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I am interested in the connection between particular Dirichlet series' abscissa of convergence and the poles of L-functions.

Let $D(z) = \sum_{n=1}^\infty\frac{a_n}{n^z}$ be a Dirichlet series admitting meromorphic continuation, with abscissa of absolute convergence (resp. conditional convergence) $\sigma_a \in \mathbb{R}$ (resp. $\sigma_c$), and let $L(z)$ be the corresponding L-function to $D(z)$. I would like to know the connection between $\sigma_c$ and the poles of $L$.

It is known that if $a_n \geq 0 \space\forall n$, then $L$ will have a pole on the real axis at $\sigma_a$ (for example, Hardy/Riesz, General Theory of Dirichlet Series, Thm 10). It is also known that $\sigma_c \leq \sigma_a \leq \sigma_c +1$. I want to know what happens to $L$ when the second inequality is not strict equality.

It is obvious if $\sigma_c = \sigma_a - 1$, that $L$ cannot have any poles in the region $S = \{z \in \mathbb{C} \space \vert \space \sigma_a-1 <\Re(z)\leq \sigma_a \}$ (It can even happen that $L$ is entire in this case - for example, Dirichlet $\eta$ function).

Define $T = \{ \Re(z) \space \vert \space L \space \text{has a pole at z} \in \mathbb{C} \}$.

My question is as follows:

If $\sigma_c > \sigma_a - 1$, must it be the case that $\sup T = \sigma_c$? How can I show this?

If needed, assume $L$ has any combination of an Euler product, functional equation, or Ramanujan conjecture.

I have seen approaches using Perron's formula for particular functions, but this technique is far from general. Thanks for any input!

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  • $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$
    – GH from MO
    Feb 25 at 6:34
  • $\begingroup$ Cross posted: math.stackexchange.com/q/4870067/11323 $\endgroup$
    – Kimball
    Feb 25 at 13:36
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    $\begingroup$ Thank you guys, I am new to the forums and am still learning proper etiquette. $\endgroup$ Feb 25 at 15:54

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The answer to your question is "no". Let $\chi$ be a nontrivial Dirichlet character. Then for the Dirichlet series $\sum_{n=1}^\infty \chi(n)/n^{2s}$ we have $\sigma_c=0$ and $\sigma_a=1/2$, but the corresponding $L$-function $L(2s,\chi)$ is entire.

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  • $\begingroup$ Thank you for your response. Sending $2s \to s/2$ in your example will yield a series with $\sigma_c = 0, \sigma_a = 2$, violating the theorem that $\sigma_a \leq \sigma_c +1$. Thus it is clear that the assumption that $D(z)$ has the form $\sum_{n=1}^\infty \frac{a_n}{n^z}$ is important. This is sidestepping ambiguity in rescaling the complex plane. $\endgroup$ Feb 25 at 15:50
  • $\begingroup$ Indeed, $L(s/2,\chi)$ is not a Dirichlet series. But $L(2s,\chi)$ is a Dirichlet series, which shows that the answer to your question is "no". If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Feb 25 at 16:05
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    $\begingroup$ Sorry, I guess I don't understand something fundamental. How is your example different from just scaling the complex plane? edit: nevermind, I'm dumb. Thanks! $\endgroup$ Feb 25 at 16:13

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