + 3$$.

• The second question (provided that this stronger form of the theorem is true, are the sets of primes of this form the 'smallest' (in terms of asymptotic density) among the large sets (i.e., sets of positive integers whose reciprocals sum to infinity)?

(To clarify, we can take $$4n+3$$ as a sole example, because if the second part of the 'stronger form' is true, all of the primes $$an+b$$ are of the same size, i.e. asymptotic density)

There was a similar question as my first one before, but the answers are not really informative.

• If the 'usual proof' of the Dirichlet's theorem also proves the 'stronger form', then why is it called a 'stronger form'? And is it actually called that, or is it just Wikipedia authors' invention?
• yes, all you need is proving the PNT for $\zeta(s)$, and then adapt if to Dirichlet L-functions $L(s,\chi)$, using that $\sum_{p^k \equiv a \bmod q} p^{-sk} = \frac{1}{\phi(q)} \sum_{\chi \bmod q} \chi(a) \log L(s,\chi)$ (analog to the discrete Fourier transform on the set $\{ 1 \le n < q \ \mid \ gcd(n,q) = 1\}$) Aug 20 '16 at 9:39

It is just too much to include all details in the answer, so the comment and link for the other question detour the questioner toward well-known literature. While I think they are great, I prefer this one:

Montgomery, Vaughan 'Multiplicative Number Theory', chapter 4.

I think 'strong' is a relative term that the wikipedia article used. According to it,

Dirichlet's Theorem is

If $(a,q)=1$, then there are infinitely many primes $p\equiv a \ \mathrm{mod} \ q$.

The strong form of Dirichlet's Theorem is

If $\chi$ is a non-principal character modulo $q$, then $$L(1,\chi)\neq 0.$$

What is referred as Dirichlet's theorem is a corollary of this. Also, there are many other corollaries. Some of them include

If $(a,q)=1$, then $$\sum_{p\leq x, \ p\equiv a \ \mathrm{mod} \ q} \frac{\log p}p = \frac1{\phi(q)} \log x + O_q(1),$$ and $$\sum_{p\leq x, \ p\equiv a \ \mathrm{mod} \ q} \frac1p = \frac1{\phi(q)} \log\log x + b(a,q) + O_q\left(\frac1{\log x}\right).$$

Your question about approximately the same proportion is known as Siegel-Walfisz theorem. This is for $(a,q)=1$, the number of primes $p\leq x$ with $p\equiv a \ \mathrm{mod} \ q$, denoted by $\pi(x;q,a)$ satisfies:

Given any positive real number $N$, there exists $c_N>0$ such that $$\pi(x;q,a)=\frac{\mathrm{Li}(x)}{\phi(q)} + O\left(x\exp(-c_N \sqrt{\log x})\right).$$ This result suggests that primes in different arithmetic progression with the same moduli, have the same density $1/\phi(q)$.

Euclid's result that there are infinitely many primes also has a "strong" form, namely that $$\sum_{p\le x}\frac{1}{p}=\log (\log(x))+O(1),$$ or also $$\sum_{p\le x}\frac{\log(p)}{p}=\log(x)+O(1).$$ However, here the proof of Euclid's result is significantly simpler than the proof for the stronger form.

For Dirichlet's result, that there are infinitely many primes in every arithmetic progression $ax+b$ with coprime $a$ and $b$, the proof is almost as difficult as for the strong form, $$\sum_{p\le x, p\equiv b \bmod a}\frac{\log(p)}{p}=\frac{1}{\phi(a)}\log(x)+O(1).$$ In this sense, the strong form is really just the standard result for "Dirichlet's theorem".