Abscissa of Absolute Convergence
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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 $\sigma_{a}$ such that $f(s)$ 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 $f(s)$ converges absolutely.
By hypothesis, there is some $s_0 = \sigma_0 + it_0 \in \C$ such that $f(s_0)$ converges absolutely, so $S$ is not empty.
Moreover, $S$ is bounded below, for otherwise it follows from Dirichlet Series Absolute Convergence Lemma that $f(s)$ converges absolutely for all $s \in \C$, a contradiction of our assumptions.
Therefore the infimum
- $\displaystyle \sigma_a = \inf \left\{ \sigma : s = \sigma + i t \in S \right\} \in \R$
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 $f(s')$ is absolutely convergent.
Then it follows from Dirichlet Series Absolute Convergence Lemma that $f(s)$ is absolutely convergent.
If $s = \sigma + it$ with $\sigma < \sigma_a$, and $f(s)$ is absolutely convergent then $s$ contradicts the definition of $\sigma_a$.
Therefore, $\sigma_a$ has the claimed properties.
$\blacksquare$
Note
It is conventional to set $\sigma_a = -\infty$ if the series $f(s)$ 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
Sources
- Tom M. Apostol: Introduction to Analytic Number Theory (1976): $\S 11.2$: Theorem $11.1$
