Section 24.3 From Riemann to Dirichlet and Euler
\begin{equation*}
\sum_{d\mid n}\frac{\mu(d)}{d}=\prod_{p\mid n}\left(1-\frac{1}{p}\right)
\end{equation*}
xxxxxxxxxxdef _(e=(1,[0..3]),f=(2,[0..3]),g=(0,[0..3])): n = 2^e*3^f*5^g pretty_print(html("You picked $%s=2^{%s}3^{%s}5^{%s}$"%(n,e,f,g))) str = '$$'+'+'.join([r'\frac{%s}{%s}'%(moebius(d),d) for d in divisors(n)])+'=%s$$'%sum([moebius(d)/d for d in divisors(n)]) str2 = '$$'+''.join([r'\left(1-\frac{1}{%s}\right)'%p for (p,e) in factor(n)])+'=%s$$'%prod([1-1/p for (p,e) in factor(n)]) pretty_print(html(str)) pretty_print(html("compare to "+str2))
\begin{equation*}
\sum_{n=1}^\infty \frac{\mu(n)}{n^s}\text{.}
\end{equation*}
Subsection 24.3.1 Dirichlet series
We give such series a name. The following definition is purely formal, considered without considering issues such as convergence. (See [E.2.8, Chapter 4.6] for an interesting formal viewpoint on the set of these series.)Definition 24.3.1. Dirichlet Series.
In general, for an arithmetic function f(n)\text{,} its Dirichlet series is
\begin{equation*}
F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}\text{.}
\end{equation*}
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\text{?}
Subsection 24.3.2 Euler products
For our purposes, the very important thing to note about such series is that they often can be expanded as infinite products.Definition 24.3.2. Euler Products.
In general, for an arithmetic function f(n)\text{,} its Dirichlet series is said to have an Euler product if the series can be written as an infinite product in the following manner:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_p (\text{ a formula involving }f(p)\text{ and }p^s)\text{.}
\end{equation*}
Example 24.3.3. Euler product for Riemann zeta function.
We have already suggested one for the zeta function:
\begin{equation*}
\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=\prod_p \left(\frac{1}{1-p^{-s}}\right)\text{.}
\end{equation*}
\begin{equation*}
\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_{p}\left(1-\frac{1}{p^s}\right)=\prod_p (1-p^{-s})
\end{equation*}
At least, we can consider this wherever it makes sense. See [E.4.6, Chapter 11.5] or [E.2.1, Chapter 9.8] for some criteria, or simply below at Theorem 24.5.4.
In the next section, we justify more of this discussion, and connect our wonderful results about Dirichlet products of finite arithmetic functions to deep properties of their Dirichlet series.