Let h_{n} = ∑_{1}^{n} 1/r. Show that n - (n - 1) n^{-1/(n-1)} > h_{n} > n(n + 1)^{1/n} - n for n > 2.

**Solution**

Consider the numbers: 1 + 1/1, 1 + 1/2, 1 + 1/3, ... , 1 + 1/n. Their arithmetic mean is (n + h_{n})/n. The geometric mean is (∏ (1 + 1/r) )^{1/n}. But ∏ (1 + 1/r) = 2/1 3/2 4/3 ... (n+1)/n = n + 1. So the geometric mean is (n + 1)^{1/n}. The numbers are not all equal, so the arithmetic mean is strictly greater than the geometric mean. That gives the right inequality.

Similarly, consider the numbers: 1 - 1/2, 1 - 1/3, ... , 1 - 1/n. [One must be careful not to include 1 - 1/1, because that would make the geometric mean 0.] The arithmetic mean is 1 - (h_{n} - 1)/(n-1). The geometric mean is n^{-1/(n-1)} (the terms telescope is a similar way). Rearranging gives the left inequality.

© John Scholes

jscholes@kalva.demon.co.uk

27 Jan 2001