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Old 2021-08-12, 10:33   #3
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"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

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Originally Posted by bur View Post
I'm struggling with this problem:

Show that the sum of the reciprocal of all primes \frac{1}{p_{1}} + \frac{1}{p_{2}} + ... + \frac{1}{p_{n}} = P_{n} is never an integer.

I know there are many solutions online, but at a quick glance they often give away the answer immediately or seem to take a different path, so I'd rather like a hint if my attempt is worthwhile:

I came up with the general form of the fraction which has primorial pn# as denominator and the numerator is the sum of the products of all possible combinations of n-1 primes, i.e. "n choose n-1" addends. For example:

\frac{1}{p_{1}} + \frac{1}{p_{2}} + \frac{1}{p_{3}} = \frac{p_{1}*p_{2}+p_{1}*p_{3}+p_{2}*p_{3}}{primorial(p_{3})}

A sum will never have a prime factor that isn't a prime factor of all addends. Since for all primes <= pn there is a addend that doesn't contain it, the sum, i.e. the numerator, will not be divisible by any of p1 ... pn. So it doesn't share any factor with the denominator. Thus, the fraction cannot be reduced to an integer.

Is that reasoning correct? And if so, I guess the proof would have to contain a proof for why the fraction is of that form?

(how do you produce a # in Tex here? \# or \text{#} didn't work...)
Since its denominator is always p1*p2*p3*...*pn and not 1
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