Vanishing of Dirichlet Series 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 $k\in\mathbb{N}$)
 A: 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:


*

*$Re(c)\le Re(a)$. 

*$|Re(a)-Re(c)|\le 1$. 

*$D$ converges absolutely at every complex number to the right of $a$ (i.e. $Re(a)\le Re(z)$). 

*$D$ converges conditionally at every complex number to the right of $c$. 

*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 $c$. 

*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. 
