Given a Dirichlet series $$\phi(s)=\sum\_{n\ge1}\frac{a\_n}{n^s}$$  

let $\sigma\_{\text{conv}}\in\bar{\mathbb{R}}$ its abscissa of convergence, then we know that $\phi(s)$ is holomorphic on the half-plan $\mathrm{Re}(s)>\sigma\_{\text{conv}},$ then if we denot by $\sigma\_{\text{hol}}$ the abscissa of holomorphy of $\phi(s)$ we have 
$$\sigma\_{\text{hol}}\le \sigma\_{\text{conv}}.$$
 My question is : *it is in general true that $\sigma\_{\text{hol}}= \sigma\_{\text{conv}}$*?


**Edit :** $\sigma\_{\text{hol}}:=\inf\{\sigma\in\mathbb{R}\; ; \; \phi(s)\;\text{has a holomorphic prolongation on}\; \mathrm{Re}(s)>\sigma \}$



What is the relationship between the abscissa of holomorphy and abscissa of convergence of a Dirichlet series
The answer to your question is no, the counterexample is $L(s, \chi) = \sum\_{n = 1}^{\infty} \chi(n) n^{-s}$ with $\chi$ any non-principal character. $L(s,\chi)$ can be continued to an entire function thus $\sigma\_{\text{hol}} = -\infty$, while $\sigma\_{\text{abs}} = 1$ (abscissa of absolute convergence) and $\sigma\_{\text{cond}} = 0$ (abscissa of conditional convergence).


