Existence of Abscissa of Convergence

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

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

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

We call $\sigma_c$ the abscissa of convergence of the Dirichlet series.

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

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

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

Therefore the infimum:


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

is well defined.

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

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

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

Therefore, $\sigma_c$ has the claimed properties.

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

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

Also see

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