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Let $n$ an odd square free number, and $p_1, \ldots , p_n$ their distinct prime factors. Ir is true that $$ \sum\limits_{i=1}^n \frac{1}{p_i} < 1? $$

Otherwise, there exists some conditions to ensure that this works?

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  • $\begingroup$ perfect numbers ... $\endgroup$ – user645636 Feb 10 at 18:56
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    $\begingroup$ Is it on purpose that $n$ is the same number as the number of the prime factors of $n$ (there is no such number, so the statement would be true)? Otherwise, then the sum doesn't have to be less than $1$ (indeed, consider the product of the first $n$ prime numbers) $\endgroup$ – Maximilian Janisch Feb 10 at 18:58
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    $\begingroup$ Take $n=30=2\times 3\times 5$. Note that $\frac 12+\frac 13+\frac 15=\frac {31}{30}$. $\endgroup$ – lulu Feb 10 at 19:03
  • $\begingroup$ Sorry, I mean an odd number Typo fixed. $\endgroup$ – 674123173797 - 4 Feb 10 at 19:15
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The sum of the reciprocals of the primes diverges, see, e.g., this question, thus the answer to your question is No.

Mind you, it diverges very slowly. The first odd counterexample is $$N= 3234846615=3\times 5\times 7\times 11\times 13\times 17\times 19\times 23\times 29 $$

For that number, the sum comes to just over $1$ (see this). But don't let the slow speed of divergence fool you. If you put in enough primes you can make the sum as large as you want.

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