Question

已知各项均为正数的数列{an}满足,a1=1,a(n+1)·an+a(n+1)-an=0，求证：{1/an}是等差数列并求an

已知各项均为正数的数列{an}满足,a1=1,a(n+1)·an+a(n+1)-an=0，（1）求证：{1/an}是等差数列并求an。（2）设bn=an·a(n+2),数列{bn}的前n项和为S你，求证1/3≤Sn＜3/4

Answer 1

∵a(n+1)·an+a(n+1)-an=0
两边除以a(n+1)*an
∴1+1/an-1/a(n+1)=0
∴1/a(n+1)-1/an=1
∴{1/an}是等差数列
公差为1
∵a1=1,1/a1=1
∴1/an=1+(n-1)=n
∴an=1/n
(2)
bn=an*a(n+2)=1/[n(n+2)]=1/2[1/n-1/(n+2)]
∴{bn}前n项和
Sn=b1+b2+.......+bn
=1/2[1-1/3+1/2-1/4+1/3-1/5+.........+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1）-1/(n+2)]
=3/4-1/2[1/(n+1)+1/(n+2)]
∵1/2[1/(n+1)+1/(n+2)]>0
∴3/4-1/2[1/(n+1)+1/(n+2)]<3/4
即Sn<3/4
∵bn>0
∴n≥2时，Sn-S(n-1)=bn>0
∴Sn递增
∴Sn≥S1=b1=1/3
∴1/3≤Sn＜3/4