Let $\phi(s)=\sum_{n\ge 1}a_n n^{-s},$ a Dirichlet series. Suppose that $\phi(s)$ is holomorphic at $s_0\in \mathbb{C}$.
My question is: Can we infer that $\phi(s)$ is holomorphic on the half plan $\Re(s)>\Re(s_0)$ ?
Thanks.
Let $\phi(s)=\sum_{n\ge 1}a_n n^{-s},$ a Dirichlet series. Suppose that $\phi(s)$ is holomorphic at $s_0\in \mathbb{C}$.
My question is: Can we infer that $\phi(s)$ is holomorphic on the half plan $\Re(s)>\Re(s_0)$ ?
Thanks.
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If you meant $\sum_{n=1}^\infty a_n n^{-s}$ converges at $s = s_0$, then yes, see what's on wikipedia : summing by parts $\sum_{n=1}^\infty a_n n^{-s}$ converges for every $Re(s) > Re(s_0)$ and is holomorphic.
Otherwise no, take $a_n = 1$ and $s_0= 1/2$, then $\zeta(s) = \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ is holomorphic at $s_0$ but it has a pole at $s=1$