Dirichlet Series Convergence Lemma

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 $\sigma > \sigma_0$.

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 $\sigma > \sigma_0$:


 * $\displaystyle \left\vert{\sum_{n \mathop = m}^N a_n n^{-s} }\right\vert \le 4 M m^{\sigma_0 - \sigma}$

Proof of Lemma
We have the Summation by Parts formula:


 * $\displaystyle \sum_{n \mathop = m}^N f_n g_n = f_N G_N - f_n G_{n-1} - \sum_{n \mathop = m}^{N - 1} G_n \left({f_{n+1} - f_n}\right)$

We let $g_n = a_n n^{-s_0}$ and $f_n = n^{s_0 - s}$.

For $N \ge 1$, the quantities $G_N$ are the partial sums $(1)$

Thus $G_N \le M$ for all $N \ge 1$.

We have:

Finally, because $\frac N m > 1$ and $\sigma_0 - \sigma < 0$, we can estimate:


 * $\left\vert{\left({\dfrac N m}\right)^{s_0 - s} - 1}\right\vert \le \left({\dfrac N m}\right)^{\sigma_0 - \sigma} + 1 \le 2$

Therefore:


 * $\displaystyle \left\vert{\sum_{n \mathop = m}^N \frac {a_n} {n^s} }\right\vert \le 4 M m^{\sigma_0 - \sigma}$

Hence the result.

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

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

The lemma shows that for a constant $C$ independent of $m$:


 * $\displaystyle \left\vert{\sum_{n \mathop = m}^N a_n n^{-s} }\right\vert \le C m^{\sigma_0 - \sigma}$

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

Therefore, it follows from Cauchy's Convergence Criterion that the sum converges.