Dirichlet Series Convergence Lemma/General

Theorem
Let $\displaystyle f \left({s}\right) = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n \left({s}\right)}$ be a general Dirichlet series.

Let $f \left({s}\right)$ converge at $s_0 = \sigma_0 + i t_0$.

Then $f \left({s}\right)$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$.

Proof
Let $s=\sigma+it$

Let $s_0$ be such that $f \left({s_0}\right)$ converges.

Let $S\left(m,n\right) = \displaystyle \sum_{k \mathop = n}^m a_k e^{-\lambda_ks_0}$

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

Thus it suffices to show $g \left({s}\right)$ 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 $m,n>N$:
 * $ \left\vert \displaystyle \sum_{k \mathop = n}^m a_n e^{-\lambda_ks_0} e^{-\lambda_ks} \right\vert < \epsilon$

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

Because $S\left(m,n\right)$ is the difference of partial sums of a convergent, and thus cauchy, sequence, its modulus, $\left\vert S\left(m,n\right) \right\vert$, 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