已知Sn=(2^n)+n-1,bn=n/(Sn-n+2) 设bn的前n项和为Tn,证明 Tn

差不多有题目,参考一下:
已知数列{an}的前N项和为Sn 且an+1=Sn-n+3,a1=2,设Bn=n/Sn-n+2前N项和为Tn 求证Tn 小于4/3
Sn+1 —Sn=an+1=Sn—n+3,
即Sn+1=2Sn-n+3,
所以Sn+1 -(n+1)+2=2(Sn-n+2)
又S1 -1+2=3,所以Sn-n+2=3*2^n-1,
所以bn=n/(3*2^n-1),
Tn=1/3 (1/2^0 +2/2 ^1+3/2^2+4/2^3+.+n/2^n-1) .1
0.5Tn=1/3 ( 1/2^1+2/2^2+.+n-1/2^n-1 +n/2^n).2
1式-2式 得0.5Tn=1/3(1 - n/2^n + 1/2+1/2^2+1/2^3.+1/2^n-1)
Tn=2/3(1 - n/2^n + 1-1/2^n-1)
=2/3{2-(2n+1)/2^n-1} 小于4/3