Section24.3From Riemann to Dirichlet and Euler
In order to see this (the convergence of the infinite product), let's instead show that our other main example of a sum over divisors equalling a product over primes works.
This is the same setup, but for the sum/product combination \[\sum_{d\mid n}\frac{\mu(d)}{d}=\prod_{p\mid n}\left(1-\frac{1}{p}\right)\, .\] It certainly seems to work. This would lead to the series \[\sum_{n=1}^\infty \frac{\mu(n)}{n^s}\; .\]
Subsection24.3.1Dirichlet series
We call a series like this a Dirichlet series. This is defined formally, irrespective of things like convergence.
Answer the following three questions to see if you understand this definition.
- For what arithmetic function is the Riemann zeta function the Dirichlet series?
- What would the Dirichlet series of \(N\) be?
- What about the Dirichlet series of \(I\)?
Subsection24.3.2Euler Products
For our purposes, the very important thing to note about such series is that they often can indeed be expanded as infinite products as well. In a very general way, it looks like \[\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_p (\text{ something involving }f(p)\text{ and }p^s)\] If this is possible, then we say that the series has an Euler product form.
We have already suggested one for the zeta function: \[\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=\prod_p \left(\frac{1}{1-p^{-s}}\right)\]
But another potential new Euler product, then, is for the Dirichlet series of the Moebius function: \[\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_{p}\left(1-\frac{1}{p^s}\right)=\prod_p (1-p^{-s})\] At least, we can consider this wherever it makes sense.
In the next section, we justify more of this, and connect our wonderful results about Dirichlet products of finite arithmetic functions to deep properties of their Dirichlet series.