# Dirichlet Series Convergence Lemma

## Theorem

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

Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.

Then $\map f s$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$.

### General Dirichlet Series

Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\map {\lambda_n} s}$ be a general Dirichlet series.

Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.

Then $\map f s$ 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 $\sigma > \sigma_0$:

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

$\Box$

Suppose that $\map f s$ converges at $s_0 = \sigma_0 + it_0$.

Then by Convergent Sequence in Metric Space is Bounded, $\size {\ds \sum_{k \mathop = 1}^n \frac {a_k} {k^{s_0} } }$ 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 $m>n>N$

- $\ds \size {\sum_{k \mathop = 1}^m \frac {a_k} {k^s} - \sum_{k \mathop = 1}^n \frac {a_k} {k^s} } = \size {\sum_{k \mathop = n + 1}^m \frac {a_k} {k^s} } < \epsilon$

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

- $\ds \size {\sum_{k \mathop = n + 1}^n a_n n^{-s} } \le \map C {n + 1}^{\sigma_0 - \sigma}$

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 $n > N$.

- $\ds \paren {n + 1}^{\sigma_0 - \sigma} < \dfrac \epsilon C$

which gives us:

- $\ds \size {\sum_{k \mathop = n + 1}^n a_n n^{-s} } < \epsilon$

$\blacksquare$

## Sources

- 1976: Tom M. Apostol:
*Introduction to Analytic Number Theory*: $\S 11.6$: Lemma $2$, Theorem $11.8$