The problem was to show that
Hn=n∑k=11k=1+12+⋯+1n
is never an integer for n>1.
Solution
There are multiple solutions. One of them uses Bertrand's postulate, which says that there is a prime between n and 2n for any n>1.
Now consider the largest prime p in 1,2,…,n.
The denominator of Hn is divisible by p. The numerator has terms divisible by p, plus the term 1×2×⋯×(p−1)×(p+1)×⋯×n
By Bertrand's postulate, we have that 2p>n (otherwise p won't be the largest prime).
Thus the numerator is not divisible by p. Thus Hn is not an integer.
Hn=n∑k=11k=1+12+⋯+1n
is never an integer for n>1.
Solution
There are multiple solutions. One of them uses Bertrand's postulate, which says that there is a prime between n and 2n for any n>1.
Now consider the largest prime p in 1,2,…,n.
The denominator of Hn is divisible by p. The numerator has terms divisible by p, plus the term 1×2×⋯×(p−1)×(p+1)×⋯×n
By Bertrand's postulate, we have that 2p>n (otherwise p won't be the largest prime).
Thus the numerator is not divisible by p. Thus Hn is not an integer.