Existence of Abscissa of Convergence/General

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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.

$\blacksquare$