Sn=n^2,令bn=1/anan+1,Tn是数列bn的前n项和,试证明Tn

Sn=n^2n>=2时,An=Sn-S(n-1)=n^2-(n-1)^2=2n-1n=1时,A1=S1=1也满足上式Bn=1/AnA(n+1)=1/[(2n-1)(2n+1)]=(1/2)[(2n+1)-(2n-1)]/[(2n-1)(2n+1)]=(1/2)[1/(2n-1)-1/(2n+1)]Tn=B1+B2+……+Bn=(1/2)[(1/1-1/3)+(1/3-1/5)+...