# sum of the reciprocal of the prime factors of a square free number

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?

• perfect numbers ... – user645636 Feb 10 at 18:56
• 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) – Maximilian Janisch Feb 10 at 18:58
• Take $n=30=2\times 3\times 5$. Note that $\frac 12+\frac 13+\frac 15=\frac {31}{30}$. – lulu Feb 10 at 19:03
• Sorry, I mean an odd number Typo fixed. – 674123173797 - 4 Feb 10 at 19:15

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.