# Convergence of a sequence of continuable Dirichlet series

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!

• Do you mean to say that g_n and f match on every term up to the m-th? – Matt Young Feb 20 '10 at 2:30
• Of course. Corrected. – Jared Weinstein Feb 20 '10 at 2:46
• If one imposes some conditions on the functional equations of the g_n's (in particular, bounding the conductors to avoid FC's counterexample) then the answer could potentially become "yes." – Matt Young Feb 20 '10 at 3:59
• FC- I don't understand why you wrote that as a comment rather than an answer. In particular, it gives a misleading impression on the front page. – Ben Webster Feb 20 '10 at 4:24
• If you don't want to play any reputation games, you could instead make all your new answers "community wiki." This allows the site to function correctly while not accruing any reputation to you. – Noah Snyder Feb 20 '10 at 8:15