# Existence of Abscissa of Convergence

## Contents

## Theorem

Let $\displaystyle f \left({s}\right) = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a Dirichlet series.

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

Then there exists a real number $\sigma_c$ such that $f \left({s}\right)$ 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.

### General

Let $s = \sigma + i t$

Let $\displaystyle f \left({s}\right) = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_ns}$ be a general Dirichlet series.

Then there exists a extended real number, $\sigma_0$, such that

- $(1): \quad$ For $\sigma<\sigma_0$, $\displaystyle f \left({s}\right)$ diverges
- $(2): \quad$ For $\sigma>\sigma_0$, $\displaystyle f \left({s}\right)$ converges.

## Proof

Let $S$ be the set of all complex numbers $s$ such that $f(s)$ converges.

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

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

Therefore the infimum:

- $\sigma_c = \inf \left\{{\sigma: s = \sigma + i t \in S}\right\} \in \R$

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 $f \left({s'}\right)$ is convergent.

Then it follows from Dirichlet Series Convergence Lemma that $f \left({s}\right)$ is convergent.

If $s = \sigma + it$ with $\sigma < \sigma_c$, and $f \left({s}\right)$ is convergent then $s$ contradicts the definition of $\sigma_c$.

Therefore, $\sigma_c$ has the claimed properties.

$\blacksquare$

## Note

It is conventional to set $\sigma_c = -\infty$ if the series $f \left({s}\right)$ 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

## Sources

- 1976: Tom M. Apostol:
*Introduction to Analytic Number Theory*: $\S 11.6$: Theorem $11.9$