Existence of Abscissa of Absolute Convergence

Theorem
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a Dirichlet series.

Let the series $\ds \sum_{n \mathop = 1}^\infty \size { a_n n^{-s} }$ not converge for all $s \in \C$, or diverge for all $s \in \C$.

Then there exists a real number $\sigma_a$ such that $\map f s$ converges absolutely for all $s = \sigma + it$ with $\sigma > \sigma_a$, and does not converge absolutely for all $s$ with $\sigma < \sigma_a$.

We call $\sigma_a$ the abscissa of absolute convergence of the Dirichlet series.

Proof
Let $S$ be the set of all complex numbers $s$ such that $\map f s$ converges absolutely.

By hypothesis, there is some $s_0 = \sigma_0 + it_0 \in \C$ such that $\map f {s_0}$ converges absolutely, so $S$ is not empty.

Moreover, $S$ is bounded below, for otherwise it follows from Dirichlet Series Absolute Convergence Lemma that $\map f s$ converges absolutely for all $s \in \C$, a contradiction of our assumptions.

Therefore the infimum:


 * $\sigma_a = \inf \set {\sigma: s = \sigma + i t \in S} \in \R$

is well defined.

Now if $s = \sigma + it$ with $\sigma > \sigma_a$, then there is $s' = \sigma' + i t' \in S$ with $\sigma' < \sigma$, and $\map f {s'}$ is absolutely convergent.

Then it follows from Dirichlet Series Absolute Convergence Lemma that $\map f s$ is absolutely convergent.

If $s = \sigma + it$ with $\sigma < \sigma_a$, and $\map f s$ is absolutely convergent then $s$ contradicts the definition of $\sigma_a$.

Therefore, $\sigma_a$ has the claimed properties.

Note
It is conventional to set $\sigma_a = -\infty$ if the series $\map f s$ is absolutely convergent for all $s \in \C$, and $\sigma_a = \infty$ if the series converges absolutely for no $s \in \C$.

Therefore, allowing $\sigma_a$ to be an extended real number, $\sigma_a$ is defined for all Dirichlet series.

Also see

 * Abscissa of Convergence
 * Definition:Dirichlet Series
 * Dirichlet Series Absolute Convergence Lemma