Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
3  
If p is the largest prime, the numerator can never be divisible by p. –  Qiaochu Yuan Jul 13 '10 at 22:33
4  
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.