Divergence of Dirichlet series Suppose $s$ is a complex number with $\Re(s) \in (0,1]$ and $\{a_n\}$ is a complex sequence converging to $a \neq 0$. Must the Dirichlet series $$\sum_{n=1}^\infty\frac{a_n}{n^s}$$ diverge?
 A: I have worked out a partial answer to my own question.
Define $A(s) = \sum\limits_{n=1}^\infty\frac{a_n}{n^s}$.
First, assume that $s = 1$.  WLOG we can also assume that $a = 1$.  Observe that for some $N$ and all $n > N$, $\Re(a_n) > 1/2$.  It follows that $A(1)$ diverges.
I found a pertinent result in Knopp, Konrad. Infinite Sequences and Series. New York: Dover, 1956. Item 4 on p. 138 is the result that if the partial sums of the Dirichlet series $A(s_0)$ are bounded, then $A(s)$ converges provided $\Re(s) > \Re(s_0)$.  Since $A(1)$ diverges, we conclude from this result that $A(s_0)$ diverges for $s_0 < 1$.
Thus, we see that $A(s)$ must diverge for $\Re(s) \in (0,1]$, unless $s = 1 + it$, where $t \neq 0$.
Please see my question on MO for a demonstration of divergence in the remaining case.
A: I'm going to post this without having all the facts yet. There is no ambiguity possible for the real part of $s$ strictly between 0 and 1, as in the comment by anon.
However, I am not sure at this point about the sum
$$         \sum_{n=1}^\infty \frac{1}{n^s}$$
for $s = \sigma + i t$ and $\sigma = 1,$ that is $s = 1 + i t.$ The sum diverges for $t=0,$ the Riemann zeta function has a pole there.
It turns out that the Euler product is conditionally convergent at  $s = 1 + i t$ with real $t\neq 0.$ So, it is at least possible that $\sum_{n=1}^\infty \frac{1}{n^{1 + i t}}$ is conditionally convergent. If so, the question for your sequence $a_n$ becomes extremely subtle when
$s = 1 + i t$ with real $t\neq 0.$
