Dirichlet Series Convergence Lemma/Lemma

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

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


 * $(1): \quad \ds \size {\sum_{n \mathop = 1}^N a_n n^{-s_0} } \le M$

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

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


 * $\ds \size {\sum_{n \mathop = m}^N a_n n^{-s} } \le 2 M m^{\sigma_0 - \sigma} \paren {1 + \frac {\size {s - s_0} } {\sigma - \sigma_0} }$

Proof
From the Summation by Parts formula:


 * $\ds \sum_{n \mathop = m}^N f_n g_n = f_N G_N - f_m G_{m-1} - \sum_{n \mathop = m}^{N - 1} \map {G_n} {f_{n + 1} - f_n}$

Let:
 * $g_n = a_n n^{-s_0}$
 * $f_n = n^{s_0 - s}$

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 $N \ge m$ and $\sigma_0 - \sigma < 0$, we have:


 * $N^{\sigma_0 - \sigma} +  m^{\sigma_0 - \sigma} \le 2 m^{\sigma_0 - \sigma}$

Therefore:

Hence the result.