Uniform Convergence of General Dirichlet Series

Theorem
Let $\arg\left(z\right)$ be the argument of z

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 $f \left({s}\right)$ converges uniformly for all $s$ such that $\left\vert \arg\left(s-s_0\right)\right\vert < a < \frac \pi 2$

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 uniformly for for $\left\vert \arg\left(s\right)\right\vert < a < \frac \pi 2$

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 $m,n>N$:
 * $ \left\vert \displaystyle \sum_{k \mathop = n}^m a_n e^{-\lambda_ks_0} e^{-\lambda_ks} \right\vert < \epsilon$

By Abel's Lemma we may write:

Because $S\left(k,j\right)$ is the difference of two terms of a convergent, and thus cauchy, sequence, we may pick $N$ large enough so that for $j>N$
 * $\displaystyle \left\vert S\left(k,j\right) \right\vert < \frac {\epsilon \cos\left(a\right)}{3}$

which gives us:

We see that:

From Shape of Secant Function, we have that on the interval $\left( -\frac \pi 2, \frac \pi 2\right)$:
 * $ \left\vert \arg\left(s\right) \right\vert < a \implies \sec\left(\arg\left(s\right)\right) < \sec\left(a\right)$

which gives us:
 * $ \sec\left(\arg\left(s\right)\right) \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right) < \sec\left(a\right) \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right)$

Hence:

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

Which gives us