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.

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.