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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 $L$?

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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 $\lim_{s\downarrow 1}f(s)=\zeta(1+it)$ exist, 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).

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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? –  senti_today Jul 22 '13 at 22:06
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@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 –  Kevin Ventullo 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 $s=1$. –  senti_today 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. –  GH from MO 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. –  senti_today Jul 23 '13 at 0:20
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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 $z \not= 0$, 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

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