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