Timeline for Why shouldn't this prove the Prime Number Theorem?

Current License: CC BY-SA 4.0

13 events

when toggle format	what		by	license	comment
Apr 22, 2019 at 13:04	vote	accept	Q_p
Apr 22, 2019 at 1:03	comment	added	LSpice		@YemonChoi, is that a new nickname for @‍Gro-Tsen? :-)
Apr 21, 2019 at 4:44	history	edited	Martin Sleziak		edited tags
Apr 21, 2019 at 2:51	history	became hot network question
Apr 21, 2019 at 2:48	answer	added	kodlu		timeline score: 20
Apr 20, 2019 at 23:49	history	reopened	Joseph Van Name Gro-Tsen Yemon Choi Sean Lawton Gerry Myerson
Apr 20, 2019 at 22:46	review	Reopen votes
Apr 20, 2019 at 23:50
Apr 20, 2019 at 22:43	comment	added	Yemon Choi		I agree with Fourton and have voted accordingly
Apr 20, 2019 at 22:41	comment	added	Gro-Tsen		I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
Apr 20, 2019 at 21:53	history	closed	Peter Humphries Anthony Quas Wojowu Tony Huynh Felipe Voloch		Not suitable for this site
Apr 20, 2019 at 21:48	comment	added	Wojowu		In general, limit of sums of series $\neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
Apr 20, 2019 at 21:48	comment	added	Peter Humphries		It is true that the PNT is equivalent to $\sum_{n \leq x} \frac{\mu(n)}{n} = o(1)$. It is also relatively easy to prove that $\lim_{s \searrow 1} \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = 0$. The hard part is proving that $\lim_{s \searrow 1} \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = \lim_{x \to \infty} \sum_{n \leq x} \frac{\mu(n)}{n}$. This is highly nontrivial!
Apr 20, 2019 at 21:45	history	asked	Q_p	CC BY-SA 4.0