Question

若数列an的通项an=2n-1设数列bn的通项bn=1+1/an记Tn是数列bn前n项积(1)求T1,T2,T3的值(2

Answer 1

(1)bn=2n/(2n-1) T1=2 ,T2=8/3 ,T3=16/5

(2)注意:2k/(2k-1)=1+1/(2k-1)>1+1/2k=(2k+1)/2k

于是:Tn^2=(2n/(2n-1))(2n/(2n-1))((2n-2)/(2n-3))((2n-2)/(2n-3))……(4/3)(4/3)(2/1)(2/1)

>((2n+1)/2n)(2n/(2n-1))((2n-1)/(2n-2))((2n-2)/(2n-3))……(5/4)(4/3)(3/2)(2/1)

=2n+1

所以:Tn>√(2n+1)=√a(n+1)