# Dirichlet Series Absolute Convergence Lemma

## Theorem

Let $\displaystyle \map f s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ 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 $\sigma \ge \sigma_0$.

## Proof

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

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

 $\ds \size {\frac {a_n} {n^s} }$ $=$ $\ds \frac {\size {a_n} } {n^\sigma}$ $\ds$ $\le$ $\ds \frac {\size {a_n} } {n^{\sigma_0} }$ $\ds$ $=$ $\ds \size {\frac {a_n} { n^{s_0} } }$

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

$\blacksquare$