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It is trivial to show that the sum of the reciprocals of the first $k$ counting numbers can never be an integer. Is the statement reducible to a similar fact about primes?

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If p is the largest prime, the numerator can never be divisible by p. – Qiaochu Yuan Jul 13 '10 at 22:33
For any finite sum $S = \frac{1}{n_1} + \ldots + \frac{1}{n_k}$ with $n_1 = p$ a primeand all $n_2,\ldots,n_k$ prime to $p$, one has $\operatorname{ord}_p(S) = -1$ and therefore $S$ is not an integer. This is an easier argument than that of the partial sums of the harmonic series. – Pete L. Clark Jul 13 '10 at 22:41

Write $n = \frac{1}{p_1} + \frac{1}{p_2} + \cdots + \frac{1}{p_k}$ with the $p_i$ distinct primes. If you multiply each side by $p_1p_2\cdots p_k$, you will see that if $n \in \mathbb{N}$, then each $p_i$ divides the left hand side but not the right hand side, a contradiction.

There are genuinely difficult questions that look a little bit like this, though. See the Wikipedia entry on Egyptian fractions and the Erdős-Straus conjecture.

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