# Classification of L functions and Dirichlet series by poles

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!

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

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.
• 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. Feb 25 at 15:50
• 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! Feb 25 at 16:05