Analytic Number Theory/Dirichlet series

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

Convergence considerations

Proof:

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

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

diverges. Due to the assumption, this set is neither empty nor equal to $ℂ$. Further, if ${\sigma }_0+i{t}_0\notin S$, then for all $\sigma >{\sigma }_0$ and all $t$ $\sigma +it\notin S$, since

$\left|\frac{f(n)}{n^{s_0}}\right|=\frac{|f(n)|}{n^{{\sigma }_0}}\ge \frac{|f(n)|}{n^{\sigma }}=\left|\frac{f(n)}{n^s}\right|$

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 ${\sigma }_a$.$◻$

Formulas

Proof:

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