Analytic Number Theory/Dirichlet series

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

Convergence considerations
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 $$\sigma_0 + i t_0 \notin S$$, then for all $$\sigma > \sigma_0$$ and all $$t$$ $$\sigma + i t \notin S$$, 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.