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!
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.