Question

已知数列｛an｝满足a1=4/3,且an+1=〔4(n+1)an〕/(3an+n)(n∈N*).

(1)求1/a1+2/a2+…+n/an的值；

（2）设bn=an/n(n∈N*).，用数学归纳法证明b1b2b3.......bn<2

Answer 1

(1)
a(n+1)=[4(n+1)an]/(3an+n)
a(n+1)/(n+1) = 4an/(3an+n)
(n+1)/a(n+1) = (3an+n)/(4an)
= (1/4)(n/an) + 3/4
(n+1)/a(n+1)-1 = (1/4)[ n/an - 1 ]
{ n/an - 1 } 是等比数列, q= 1/4
n/an - 1 = (1/4)^n .(1/a1 - 1)
=-(1/4)^n
n/an = 1- (1/4)^n
1/a1+2/a2+...+n/an = n - (1/3)[ 1- (1/4)^n ]

追问

（2）设bn=an/n(n∈N*).，用数学归纳法证明b1b2b3.......bn<2
第一问我会，我想知道第二问怎么写

Answer 2

(2)先证明b1b2b3.......bn<2(4^n-1)/4^n,这个用数学归纳法好证。