Let's say $f$ is a Dirichlet series which converges on the half-plane $\text{Re }s>\sigma$ to a function $f(s)$. Suppose further that $f(s)$ admits an analytic continuation to an entire function, together with the standard sort of functional equation. Let $g\_n$ be a sequence of Dirichlet series, also convergent on $\text{Re }s>\sigma$, which each admit an analytic continuation and functional equation, though their precise FEs may vary. We assume that $g\_n$ converges to $f$ in the following sense: for every $m>0$ there exists an $N$ for which the series $g\_n$ and $f$ match on every term up to the $m$th, for all $n>N$. Note this implies that $g\_n(s)$ converges to $f(s)$ for every $\text{Re }s>\sigma$.


Can it be said that $g\_n(s)$ converges to $f(s)$ for any $s$ outside the domain of convergence?


Perhaps that's too much to hope for, and you can't even expect that $g\_n(s)$ converges to $f(s)$ even for the point $s=\sigma$. I'd certainly be interested in a counterexample which does this!



Convergence of a sequence of continuable Dirichlet series
(This answer is a community wiki version of a comment above by FC which answered the question.)


For any integer M, there exists a prime p such that chi\_p(n) = (n/p) = 1 for all n = 1...M. This means that the Dirichlet series L(s,V chi\_p) (for any representation V) "converges" in your sense to L(s,V). but they do not converge at s = 0. If V is trivial, then we are comparing zeta(0) = -1/2 with L(0,chi\_p) which grows without bound by Brauer-Siegel. I think in this class of examples one does get convergence at the critical point.


