Dirichlet Series Convergence Lemma/General

Theorem
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
Let $s = \sigma + i t$

Let $s_0 \in \C$ be such that $\map f {s_0}$ converges.

Let $\map S {m, n} = \ds \sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0}$

We may create a new Dirichlet series that converges at 0 by writing:

Thus it suffices to show $\map g s$ converges for $\sigma > 0$.

By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon>0$ there exists an $N$ such that for all $m, n > N$:
 * $\ds \cmod {\sum_{k \mathop = n}^m a_n e^{-\lambda_k s_0} e^{-\lambda_k s} } < \epsilon$

By Abel's Lemma: Formulation 2 we may write:

Because $\map S {m, n}$ is the difference of partial sums of a convergent, and thus cauchy, sequence, its modulus, $\cmod {\map S {m, n} }$, is bounded, say by $Q$.

Thus we have:

We see that:

Thus we have:

Because $\lambda_n$ tends to infinity, both summands tend to $0$ as $n$ goes to $\infty$ if $\sigma > 0$.

Thus we can pick $N$ large enough such that both terms are less than $\dfrac \epsilon 2$ for $n, m > N$, giving us the desired result.