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I was working on a problem, and I had made the assumption that given a Dirichlet series $$ L(s,f)=\sum_{n\geq 1}\frac{f(n)}{n^s} $$ If I have some $\sigma\in\mathbb{C}$ such that $L(\sigma,f)$ diverges, then I had assumed that $L(\sigma,f)\rightarrow\infty$. However, after thinking about it, I realized that this is false as I had considered the case of $f(n)=(-1)^n$ as then $$ L(-1,f)=\sum_{n\geq 1}(-1)^nn $$ and in this case, we have that this series does not diverge to infinity (thinking of it as a series of real numbers, but I guess if you think of it on the Riemann sphere it does approach infinity). Thus, I then thought about if it is always the case that if $L(\sigma,f)$ diverges, then $\vert L(\sigma,f)\vert\rightarrow\infty$. However, I then realized that this is also too much to ask for since with my above example we have that $$ L(0,f)=\sum_{n\geq 1}(-1)^n $$ which diverges, but does not go off to infinity.

Thus, my question is whether or not there is some criteria on $\sigma$ or $f(n)$ that says if $L(\sigma,f)$ diverges then $\vert L(\sigma,f)\vert\rightarrow\infty$ or even $L(\sigma,f)\rightarrow\infty$? Also, if $\vert L(\sigma,f)\vert\not\rightarrow\infty$, can we conclude that there is a removable singularity at $s=\sigma$.

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  • $\begingroup$ If you're asking about the divergence of the series $\sum_n f(n)n^{-s}$ or about the behaviour of its partial sums, it's better to discuss that object explicitly—it's not the same object as $L(s,f)$ in the regions where the series doesn't converge. For example, "$L(\sigma,f)\to\infty$" makes no sense when $\sigma$ is a fixed number. $\endgroup$
    – Greg Martin
    Commented Feb 7, 2023 at 18:05
  • $\begingroup$ That's is a good point, I believe that I meant the function defined over $\mathbb{C}$ given by this series. I know that there is some half plane of absolute convergence, and I guess I am asking about the limiting behavior of the partial sums if there is a criteria to say when this limiting behavior goes to infinity $\endgroup$
    – Steven Creech
    Commented Feb 8, 2023 at 0:55

2 Answers 2

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There is some confusion between the Dirichlet series and its analytic continuation (which doesn't have to exist..)

For the Dirichlet series things are relatively simple. If $A_{s_0}(m)=\sum_{n\le m} a_n n^{-s_0}$ is bounded then $$\sum_{n\ge 1}^\infty a_n n^{-s} = \lim_{N\to \infty} A_{s_0}(N) N^{s_0-s} +\sum_{m=1}^{N-1} A_{s_0}(m) (m^{s_0-s}-(m+1)^{s_0-s})$$ converges for all $\Re(s) > \Re(s_0)$.

So if $\Re(s_0)$ is smaller than the abscissa of convergence then $$\sup_m|A_{s_0}(m)|=\infty$$

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For the second question take the Dirichlet series $f(s)=\sum_{n \ge 2}\frac{1}{\log^2 n}n^{-s}$ which clearly satisfies $f''(s)=\zeta(s)-1$ for $\Re s >1$ so has a singularity at $1$, while clearly $f$ is continuous on the line $\Re s =1$

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  • $\begingroup$ I'm going to be completely honest, I have no idea how your function satisfies $f''(s)=\zeta(s)-1$, since isn't $f'(s)=\sum_{n\geq 2}\frac{-s}{\log^2n}n^{-s-1}$, so we get $f''(s)=\sum_{n\geq 2}\frac{s^2+s}{\log^2n}n^{-s-2}$ and it isn't clear to me how the log term goes away when taking the derivative $\endgroup$
    – Steven Creech
    Commented Feb 8, 2023 at 1:14
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    $\begingroup$ the derivative of $n^{-s}=e^{-s\log n}$ is $-n^{-s}\log n$ $\endgroup$
    – Conrad
    Commented Feb 8, 2023 at 1:17
  • $\begingroup$ Thank you, I was being silly with taking the derivative $\endgroup$
    – Steven Creech
    Commented Feb 8, 2023 at 1:19
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    $\begingroup$ no problem - for a Dirichlet series the only place where things (partial sums bounded etc) are not known is on the line of convergence - to the right all good, to the left partial sums are unbounded; on the line of convergence pretty much anything can happen as convergence and partial sums go and there is little to say about the relation between that and analytic continuation to the left (except for obvious things like if the function goes to infinity we cannot have continuation); not unlike power series and the circle of convergence, except that here we may not even have a singularity $\endgroup$
    – Conrad
    Commented Feb 8, 2023 at 1:23

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