Uniform Convergence of General Dirichlet Series

Theorem
Let $\map \arg z$ denote the argument of the complex number $z \in \C$.

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 $\map f s$ converges uniformly for all $s$ such that:
 * $\cmod {\map \arg {s - s_0} } \le a < \dfrac \pi 2$

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 uniformly for for $\cmod {\map \arg s} \le a < \frac \pi 2$

By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon > 0$ there exists an $N$ independent of $s$ 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 $\map S {k, j}$ is the difference of two terms of a convergent, and thus cauchy, sequence, we may pick $N$ large enough so that for $j > N$:


 * $\ds \cmod {\map S {k, j} } < \frac {\epsilon \cos a} 3$

which gives us:

We see that:

From Shape of Secant Function, we have that on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:


 * $\cmod {\map \arg s} \le a \implies \map \sec {\map \arg s} \le \sec a$

which gives us:


 * $\map \sec {\map \arg s} \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} } \le \sec a \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }$

Hence:

Because $\sigma>0$, we have that $ - \lambda_k \sigma <0$ and hence:
 * $e^{-\lambda_k \sigma} < 1 \le \sec a$

which gives us: