• The existence of dark numbers proved by the thinned out harmonic series

    From WM@21:1/5 to All on Fri Mar 7 10:37:08 2025
    The harmonic series diverges. Kempner has shown in 1914 that when all
    terms containing the digit 9 are removed, the series 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 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.

    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.

    Regards, WM

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