Dirichlet Series Convergence Lemma/Lemma
Dirichlet Series Convergence Lemma/Lemma 0 53046 357559 350951 2018-05-28T17:18:39Z AliceInNumberland 3357 Fixed the proof, it follows the broad strokes of the cited source, with a few modifications to make it more elementary, as that is the objective of proving it in the ordinary sense wikitext text/x-wiki == 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: {{begin-eqn}} {{eqn | l = \left\vert{\sum_{n \mathop = m}^N \frac {a_n} {n^s} }\right\vert       | r = \left\vert{\sum_{n \mathop = m}^N f_n g_n}\right\vert }} {{eqn | r = \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       | o = \le       | c = using partial summation and the Triangle Inequality }} {{eqn | r = 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       | o = \le       | c = using the given bound on the partial sums }} {{eqn |r = 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 }} {{eqn |o= \le       |r = 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       |c = Modulus of Complex Integral }} {{eqn |r =  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} }} {{eqn |r =  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} }} {{eqn |r =  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) }} {{eqn |o= \le       |r =  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) }} {{end-eqn}} 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: {{begin-eqn}} {{eqn |l= \left\vert{\sum_{n \mathop = m}^N a_n n^{-s} }\right\vert       |o= \le       |r= 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) }} {{eqn |o=\leq       |r=   2 M m^{\sigma_0 - \sigma} + 2 M \frac{\left\vert{ s - s_0 }\right\vert}{\sigma - \sigma_0} m^{\sigma_0 - \sigma} }} {{eqn |r = 2 M m^{\sigma_0 - \sigma} \left(1 + \frac{\left\vert{ s - s_0 }\right\vert}{\sigma - \sigma_0} \right) }} {{end-eqn}} Hence the result. {{qed|lemma}} == Sources == {{SourceReview|wrt Dirichlet Series Convergence Lemma}} * {{BookReference|Introduction to Analytic Number Theory|1976|Tom M. Apostol}}: $\S 11.6$: Lemma $2$, Theorem $11.8$