# Analytic Number Theory/Dirichlet series

For the remainder of this book, we shall use Riemann's convention of denoting complex numbers:

${\displaystyle s=\sigma +it}$

## Definition

Definition 5.1:

Let ${\displaystyle f}$ be an arithmetic function. Then the Dirichlet series associated to ${\displaystyle f}$ is the series

${\displaystyle \sum _{n=1}^{\infty }{\frac {f(n)}{n^{s}}}}$,

where ${\displaystyle s}$ ranges over the complex numbers.

## Convergence considerations

Theorem 5.2 (abscissa of absolute convergence):

Let ${\displaystyle f}$ be an arithmetic function such that the series of absolute values associated to the Dirichlet series associated to ${\displaystyle f}$

${\displaystyle \sum _{n=1}^{\infty }\left|{\frac {f(n)}{n^{s}}}\right|}$

neither diverges at all ${\displaystyle s\in \mathbb {C} }$ nor converges for all ${\displaystyle s\in \mathbb {C} }$. Then there exists ${\displaystyle \sigma _{a}\in \mathbb {R} }$, called the abscissa of absolute convergence, such that the Dirichlet series associated to ${\displaystyle f}$ converges absolutely for all ${\displaystyle \sigma +it}$, ${\displaystyle \sigma >\sigma _{a}}$ and it's associated series of absolute values diverges for all ${\displaystyle \sigma +it}$, ${\displaystyle \sigma <\sigma _{a}}$.

Proof:

Denote by ${\displaystyle S}$ the set of all real numbers ${\displaystyle \sigma }$ such that

${\displaystyle \sum _{n=1}^{\infty }\left|{\frac {f(n)}{n^{s}}}\right|}$

diverges. Due to the assumption, this set is neither empty nor equal to ${\displaystyle \mathbb {C} }$. Further, if ${\displaystyle \sigma _{0}+it_{0}\notin S}$, then for all ${\displaystyle \sigma >\sigma _{0}}$ and all ${\displaystyle t}$ ${\displaystyle \sigma +it\notin S}$, since

${\displaystyle \left|{\frac {f(n)}{n^{s_{0}}}}\right|={\frac {|f(n)|}{n^{\sigma _{0}}}}\geq {\frac {|f(n)|}{n^{\sigma }}}=\left|{\frac {f(n)}{n^{s}}}\right|}$

and due to the comparison test. It follows that ${\displaystyle S}$ has a supremum. Let ${\displaystyle \sigma _{a}}$ be that supremum. By definition, for ${\displaystyle \sigma >\sigma _{a}}$ we have convergence, and if we had convergence for ${\displaystyle \sigma <\sigma _{a}}$ we would have found a lower upper bound due to the above argument, contradicting the definition of ${\displaystyle \sigma _{a}}$.${\displaystyle \Box }$

Theorem 5.3 (abscissa of conditional convergence):

## Formulas

Theorem 8.4 (Euler product):

Let ${\displaystyle f}$ be a strongly multiplicative function, and let ${\displaystyle s\in \mathbb {C} }$ such that the corresponding Dirichlet series converges absolutely. Then for that series we have the formula

${\displaystyle \sum _{n=1}^{\infty }{\frac {f(n)}{n^{s}}}=\prod _{p{\text{ prime}}}{\frac {1}{1-{\frac {f(p)}{p^{s}}}}}}$.

Proof:

This follows directly from theorem 2.11 and the fact that ${\displaystyle f}$ strongly multiplicative ${\displaystyle \Rightarrow }$ ${\displaystyle {\frac {f(n)}{n^{s}}}}$ strongly multiplicative.${\displaystyle \Box }$