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I am not a mathematician. I am reading David Applebaum's book "Limits, limits everywhere ", in which he gives a proof of the divergence of the series of square-free numbers on page 91. I don't understand where he gets the inequality that he uses to prove the theorem. I also don't understand whether the three sums in the inequality run over the same limits. Any light shed on this issue will be greatly appreciated.

terahertz
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    Even the sum of the reciprocals of the primes diverge. See, e.g., this question – lulu Jun 05 '20 at 14:54
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    Does this answer your question? Divergence of sum of reciprocals of square-free numbers – lulu Jun 05 '20 at 14:59
  • @lulu Very good point, we can even strenghten this (although for the question the primes are enough) : The sum of the reciprocals of the primes of the form $an+b$ with $n$ posisitive integer diverges whenever $a$ and $b$ are coprime positive integers. – Peter Jun 05 '20 at 15:00
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    @lulu These are in a sense equivalent. The sum of reciprocals of the square-free numbers is the product of $1 + 1/p$ over primes $p$. The infinite product $\prod_n (1+a_n)$ (where $a_n > 0$) diverges iff $\sum_n a_n$ diverges. – Robert Israel Jun 05 '20 at 15:02
  • @RobertIsrael Oh, agreed. I should note that the duplicate I linked to specifically seeks to avoid using the fact that $\sum \frac 1p$ diverges...thinking that the OP might find those arguments more elementary. – lulu Jun 05 '20 at 15:03

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