题目

已知数列(An)中,A1=1/3,AnA（n-1）=A（n-1）-An(n>=2),数列Bn满足Bn=1/An
求数列Bn的通项公式
求数列{an/n}的前n项和Tn,并证明Tn

提问时间：2020-11-08

答案

a(n+1)a(n) = a(n) - a(n+1),
若a(n+1)=0,则a(n)=0,...,a(1)=0,与a(1)=1/3矛盾.
因此,a(n)不为0.
1 = 1/a(n+1) - 1/a(n),
1/a(n+1) = 1/a(n) + 1,
{b(n)=1/a(n)}是首项为1/a(1)=3,公差为1的等差数列.
b(n)=1/a(n) = 3 + (n-1) = n+2.
a(n) = 1/(n+2),
c(n) = a(n)/n = 1/[n(n+2)] = (n+1)/[n(n+1)(n+2)] = [(n+2) - 1]/[n(n+1)(n+2)] = 1/[n(n+1)] - 1/[n(n+1)(n+2)],
=1/n - 1/(n+1) - (1/2){ 1/[n(n+1)] - 1/[(n+1)(n+2)] }
t(n) = c(1)+c(2)+...+c(n-1)+c(n)
= 1/1-1/2 + 1/2-1/3 + ...+ 1/(n-1)-1/n + 1/n-1/(n+1)-(1/2){1/[1*2]-1/[2*3] + 1/[2*3]-1/[3*4] + ...+ 1/[(n-1)n]-1/[n(n+1)] + 1/[n(n+1)] - 1/[(n+1)(n+2)] }
= 1/1 - 1/(n+1) - (1/2){ 1/[1*2] - 1/[(n+1)(n+2)] }
=1 - 1/(n+1) - 1/4 + 1/[2(n+1)(n+2)]
= 3/4 + [1-2(n+2)]/[2(n+1)(n+2)]
= 3/4 - (2n+3)/[2(n+1)(n+2)]
< 3/4 - (2n+2)/[(2n+2)(n+2)]
= 3/4 - 1/(n+2)