已知数列{an}中,a1=3\5,an.a(n-1)=2a(n-1)(n≥2,n∈N).数列{bn}是等差数列且满足bn=1\(an-1)(n∈N).求数列{an}中的最小值与最大值

(1)an*a(n-1)+1=2a(n-1)an=[2a(n-1)-1]/a(n-1)an -1= [2a(n-1)-1]/a(n-1) -1= [a(n-1)-1]/a(n-1)1/(an -1) = a(n-1)/[a(n-1)-1] = {[a(n-1)-1]+1} /[a(n-1)-1]=1+ 1/[a(n-1) -1]1/(an -1) -1/[a(n-1) -1] =1即 bn －...