Difference of Abscissae of Convergence

Theorem

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

Suppose that $f(s)$ has finite Abscissa of Convergence $\sigma_c$.

Then the Abscissa of Absolute Convergence $\sigma_a$ is finite, and

$0 \leq \sigma_a - \sigma_c \leq 1$

Proof

It is trivial that $\sigma_a \geq \sigma_c$.

Suppose $s_0 = \sigma_0 + i t_0 \in \C$ such that $f(s_0)$ converges.

It is sufficient to show that $f(s)$ converges absolutely for all $s = \sigma + it$ with $\sigma > \sigma_0 + 1$.

Pick an upper bound $M$ for the real numbers $\left| a_n n^{-s_0} \right|$.

Then for $s = \sigma + it$ with $\sigma > \sigma_0 + 1$:

$\displaystyle \left| \frac{a_n}{n^s} \right| = \left| \frac{a_n}{n^{s_0} n^{s-s_0}} \right| = \frac M {n^{\sigma-\sigma_0}}$

Then by the Comparison Test, $\displaystyle \sum_{n=1}^\infty \left| a_n n^{-s} \right|$ converges.

$\blacksquare$

Sources

• Tom M. Apostol: Introduction to Analytic Number Theory (1976): $\S 11.6$: Theorem $11.10$