Dirichlet's Theorem

§1The Theorem

The statement:

More specifically, we can say the sum of reciprocals of such primes diverges.

Even more specifically, the following statement holds, where $S$ is the set of primes of the form $an+b$: \[\lim_{s\to1^+}\frac{\sum_{p\in S}p^{-s}}{-\log(s-1)}=\frac{1}{\phi(a)},\] where $\phi$ is the Euler Totient Function.

The expression on the left is called the Dirichlet density of $S$.

§2Euler Products

The harmonic series can be "factored" as follows: \[\sum_{n=1}^{\infty}\frac1n=\left(1+\frac12+\frac14+\frac18+\cdots\right)\left(1+\frac13+\frac19+\cdots\right)\left(1+\frac15+\cdots\right)\left(1+\frac17+\cdots\right)\cdots\]

The left side is well-known to be infinite, and applying the geometric series formula to each factor in the right side yields \[\infty=\prod_{p\text{ prime}}\frac1{1-\frac1p}=\prod_{p\text{ prime}}\frac p{p-1}.\]

This is sufficient to prove that there are infinitely many primes. Taking the logarithm of the product, we can see: \[\infty = \sum_{p\text{ prime}}\log\left(\frac{p}{p-1}\right).\]

Then, using the fact that $\log(1+x)\leq x$, we see that \[\log\left(\frac{p}{p-1}\right) = \log\left(1+\frac{1}{p-1}\right)<\frac{1}{p-1}\leq\frac2p,\] showing that the sum of reciprocals of primes diverges.

There are many other sums which are useful to factor similarly.

Then, for every multiplicative function $f$, the following is a formal factorization for the infinite series of $f(n)$: \[\sum_{n\in\N}f(n)=\prod_{p\text{ prime}}(1+f(p)+f(p^2)+\cdots).\]

However, we must be very careful not to say the two sides are equal, as the infinite product rearranges the order in which terms are being added. When $f$ is always positive or when $\sum_{n\in\N}f(n)$ converges absolutely, we have equality.

§3A try for $3n+1$ and $3n+2$

Let $A$ be the set of primes of the form $3n+1$, and $B$ the set of primes of the form $3n+2$. Consider \[\sum_{n=0}^{\infty}\frac1{3n+1}-\frac1{3n+2}=\frac11-\frac12+\frac14-\frac15+\frac17-\frac18+\cdots\]

Note that unlike before, this sum is finite! This one factors as \[\left(\prod_{p\in A}\frac p{p-1}\right)\left(\prod_{p\in B}\frac p{p+1}\right),\] and taking logarithms would seem to leave us with $\sum_{n\in A}\frac1n-\sum_{n\in B}\frac1n$. From there, using the fact that the sum is infinite but the difference is finite, we could conclude…