Dirichlet Series Absolute Convergence Lemma

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Theorem

Let $\displaystyle f(s)= \sum_{n=1}^\infty \frac{a_n}{n^s}$ 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 $\sigma \geq \sigma_0$.

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

If $\sigma \geq \sigma_0$, then

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \left\vert { \frac{a_n}{ n^s } } \right\vert$	$=$	$\displaystyle \frac{ \left\vert { a_n } \right\vert }{ n^\sigma }$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\leq$	$\displaystyle \frac{ \left\vert { a_n } \right\vert }{ n^{\sigma_0} }$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \left\vert { \frac{a_n}{ n^{s_0} } } \right\vert$	$\displaystyle $	$\displaystyle $	$\displaystyle $

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

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert { \frac{a_n}{ n^s } } \right\vert\)	\(=\)	\(\displaystyle \frac{ \left\vert { a_n } \right\vert }{ n^\sigma }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \frac{ \left\vert { a_n } \right\vert }{ n^{\sigma_0} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\vert { \frac{a_n}{ n^{s_0} } } \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)