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 2021-08-12, 12:43 #4 Dr Sardonicus     Feb 2017 Nowhere 3·1,931 Posts The fact that, for each prime p, only *one* of the fractions has a denominator divisible by p is indeed the key here. A slightly more sophisticated argument shows that for n > 1, the "harmonic numbers" $H_{n}\;=\;\sum_{i=1}^{n}1/i$ all have denominators divisible by 2 - and, in fact, the power of 2 dividing the denominator never goes down, and increases every time n is a power of 2. The sum of fractions with a common factor in the denominator can be a fraction without that factor in the denominator, e.g. 1/3 + 1/6 = 1/2. In fact, the common factor can show up in the numerator of the sum! We have the following well known result: If p > 3 is prime, then Hp-1 has a numerator divisible by p2 ["Wolstenholme's Theorem"]. It follows that if p > 3 is prime, then 1/p + 1/(2*p) + ... + 1/((p-1)*p) has a numerator divisible by p.