Dirichlet Series Convergence Lemma

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Theorem

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

Let $f \left({s}\right)$ converge at $s_0 = \sigma_0 + i t_0$.


Then $f \left({s}\right)$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$.


General Dirichlet Series

Let $\displaystyle f \left({s}\right) = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n \left({s}\right)}$ be a general Dirichlet series.

Let $f \left({s}\right)$ converge at $s_0 = \sigma_0 + i t_0$.


Then $f \left({s}\right)$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$.


Proof

We begin with a lemma.

Lemma

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

Suppose that for some $s_0 = \sigma_0 + i t_0 \in \C$, $f \left({s_0}\right)$ has bounded partial sums:

$(1): \quad \displaystyle \left\vert{\sum_{n \mathop = 1}^N a_n n^{-s_0} }\right\vert \le M$

for some $M \in \R$ and all $N \ge 1$.


Then for every $s = \sigma + i t \in \C$ with $\sigma > \sigma_0$:

$\displaystyle \left\vert{\sum_{n \mathop = m}^N a_n n^{-s} }\right\vert \le 2 M m^{\sigma_0 - \sigma} \left( 1+ \frac {\left\vert s - s_0\right\vert} {\sigma-\sigma_0} \right)$


Proof of Theorem

Suppose that $f \left({s}\right)$ converges at $s_0 = \sigma_0 + it_0$.

Then by Convergent Sequence in Metric Space is Bounded, $\left\vert \displaystyle \sum_{k\mathop=1}^n \frac {a_k} {k^{s_0}} \right\vert$ is bounded. Thus the results of the lemma hold.

Choose any $s = \sigma + it$ with $\sigma > \sigma_0$.

By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon>0$, there exists an $N>0$ such that for all $m>n>N$

$\displaystyle \left\vert \sum_{k \mathop = 1}^m \frac {a_k} {k^s} - \sum_{k \mathop = 1}^n \frac {a_k} {k^s} \right\vert = \left\vert \sum_{k \mathop = n+1}^m \frac {a_k} {k^s} \right\vert < \epsilon$


The lemma shows that for a given $s$ there exists a constant $C$ independent of $N$ such that:

$\displaystyle \left\vert{\sum_{k \mathop = n+1}^n a_n n^{-s} }\right\vert \le C \left({n+1}\right)^{\sigma_0 - \sigma}$

Since $\sigma_0 - \sigma <0$, the right hand side tends to zero as $n \to \infty$.

Thus we may choose $N$ large enough so that for $n>N$

$\displaystyle \left({n+1}\right)^{\sigma_0 - \sigma} < \dfrac \epsilon C$

Which gives us $\displaystyle \left\vert{\sum_{k \mathop = n+1}^n a_n n^{-s} }\right\vert < \epsilon$


$\blacksquare$


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