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 $\sigma > \sigma_0$.

Proof
We begin with a lemma:

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