Sujet : Re: Dark numbers (should be Kempner series)
De : noreply (at) *nospam* example.org (joes)
Groupes : sci.mathDate : 13. Apr 2025, 09:43:58
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <08b0c3b7e3c8a8faaf72785e0d4ce0b066e2f4f6@i2pn2.org>
References : 1
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Tue, 08 Apr 2025 20:09:48 +0200 schrieb WM:
The harmonic series diverges. Kempner has shown in 1914 that when all
terms containing the digit 9 are removed, the serie converges. Here is a
simple derivation: https://www.hs-augsburg.de/~mueckenh/HI/ p. 15.
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.
>
We can continue and remove all terms containing 1, 2, 3, 4, 5, 6, 7, 8,
9, 0 in the denominator without changing this. That means that only the
terms containing all these digits together constitute the diverging
series.
There are no naturals that contain none of these digits.
But that's not the end! We can remove any number, like 2025, and the
remaining series will converge. For proof use base 2026. This extends to
every definable number. Therefore the diverging part of the harmonic
series is constituted only by terms containing a digit sequence of all
definable numbers.
I.e. infinite numbers, so not naturals.
-- Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:It is not guaranteed that n+1 exists for every n.