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  • Dark numbers

    From WM@21:1/5 to All on Tue Apr 8 20:09:48 2025
    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.

    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.

    The terms are tiny but that part of the series diverges. This is a proof
    of the huge set of undefinable or dark numbers.

    Regards, WM

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  • From Alan Mackenzie@21:1/5 to wolfgang.mueckenheim@tha.de on Tue Apr 8 18:56:25 2025
    WM <wolfgang.mueckenheim@tha.de> wrote:
    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.

    [ .... ]

    This is the same as a previous thread you started a few weeks ago. It
    is thus spam.

    Please stop spamming this newsgroup.

    Regards, WM

    --
    Alan Mackenzie (Nuremberg, Germany).

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  • From WM@21:1/5 to Alan Mackenzie on Tue Apr 8 21:28:10 2025
    On 08.04.2025 20:56, Alan Mackenzie wrote:
    WM <wolfgang.mueckenheim@tha.de> wrote:
    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.

    Correction:

    https://www.hs-augsburg.de/~mueckenh/HI/HI02.PPT, p. 15.
    [ .... ]

    This is the same as a previous thread you started a few weeks ago. It
    is thus spam.

    It has not yet been understood by any of you. Try it again. Maybe you
    will succeed to understand one of the most important mathematical
    discoveries since Hippasos of Metapont.

    Regards, WM

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  • From Moebius@21:1/5 to All on Tue Apr 8 21:29:08 2025
    Am 08.04.2025 um 20:56 schrieb Alan Mackenzie:
    WM <wolfgang.mueckenheim@tha.de> wrote:

    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.

    [ .... ]

    This is the same as a previous thread you started a few weeks ago. It
    is thus spam.

    You think?

    Please stop spamming this newsgroup.

    Regards, WM


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  • From joes@21:1/5 to All on Sun Apr 13 08:43:58 2025
    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.

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  • From WM@21:1/5 to joes on Sun Apr 13 17:42:07 2025
    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.

    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.

    Regards, WM

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  • From joes@21:1/5 to All on Mon Apr 14 08:48:58 2025
    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.

    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.

    --
    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.

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  • From WM@21:1/5 to joes on Mon Apr 14 13:52:44 2025
    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.

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

    Regards, WM


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  • From joes@21:1/5 to All on Mon Apr 14 12:20:09 2025
    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.

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