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Oct 11, 2021 at 18:37 answer added KConrad timeline score: 16
Oct 11, 2021 at 18:36 comment added Pace Nielsen This also appears in Titchmarsh's book "The Theory of the Riemann Zeta-function".
Oct 11, 2021 at 17:57 comment added Joseph Van Name You are right. I was not being rigorous.
Oct 11, 2021 at 17:57 review Close votes
Oct 11, 2021 at 18:47
Oct 11, 2021 at 17:46 comment added Wojowu @JosephVanName This argument (together with an appropriate version of Abel's theorem for Dirichlet series) shows that if the limit exists, then it is equal to $0$. However, convergence is far from clear. Questions like this are subtle and convergence depends on distribution of zeros of zeta. For instance, convergence for all $s>1/2$ is equivalent to RH.
Oct 11, 2021 at 17:42 comment added Anurag Sahay @Joseph Van Name: Unless I'm missing something, it's a lot subtler than the fact that $\zeta$ has a pole at $s=1$. You need to be able to justify the interchange of limits (otherwise, for example, the prime number theorem would be much easier than it is).
Oct 11, 2021 at 17:33 history edited Alex M. CC BY-SA 4.0
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Oct 11, 2021 at 17:29 comment added Joseph Van Name More generally, $\sum_{k=1}^{\infty}\frac{\mu(k)}{k^{s}}=\frac{1}{\zeta(s)}$. Your result follows from the fact that $s=1$ is a pole of the Riemann zeta function.
Oct 11, 2021 at 17:17 comment added Wojowu math.stackexchange.com/q/1169675/127263
Oct 11, 2021 at 16:47 history edited LSpice CC BY-SA 4.0
Proofreading and [tag:reference-request]; deleted "thanks"
Oct 11, 2021 at 16:45 comment added LSpice @MicahMilinovich, I think that's an answer!
Oct 11, 2021 at 16:45 comment added Micah Milinovich Landau's PhD thesis: arxiv.org/abs/0803.3787
S Oct 11, 2021 at 16:37 review First questions
Oct 11, 2021 at 17:34
S Oct 11, 2021 at 16:37 history asked Coxi CC BY-SA 4.0