Re: Kempner series (was: Dark numbers)

Am Mon, 14 Apr 2025 13:52:44 +0200 schrieb WM:

On 14.04.2025 10:48, joes wrote:

Am Sun, 13 Apr 2025 17:42:07 +0200 schrieb WM:

On 13.04.2025 10:43, joes wrote:

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 there are naturals which contain all of these digits and in
addition all definable sequences of digits.

There are also no naturals that contain all sequences of digits, as
those would have to be infinite.

No. All definable sequences of digits are finite. It wouldn't be
possible to define infinitely many numbers individually.

I think it’s possible to define an infinite number sequence, like the
decimal expansion of e. Anyway, you wrote above there were naturals
„which contain all definable sequences of digits”. Thanks for retracting
that statement.

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.

Wrong. The denominators of the harmonic sequence are finite numbers
but the diverging part consists of numbers which are larger than all
definable numbers.

Sounds pretty infinite to me.

None is infinite. If you cannot comprehend the meaning of "definable",
then first try "defined". There are only finitely many numbers defined.

That is wrong already. The set N is infinite.

Since this remains so forever, there are also only finitely many numbers
definable.

Does not follow if you have infinite time.

--
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.