Given a sequence of complex numbers $(a_n)$ one defines the corresponding Dirichlet series as $$f(s) = \sum_{n = 1}^\infty \frac{a_n}{n^s}.$$ I know that there is some $x_0 \in \mathbb{R} \cup \{\pm \infty\}$ such that $f(s)$ converges absolutely whenever the real part of $s$ is greater than $x_0$.
Suppose that $x_0 < +\infty$ (so $f$ converges absolutely on some half-plane). Is it true that $\lim_{x \to \infty} f(x) = a_1$? If not, does it become true if we know that the sequence $a_n$ does not grow 'too fast' in some sense (for instance exponentially bounded or polynomially bounded)?
Notice that this is a case of switching the order of the limits, because when $x$ goes to infinity, every term in the definition of $f$ goes to zero except the $n = 1$ term. But I'm not sure if switching the limits is allowed in this case.
Motivation: Series like this show up in the theory of finitely generated profinite groups. Specifically, if we calculate the probability that $k$ randomly chosen elements of a fixed finitely generated profinite group $G$ generate this group topologically, we find the answer is given by a Dirichlet series $P_G$ in $k$ with constant term 1. For instance, if $G = \hat {\mathbb{Z}}$ is the group of profinite integers, one finds that $P_G(k)$ is equal to $1/\zeta(k)$, with $\zeta$ the usual Riemann zeta function. If $G$ has a lot of subgroups of finite index, it may happen that $P_G$ fails to converge (i,e, $x_0 = +\infty$ in the notation above). However, if $G$ behaves well enough to guarantee that $P_G$ exists on some half-plane, it would be nice to be able to conclude that $P_G(k)$ goes to $1$ for $k \to \infty$.