Vanishing of Dirichlet Series

Question

Suppose the function $\sum_{n=1}^{\infty}{a_{n}n^{-s}}$ is $0$ on some open set $U\subset\mathbb{C}$. (Can assume the sum converges absolutely on $U$.)

Is it true that $a_{n}=0$ for all $n$?

(This feels like it should be true, but we aren't sure how to see it. As a trivial observation, taking derivatives gives us $\sum_{n=1}^{\infty}{\ln(n^{-1})^{k}a_{n}n^{-s}}=0$ for all $k\in\mathbb{N}$)

Answer 1

This is not a trivial problem at all. However, Apostol's Intoduction to Analytic Number Theory - Chapter 11 gives many interesting results.

Theorem(s). For Dirichlet series, $D$, there are numbers $a,c\in\mathbb{C}\cup\{\pm\infty\}$ such that:

1. $Re(c)\le Re(a)$.
2. $|Re(a)-Re(c)|\le 1$.
3. $D$ converges absolutely at every complex number to the right of $a$ (i.e. $Re(a)\le Re(z)$).
4. $D$ converges conditionally at every complex number to the right of $c$.
5. The convergences is uniform on every compact subset in the half plane to the right of $c$. Thus, $D$ is equal to some analytic function on the half plane to the right of $c$.
6. If $D(p_n)=0$, where $Re(p_n)\to+\infty$, then the coefficients of the Dirichlet series are the zero coefficients. That is, non-zero Dirichlet series must have some right half plane where the Dirichlet series is never zero.

These statements allow us to answer your question. We have a NO then YES.

NO we can't assume that if a Dirichlet series converges then the series converges absolutely. However, if it does converge to zero on some $U$, then the function $D$ must converge to the analytic zero function. But that means that $D$ agrees with zero on the whole right half plane to the right of $U$. But that means that the Dirichlet coefficients of $D$ are zeros. Such $D$ converges absolutely and uniformly in the whole complex plane. So, YES the convergence on $U$ will be absolute.