Dirichlet Series Convergence Lemma/Lemma

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Lemma to Dirichlet Series Convergence 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 $\sigma > \sigma_0$:

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


Proof

We have the Summation by Parts formula:

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

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:

\(\displaystyle \left\vert{\sum_{n \mathop = m}^N \frac {a_n} {n^s} }\right\vert\) \(=\) \(\displaystyle \left\vert{\sum_{n \mathop = m}^N f_n g_n}\right\vert\)
\(\displaystyle \) \(\le\) \(\displaystyle \left\vert{f_N G_N}\right\vert + \left\vert{f_m G_{m-1} }\right\vert + \sum_{n \mathop = m}^{N - 1} \left\vert{G_n \left({f_{n+1} - f_n}\right)}\right\vert\) using partial summation and the Triangle Inequality
\(\displaystyle \) \(\le\) \(\displaystyle M \left\vert{N^{s_0 - s} }\right\vert + M \left\vert{m^{s_0 - s} }\right\vert + M \sum_{n \mathop = m}^{N - 1} \left\vert{\left({\left({n + 1}\right)^{s_0 - s} - n^{s_0 - s} }\right) }\right\vert\) using the given bound on the partial sums
\(\displaystyle \) \(=\) \(\displaystyle M N^{\sigma_0 - \sigma} + M m^{\sigma_0 - \sigma} + M \sum_{n \mathop = m}^{N - 1} \left\vert{ \left(s - s_0 \right) \int_n^{n+1} t^{s_0 - s -1} }\right\vert\)
\(\displaystyle \) \(\le\) \(\displaystyle M N^{\sigma_0 - \sigma} + M m^{\sigma_0 - \sigma} + M \sum_{n \mathop = m}^{N - 1} \left\vert{ s - s_0 }\right\vert \int_n^{n+1} \left\vert{ t^{s_0 - s -1} }\right\vert\) Modulus of Complex Integral
\(\displaystyle \) \(=\) \(\displaystyle M N^{\sigma_0 - \sigma} + M m^{\sigma_0 - \sigma} + M \sum_{n \mathop = m}^{N - 1} \left\vert{s - s_0 }\right\vert \int_n^{n+1} t^{\sigma_0 - \sigma -1}\)
\(\displaystyle \) \(=\) \(\displaystyle M N^{\sigma_0 - \sigma} + M m^{\sigma_0 - \sigma} + M \left\vert{ s - s_0 }\right\vert \int_m^{N} t^{\sigma_0 - \sigma -1}\)
\(\displaystyle \) \(=\) \(\displaystyle M N^{\sigma_0 - \sigma} + M m^{\sigma_0 - \sigma} + M \frac{\left\vert{ s - s_0 }\right\vert}{\sigma - \sigma_0} \left( m^{\sigma_0 - \sigma}- N^{\sigma_0 - \sigma} \right)\)
\(\displaystyle \) \(\le\) \(\displaystyle M N^{\sigma_0 - \sigma} + M m^{\sigma_0 - \sigma} + M \frac{\left\vert{ s - s_0 }\right\vert}{\sigma - \sigma_0} \left( m^{\sigma_0 - \sigma} + N^{\sigma_0 - \sigma} \right)\)

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:

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

Hence the result.

$\Box$


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