When grouping by 9, you are only adding 1/8th of the numbers so S from 1 to infinity is not equal to adding only the factors of 9. You skip the 1/8th of the numbers in between.

This post was edited on 7/23/21 at 12:20 pm

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Posted by eatpie

Kentucky

Member since Aug 2018

1123 posts

Posted on 7/23/21 at 12:45 pm to GumboPot

This is a classic scenario:

Simply put, the assumed power of 9 is not constant something to do with the decline in frequency of prime numbers this higher you get.

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Posted by TigerstuckinMS

Member since Nov 2005

33687 posts

Posted on 7/23/21 at 12:50 pm to GumboPot

The series inside the parenthesis in the fourth line is NOT the same series as the series in the second line. The two sets don't map 1:1 to each other. Though each index in the set is equal (i.e. identical terms in each set are the same value, so the third term is 3 in both, the fourth 4, and so on) the size of the two sets is not the same. As more terms are added to the first set, the second set grows slower to keep the equality in the fifth line true. For instance, when you add three terms to the first set, the second set would only add one term to maintain the equality in the final equation. If you add "11, 12, 13" to the first set, the second adds "4". If you add "14, 15, 16" to the first, the second only gets "5".

It would be more proper to call the first set S0 and the second S1.
1 + 9(S1) = S0
S1 = (S0-1)/9

Recognizing that the two infinite sets can have identical terms at each index, yet not be identical to each other because they are different sizes is the key.

This post was edited on 7/23/21 at 1:42 pm

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