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I was wondering if there is any known sum of reciprocals of distinct odd prime numbers such that $$\sum_{k=1}^{n}\frac{1}{p_k}=1$$ Could someone give an example of one, or tell if there is none known? Or maybe it is impossible to find one, then it would be great to know the proof.

Thanks!

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    $\begingroup$ Hint: show that, in any such sum, the denominator of the sum is divisible by $p_1$ but the numerator is not. $\endgroup$
    – lulu
    Jul 2, 2020 at 14:55
  • $\begingroup$ @lulu, thanks for your hint and the link! $\endgroup$ Jul 2, 2020 at 15:12

1 Answer 1

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Multiply both sides by $\prod_{j=1}^n p_j$ and you obtain $$\sum_{k=1}^n\left(\prod_{j\ne k} p_j\right)=\prod_{j=1}^n p_j\\ \prod_{j=1}^{n-1}p_j+p_n\sum_{k=1}^{n-1}\prod_{j\ne k\\ j<n}p_j=p_n\prod_{j=1}^{n-1} p_j\\\prod_{j=1}^{n-1}p_j=p_n\left(\prod_{j=1}^{n-1}p_j-\sum_{k=1}^{n-1}\prod_{j\ne k\\ j<n}p_j\right)$$

And therefore $p_n\mid\prod_{j=1}^{n-1}p_j$, which is impossible by $p_1,\cdots,p_n$ being distinct primes.

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  • $\begingroup$ Thanks @Gaee. S.! Nice answer $\endgroup$ Jul 2, 2020 at 15:13

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