(1)由题意,当n=1时,a1=S1=,则a1=1,a2=2,则a2-a1=1, 当n≥2时,an=Sn-Sn-1=-=[nan-(n-1)an-1+1]an+1=[(n+1)an+1-nan+1] 则an+1-an=[(n+1)an+1-2nan+(n-1)an-1], 即(n-1)an+1-2(n-1)an+(n-1)an-1=0, 即an+1-2an+an-1="0, " 即an+1-an=an-an-1 则数列{an+1-an}是首项为1,公差为0的等差数列. 从而an-an-1=1,,则数列{an}是首项为1,公差为1的等差数列, 所以,an=n(n∈N*) (2)bn===(- ) 所以,Tn=b1+b2+…+bn=[(1-)+(-)+…+(-)] =(1-)= 由于Tn+1-Tn=-=>0, 因此Tn单调递增,故Tn的最小值为T1= 令>,得k<19,所k的最大值为18 |