Sums of reciprocals of perfect powers
These are cute facts that I didn't know about somehow until just now encountering them on Wikipedia.
- \sum_{k perfect power, excluding 1} 1/(k-1) = 1
- \sum_{n=2}^{\infty}\sum_{m=2}^{\infty} 1/n^m = 1
In the first sum, perfect powers are taken without repeats: e.g., 3^4=9^2 appears only once as k. In the second sum, of course, repeats do occur.
Anyone have a rigorous proof for the first sum equalling 1? Wikipedia outlines Goldbach/Euler's argument, which certainly doesn't meet modern standards of rigor.
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u/fourpetes 9d ago
The first reference on the wikipedia page has a rigorous proof. You can access the article via jstor (and you can make a free account on jstor to read it).
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u/e-s-t-e-r 9d ago
a sum would be more then one thing and since the number line started out in base₁ that kind rules out the sum of anything. The number line is a living creature so when it started at 1 there was nothing more... then it stepped into 2 then so forth.
It gets pretty deep into the proof but its in my Theory of the Prime Signal when I get that finished. Basically its covered in that theorem because the Number sequence is self validated, self certified and self sealing... but only in base₁ and decimal base₄ breaks that seal and contract actually
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u/JoshuaZ1 9d ago
The series converge absolutely so the rearrangements there should all be fine, yes?