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[-] eestileib@sh.itjust.works 5 points 1 week ago

Should be anything less than a harmonic decrease (that is, the nth centaur is 1/n the size of the original).

The harmonic series is the slowest-diverging series.

[-] kogasa@programming.dev 1 points 1 week ago

The assumption is that the size decreases geometrically, which is reasonable for this kind of self similarity. You can't just say "less than harmonic" though, I mean 1/(2n) is "slower".

[-] eestileib@sh.itjust.works 2 points 1 week ago

Eh, that's just 1/2 of the harmonic sum, which diverges.

[-] kogasa@programming.dev 2 points 1 week ago

Yes, but it proves that termwise comparison with the harmonic series isn't sufficient to tell if a series diverges.

[-] eestileib@sh.itjust.works 3 points 1 week ago

Very well, today I accede to your superior pedantry.

But one day I shall return!

this post was submitted on 23 Jun 2024
251 points (95.0% liked)

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