已知数列a1=1/2且an+1=an/1+an(1)证明数列1/an是等差数列并求an的通项公式(2求数列an/n的前n项和Tn

Question

已知数列 a1=1/2 且an+1=an/1+an (1)证明数列1/an是等差数列 并求an的通项公式 (2求数列an/n的前n项和Tn

Answer 1

1.
a(n+1)=an/(1+an)
1/a(n+1)=(1+an)/an=1/an +1
1/a(n+1)-1/an=1,为定值.
1/a1=1/(1/2)=2
数列{1/an}是以2为首项,1为公差的等差数列.
1/an=2+1×(n-1)=n+1
an=1/(n+1)
2.
an/n=[1/(n+1)]/n=1/[n(n+1)]=1/n -1/(n+1)
Tn=a1/1+a2/2+...+an/n
=1/1-1/2+1/2-1/3+...+1/n -1/(n+1)
=1- 1/(n+1)
=n/(n+1)