已知函数f(x)=(2x+3)/3x,数列{an}满足a1=1,a(n+1)=f(1/an),n属于N^+ (1)求数列

(1) a(n+1)=f(1/an)=(2/an+3)/(3/an)=(3an+2)/3=an+2/3
a1=1
an=1+2(n-1)/3=(2n+1)/3
(2) ana(n+1)=(2n+1)(2n+3)/9=4/9*n^2+8/9*n+1/3
Tn=a1a2-a2a3+a3a4-a4a5+...+a(2n-1)a2n-a2na(2n+1)
=4/9*(1^2-2^2+3^2-4^2+...-(2n)^2)+8/9*(1-2+3-4+...-2n)
=4/9*(-n)(2n+1)+8/9*(-n)
=-4n(2n+3)/9
(3) bn=1/a(n-1)an=9/(2n-1)(2n+1)=9/2*[1/(2n-1)-1/(2n+1)]
Sn=b1+b2+...+bn=9/2*[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]=9n/(2n+1)
Sn=9n/(2n+1)18n/(2n+1)+2002=(18n+9-9)/(2n+1)+2002=2011-9/(2n+1)对一切n属于N^+成立
所以最小正整数m=2011