Question

已知数列(an),(bn)满足a1=2,an-1=an(an+1-1),bn=an-1.数列（bn）的前n项和为sn

Answer 1

（1）由bn=an-1得an=bn+1代入an-1=an（an+1-1）得bn=（bn+1）bn+1

整理得bn-bn+1=bnbn+1,

∵bn≠0否则an=1,与a1=2矛盾

从而得

1/(bn+1)−1/bn＝1

∵b1=a1-1=1

∴数列{1/bn}是首项为1,公差为1的等差数列

∴1/bn＝n,即bn＝1/n

（2）∵Sn＝1+1/2+1/3+.+1/n

∴Tn=S2n-Sn=1+1/2 +1/3+.+1/n+1/(n+1)+. +1/2n−(1+1/2 +1/3 +.+1/n)

=1/(n+1) +1/(n+2) +.+1/2n

∵2n+1＜2n+2

∴1/(2n+1)＞1/(2n+2)

∴Tn+1−Tn＞1/(2n+2) +1/(2n+2)−1/(n+1) ＝0

∴Tn+1＞Tn．

（3）用数学归纳法证明：

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