Let Hn = 1/1 + 1/2 + ... + 1/n.
Show that, for n > 1, Hn is not an integer.
Consider, for n > 1, Hn = 1/1 + 1/2 + ... + 1/n, expressed as a fraction, a/b, where b is the least common multiple of 2, 3, ..., n.
Then b = 2r· s, where 2r is the largest power of 2 less than or equal to n, and s is an odd number.
As an example, consider a and b, for H5:
b = 22· s. (In fact, s = 15.)
a = 22· s + 2s + 22· (s/3) + s + 22· (s/5).
Hence a is odd, as it is the sum of one odd number, s, and several even numbers.
Clearly this argument can be generalized.
For any Hn, n > 1, a is the sum of a single odd number, coming from the largest power of 2 less than or equal to n, and n − 1 even numbers.
Therefore a is odd.
Obviously, b is even.
Hence none of the twos in the prime factorization of the denominator can be cancelled from the numerator.
So, when a/b is written as a fraction in its lowest terms, c/d, d > 1.
Therefore, for n > 1, Hn is not an integer.
Similarly, 1/m + ... + 1/n is never an integer, for 1 < m < n.
Source: Mauro Maggioni. (Puzzle page since taken down.)