# Dirichlet Series Absolute Convergence Lemma

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## Theorem

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

Suppose that $f$ converges absolutely at $s_0 = \sigma_0 + i t_0 \in \C$.

Then $f$ converges absolutely at all points $s = \sigma + i t \in \C$ with $\sigma \geq \sigma_0$.

## Proof

Suppose that $f$ converges absolutely at $\sigma_0 + i t_0$.

If $\sigma \ge \sigma_0$, then:

\(\displaystyle \left\vert{\frac{a_n} {n^s} }\right\vert\) | \(=\) | \(\displaystyle \frac {\left\vert{a_n}\right\vert} {n^\sigma}\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \frac {\left\vert{a_n}\right\vert} {n^{\sigma_0} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left\vert{\frac{a_n} { n^{s_0} } }\right\vert\) |

Therefore absolute convergence of $f(s_0)$ directly implies absolute convergence of $f(s)$ for all $s = \sigma + i t$ with $\sigma > \sigma_0$.

$\blacksquare$

## Sources

- 1976: Tom M. Apostol:
*Introduction to Analytic Number Theory*: $\S 11.2$