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Consider the following Dirichlet Series:

$$ D(s) = e^{i \theta} + \frac{e^{i \theta/2}}{2^s} + \frac{e^{i \theta / 3}}{3^s} + \dots$$

Is there a nice limit for the below?

$$ \lim_{s \to 1} \frac{D(s)}{\zeta(s)} = ?$$

More Anonymous
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  • Yes, there is. Note that the coefficients of $D(s) - \zeta(s)$ tend to $0$ like $\theta/n$. – Daniel Fischer Aug 26 '20 at 13:10
  • I'm actually a physics student ... So I'm not sure what to do with that hint. – More Anonymous Aug 26 '20 at 13:11
  • Do you know a formula for the abscissa of convergence of a Dirichlet series? – Daniel Fischer Aug 26 '20 at 13:13
  • @DanielFischer Actually, no ... I know I'm way out of my depth. The reason I asked the question was I realised then I could find an asymptotic expression for $z + z^{1/2} + z^{1/3} + \dots$ where $z$ is a complex variable (after proper branch cuts) ... If I knew the answer to the above question. – More Anonymous Aug 26 '20 at 13:16

1 Answers1

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We have the

Theorem: If $f \colon \mathbb{N} \to \mathbb{C}$ is a function such that $$A = \lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^n f(k)\,,$$ then the Dirichlet series $$F(s) := \sum_{n = 1}^{\infty} \frac{f(n)}{n^s}$$ converges (at least) for $\operatorname{Re} s > 1$, and we have $$\lim_{s \to 1} \frac{F(s)}{\zeta(s)} = A\,,$$ where the limit is taken over $s$ with $\lvert \arg (s-1)\rvert \leqslant \varphi < \frac{\pi}{2}$.

This gives $$\lim_{s \to 1} \frac{D(s)}{\zeta(s)} = 1\,,$$ provided $s$ stays within such an angular wedge.

Here, we can get rid of that restriction: Since $$e^{i\theta/n} - 1 \sim \frac{i\theta}{n}$$ the Dirichlet series $$D(s) - \zeta(s) = \sum_{n = 1}^{\infty} \frac{e^{i\theta/n} - 1}{n^s}$$ converges absolutely for $\operatorname{Re} s > 0$, thus $$\frac{D(s)}{\zeta(s)} - 1$$ is holomorphic on a neighbourhood of $s = 1$.

Daniel Fischer
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