Dirichlet Series Convergence Lemma
Theorem
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series.
Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$.
General Dirichlet Series
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\map {\lambda_n} s}$ be a general Dirichlet series.
Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$.
Proof
We begin with a lemma:
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 $\sigma > \sigma_0$:
- $\ds \size {\sum_{n \mathop = m}^N a_n n^{-s} } \le 2 M m^{\sigma_0 - \sigma} \paren {1 + \frac {\size {s - s_0} } {\sigma - \sigma_0} }$
$\Box$
Suppose that $\map f s$ converges at $s_0 = \sigma_0 + it_0$.
Then by Convergent Sequence in Metric Space is Bounded, $\size {\ds \sum_{k \mathop = 1}^n \frac {a_k} {k^{s_0} } }$ is bounded.
Thus the results of the lemma hold.
Choose any $s = \sigma + it$ with $\sigma > \sigma_0$.
By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon>0$, there exists an $N>0$ such that for all $m>n>N$
- $\ds \size {\sum_{k \mathop = 1}^m \frac {a_k} {k^s} - \sum_{k \mathop = 1}^n \frac {a_k} {k^s} } = \size {\sum_{k \mathop = n + 1}^m \frac {a_k} {k^s} } < \epsilon$
The lemma shows that for a given $s$ there exists a constant $C$ independent of $N$ such that:
- $\ds \size {\sum_{k \mathop = n + 1}^n a_n n^{-s} } \le \map C {n + 1}^{\sigma_0 - \sigma}$
Since $\sigma_0 - \sigma <0$, the right hand side tends to zero as $n \to \infty$.
Thus we may choose $N$ large enough so that for $n > N$.
- $\ds \paren {n + 1}^{\sigma_0 - \sigma} < \dfrac \epsilon C$
which gives us:
- $\ds \size {\sum_{k \mathop = n + 1}^n a_n n^{-s} } < \epsilon$
$\blacksquare$
Sources
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- 1976: Tom M. Apostol: Introduction to Analytic Number Theory: $\S 11.6$: Lemma $2$, Theorem $11.8$