(1) proof: f (x) = ㏑ (1+x)-X. (x >-1). Derive: f'(x)=-x/( 1+x)。 It's easy to know, f' At - 1，f (x) ≤ f (x) max = 0。 = = >; ㏑ (1+x)-x ≤ 0. = => ㏑ (1+x) ≤ X. Where x >- 1, the equal sign is obtained only when x=0. (2)㏑2+㏑a(n+ 1)= a(n+ 1)an+f[a(n+ 1)an]=㏑[ 1+a]2a(n+ 1)= 1+a(n+ 1)an。 = = = & gtAn = n/(n+ 1)。 (n = 1, 2,3, ...). (3) Proof: According to Euler formula, ∑ai=a 1+a2+a3+...+an=( 1/2)+(2/3)+(3/4)+...+n/(n+ 1)=[ 1-( 1/2)]+[ 1-( 1/3)]+[ 1-( 1/4) ]+...+[ 1- 1/(n+ 1)]= n-[ 1/2+ 1/3+ 1/4+]...+ 1/(n+ 1)]=(n+ 1)-[ 1+ 1/ 2+1/3+1/4+...+1(n+1)] = (n+1)-[㏑ (1+) That is, ∑ ai = (n+1)-∑ (1+n)-K. difference = [n+∑ 2-∑ (n+2)]-∑ ai = difference ≥ ㏑ (5/3)+k- Therefore, A 1+A2+A3+...+An < N+.