Abscissa of Convergence

Theorem

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

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

Then there exists a real number $\sigma_{c}$ such that $f(s)$ 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 $f(s)$ converges.

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

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

Therefore the infimum

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

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 $f(s')$ is convergent.

Then it follows from Dirichlet Series Convergence Lemma that $f(s)$ is convergent.

If $s = \sigma + it$ with $\sigma < \sigma_c$, and $f(s)$ 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 $f(s)$ 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
• Dirichlet Series
• Dirichlet Series Convergence Lemma

Sources

• Tom M. Apostol: Introduction to Analytic Number Theory (1976): $\S 11.6$: Theorem $11.9$