# Convergence of Dirichlet series ("at the boundary")

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 $L$?

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).

• Thanks! Unfortunately, the assumptions of the theorem from Newman that you mentioned are too strong for me. In the situations I am interested in, the numbers $a_n$ are all real, their signs alternate, $|a_n|$ doesn't grow too quickly (slower than $n^\epsilon$ for any $\epsilon>0$), and $s_0=1$. I wonder if you have any insight into this particular case? Jul 22 '13 at 22:06
• @senti_today If the signs are alternating, it's enough for the absolute values to tend to zero: en.wikipedia.org/wiki/Alternating_series_test Jul 22 '13 at 22:13
• @Kevin: not really, because I am not making any monotonicity assumptions. For example, the $a_n$ with even $n$ could go to zero very slowly, and the $a_n$ with odd $n$ could go to zero very quickly; then the series would not converge for $s=1$. Jul 22 '13 at 22:16
• @senti_today: I don't know the answer in that particular case. I think that it is still possible that $f(1)$ diverges, but I don't have a counterexample. Jul 23 '13 at 0:11
• @GH: OK, sorry to bother you one more time, but I wonder if you know whether the theorem from Newman you mentioned is true if instead of assuming that $|a_n|$ is bounded, one merely assumed that $|a_n|=o(n^\epsilon)$ for any $\epsilon>0$? I've looked at the proof, and it seems like I $might$ be able to generalize it to this case, but I could be making a mistake, and if you know the answer off the top of your head, I would really appreciate it. Jul 23 '13 at 0:20

For intuition, think about the analogue for power series: if $$g(z) = \sum_{n \geq 0} a_nz^n$$ converges on the disc $$|z| < R$$ and at a point $$z_0$$ with $$|z_0| = R$$ the values $$g(z)$$ converge as $$z \rightarrow z_0$$ radially from the inside of that disc, does the series $$g(z_0)$$ converge? No, and the geometric series (which is an analogue of $$\zeta(s)$$, in the sense of being the power series with all coefficients = 1) is an example of this: $$G(z) = \sum_{n \geq 0} z^n$$ converges for $$|z| < 1$$ with value $$1/(1-z)$$, which is continuous for all $$z \not= 1$$, so when $$|z_0| = 1$$ and $$z_0 \not= 1$$ the the numbers $$G(z)$$ converge as $$z \rightarrow z_0$$ radially but the series $$G(z_0)$$ does not converge.

A basic result about boundary behavior for power series is Abel's theorem: if the series $$g(z_0)$$ does converge then it must be the limit of $$g(z)$$ as $$z \rightarrow z_0$$ radially (in particular, the radial limit of $$g(z)$$ exists). The proof of this carries over to the case of Dirichlet series: in your notation, if the series $$f(s_0)$$ does converge then it must be the limit of $$f(s)$$ as $$s \rightarrow s_0$$ from the right (and, in particular, the limit of $$f(s)$$ at $$s_0$$ from the right exists). Of course this doesn't help you to prove the Dirichlet series $$f(s_0)$$ actually converges; it only tells you what the value of $$f(s_0)$$ would have to be if that series converges.

A starting point for this circle of ideas is http://en.wikipedia.org/wiki/Abelian_and_tauberian_theorems