I apologize if this is something standard and/or elementary, but I was unable to find anything relevant via Google.


Consider a Dirichlet series
$$
f(s) = \sum\_{n=1}^\infty \frac{a\_n}{n^s}
$$
and assume that $s\_0$ is a real number such that $f(s)$ converges absolutely for all $s>s\_0$. In addition, assume that the limit $L=\lim\_{s\downarrow s\_0}f(s)$ exists. Does it follow that the series $f(s\_0)$ converges and that its sum equals $L$?



Convergence of Dirichlet series ("at the boundary")
The answer is no. Take, for example, $a\_n:=n^{-it}$ with any fixed $t\neq 0$. Then $f(s)$ converges absolutely to $\zeta(s+it)$ for any $s>1$, and $\lim\_{s\downarrow 1}f(s)=\zeta(1+it)$ exists, but $f(1)$ diverges.


The answer would be yes under some stronger assumptions, see e.g. the main theorem in Chapter VII of Newman: Analytic number theory (GTM 177, Springer, 1988).


