# Existence of Abscissa of Convergence/General

## Theorem

Let $s = \sigma + i t$

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

Then there exists a extended real number, $\sigma_0$, such that

- $(1): \quad$ For $\sigma<\sigma_0$, $\displaystyle f \left({s}\right)$ diverges
- $(2): \quad$ For $\sigma>\sigma_0$, $\displaystyle f \left({s}\right)$ converges.

## Proof

If there does not exist an $s_0$ such that $\displaystyle f \left({s_0}\right)$ converges, then $\sigma_0 = \infty$ and the theorem is vacuously true

If there does exist such an $s_0$, let $\sigma_0$ be the infimum of the real part of all such $s_0$, where $\sigma_0= -\infty$ if the set is not bounded from below.

- It is clear that $f \left({s}\right)$ diverges if $\sigma<\sigma_0$ by construction of $\sigma_0$ as the infimum of all convergent s

- As $\sigma_0$ is the infimum of the the real part of the set all s such that $f \left({s}\right)$ converges, for any $\epsilon>0$, there exists an s such that $f \left({s}\right)$ converges and $\sigma_0\leq \sigma < \sigma_0+ \epsilon$ as otherwise $\sigma_0+\epsilon$ would be the infimum of the set.

- Therefore, for any s such that $\sigma>\sigma_0$, we can pick $\epsilon = \sigma-\sigma_0$ and find an $q$ such that $f \left({q}\right)$ converges and $\operatorname{Re} \left({q}\right) < \sigma$.

- By Dirichlet Series Convergence Lemma, We therefore have that $f \left({s}\right)$ converges.

$\blacksquare$