考虑函数f(x) = 2x/(x+2)-ln(1+x).
有f'(x) = 4/(x+2)²-1/(1+x) = -x²/((x+2)²(x+1)).
x > 0时, f'(x) < 0, 故f(x)严格单调递减.
有f(x) < f(0) = 0, 即得2x/(x+2) < ln(1+x).
于是对任意正整数k, 成立1/ln(1+k) < (k+2)/(2k) = 1/2+1/k.
对k从1到n求和即得1/ln(2)+1/ln(3)+...+1/ln(n+1) < n/2+1+1/2+...+1/n.