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 $L$?


2 Answers 2


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).

  • $\begingroup$ 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? $\endgroup$ Jul 22, 2013 at 22:06
  • 2
    $\begingroup$ @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 $\endgroup$ Jul 22, 2013 at 22:13
  • $\begingroup$ @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 $s=1$. $\endgroup$ Jul 22, 2013 at 22:16
  • $\begingroup$ @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. $\endgroup$
    – GH from MO
    Jul 23, 2013 at 0:11
  • $\begingroup$ @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. $\endgroup$ Jul 23, 2013 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

  • $\begingroup$ Dear Prof. Conrad, this answer is really helpful. Could you please give a reference that contains a proof of the result you mentioned that if $f(s_0)$ converges, then it must be equal to $\lim_{s \downarrow s_0} f(s)$? Thank you. $\endgroup$
    – asrxiiviii
    Oct 12, 2022 at 23:53
  • $\begingroup$ @asrxiiviii find a proof of Abel’s theorem for power series (based on partial summation) and adapt it. That is what I did. $\endgroup$
    – KConrad
    Oct 13, 2022 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.