Analytic Number Theory/Dirichlet series

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

$s=\sigma +it$

Definition

Definition 5.1:

Let $f$ be an arithmetic function. Then the Dirichlet series associated to $f$ is the series

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

where $s$ ranges over the complex numbers.

Convergence considerations

Theorem 5.2 (abscissa of absolute convergence):

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

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

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

Proof:

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

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

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

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

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

Theorem 5.3 (abscissa of conditional convergence):

Formulas

Theorem 8.4 (Euler product):

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

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

Proof:

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