Dirichlet Series Absolute Convergence Lemma

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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:

\(\displaystyle \size {\frac {a_n} {n^s} }\) \(=\) \(\displaystyle \frac {\size {a_n} } {n^\sigma}\)
\(\displaystyle \) \(\le\) \(\displaystyle \frac {\size {a_n} } {n^{\sigma_0} }\)
\(\displaystyle \) \(=\) \(\displaystyle \size {\frac {a_n} { n^{s_0} } }\)

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

$\blacksquare$


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