Path: ...!news.roellig-ltd.de!open-news-network.org!weretis.net!feeder8.news.weretis.net!news.szaf.org!news.karotte.org!news.space.net!news.muc.de!.POSTED.news.muc.de!not-for-mail From: Alan Mackenzie Newsgroups: sci.math Subject: Re: The existence of dark numbers proven by the thinned out harmonic series Date: Wed, 12 Mar 2025 10:22:01 -0000 (UTC) Organization: muc.de e.V. Message-ID: References: Injection-Date: Wed, 12 Mar 2025 10:22:01 -0000 (UTC) Injection-Info: news.muc.de; posting-host="news.muc.de:2001:608:1000::2"; logging-data="31037"; mail-complaints-to="news-admin@muc.de" User-Agent: tin/2.6.4-20241224 ("Helmsdale") (FreeBSD/14.2-RELEASE-p1 (amd64)) Bytes: 3009 Lines: 59 WM wrote: > The harmonic series diverges. Kempner has shown in 1914 that when all > terms containing the digit 9 are removed, the series converges. > That means that the terms containing 9 diverge. Same is true when all > terms containing 8 are removed. That means all terms containing 8 and 9 > simultaneously diverge. That's so slovenlily worded it's hardly meaningful. Does the adverb "simultaneously" qualify "containing" or "diverge"? As I've told you before, the terms don't diverge; their sum does. What I think you're trying to say is that the sum of all terms containing both 8 and 9 diverges. It is far from clear that this is the case. > We can continue and remove all terms containing 1, 2, 3, 4, 5, 6, 7, 8, > 9, 0 in the denominator without changing this. What do you mean by "this"? What does it refer to? > That means that only the terms containing all these digits together > constitute the diverging series. (*) There are many diverging sub-series possible. I think you mean "a" diverging series. > But that's not the end! We can remove any number, like 2025, .... From what? 2025 isn't a digit. > .... and the remaining series will converge. For proof use base 2026. > This extends to every definable number. Meaningless. "Definable number" is itself undefined. > Therefore the diverging part of the harmonic series is constituted > only by terms containing a digit sequence of all definable numbers. Gibberish. > Note that here not only the first terms are cut off but that many > following terms are excluded from the diverging remainder. > This is a proof of the huge set of undefinable or dark numbers. In your dreams. > (*) At this point the diverging series starts with the smallest term > 1023456789 and contains further terms like 1203456789 or 1234567891010 > or 123456789111 or 1234567891011. Only those containing the digit > sequence 10 will survive the next step, and only those containing the > digit sequence 1234567891011 (where the order of the first nine digits > is irrelevant) will survive the next step. > Regards, WM -- Alan Mackenzie (Nuremberg, Germany).