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Suppose $s$ is a complex number with $\Re(s) \in (0,1]$ and $\{a_n\}$ is a complex sequence converging to $a \neq 0$. Must the Dirichlet series $$\sum_{n=1}^\infty\frac{a_n}{n^s}$$ diverge?

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  • $\begingroup$ Yes. $a_n\to a$ implies for any $\epsilon>0$ the sequence is eventually trapped in an $\epsilon$-radius disk around $a$; if $a\ne0$ then we may split the series into two parts based on the index when the sequence becomes stuck inside and bound the infinite-term part in magnitude from below using real/imaginary parts and comparing with $\zeta(\cdot)$ (up to a constant). (Comment resubmitted because Will referenced it.) $\endgroup$
    – anon
    Dec 13, 2011 at 22:12
  • $\begingroup$ @anon, do you know about the possibility of conditional convergence with all $a_n= 1$ and the real part of $s$ equal to 1, but $s \neq 1?$ I'm not seeing anything definitive. $\endgroup$
    – Will Jagy
    Dec 13, 2011 at 22:14
  • $\begingroup$ @Will: Erm, I hadn't thought about Re(s)=1. I'm not sure about the case. $\endgroup$
    – anon
    Dec 13, 2011 at 22:32
  • $\begingroup$ @anon, I bet there is a pretty good discussion of the case in print somewhere, just not necessarily written within the last 100 years. $\endgroup$
    – Will Jagy
    Dec 13, 2011 at 22:43
  • $\begingroup$ @anon: Put $s = \sigma + i t$. Let's assume for the moment that $\sigma \in (0,1)$ and $t \neq 0$. The complex series $D(s) = \sum_{n=1}^\infty\frac{1}{n^s}$ certainly diverges (though $\zeta(s)$ is defined). Neither the real nor the imaginary part of $D(s)$ is a positive series. Call the original series in the statement of the problem $A(s)$. How do you show that the magnitude of your "infinite-term part" of $A(s)$ is bounded from below? $\endgroup$ Dec 14, 2011 at 0:52

2 Answers 2

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I have worked out a partial answer to my own question.

Define $A(s) = \sum\limits_{n=1}^\infty\frac{a_n}{n^s}$.

First, assume that $s = 1$. WLOG we can also assume that $a = 1$. Observe that for some $N$ and all $n > N$, $\Re(a_n) > 1/2$. It follows that $A(1)$ diverges.

I found a pertinent result in Knopp, Konrad. Infinite Sequences and Series. New York: Dover, 1956. Item 4 on p. 138 is the result that if the partial sums of the Dirichlet series $A(s_0)$ are bounded, then $A(s)$ converges provided $\Re(s) > \Re(s_0)$. Since $A(1)$ diverges, we conclude from this result that $A(s_0)$ diverges for $s_0 < 1$.

Thus, we see that $A(s)$ must diverge for $\Re(s) \in (0,1]$, unless $s = 1 + it$, where $t \neq 0$.

Please see my question on MO for a demonstration of divergence in the remaining case.

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  • $\begingroup$ Richard, very good. I never really figured out what to expect when $a_n \rightarrow a.$ When $a_n \rightarrow 0,$ we can pick a favorite point and either force convergence there or divergence. I'm not sure what happens away from the chosen point. $\endgroup$
    – Will Jagy
    Dec 18, 2011 at 4:21
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I'm going to post this without having all the facts yet. There is no ambiguity possible for the real part of $s$ strictly between 0 and 1, as in the comment by anon.

However, I am not sure at this point about the sum $$ \sum_{n=1}^\infty \frac{1}{n^s}$$ for $s = \sigma + i t$ and $\sigma = 1,$ that is $s = 1 + i t.$ The sum diverges for $t=0,$ the Riemann zeta function has a pole there.

It turns out that the Euler product is conditionally convergent at $s = 1 + i t$ with real $t\neq 0.$ So, it is at least possible that $\sum_{n=1}^\infty \frac{1}{n^{1 + i t}}$ is conditionally convergent. If so, the question for your sequence $a_n$ becomes extremely subtle when $s = 1 + i t$ with real $t\neq 0.$

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  • $\begingroup$ $\sum_{n=1}^\infty\frac{1}{n^s}$ diverges for $\sigma \le 1$ and any value of $t$, but if $t \neq 0$ neither its real nor imaginary parts is a positive series. How does one show that the original series diverges for $\sigma \in (0,1]$? (Please see my note to anon above.) $\endgroup$ Dec 14, 2011 at 3:59
  • $\begingroup$ @RichardHevener , I will need to work on it, but my immediate feeling is that we may divide through by your $a,$ getting numerators $1 + b_n$ with $b_n \rightarrow 0,$ eventually the numerators are close to 1 in both modulus and argument. $\endgroup$
    – Will Jagy
    Dec 14, 2011 at 5:56

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