a(n+1)=f[a(n)]=[3a(n)+2]/[a(n)+2],
a(n+1)+1=[3a(n)+2]/[a(n)+2] + 1 = 4[a(n)+1]/[a(n)+2],
若a(n+1)=-1,則a(n)=-1,...,a(1)=-1與a(1)=1/2矛盾,因此,a(n)不為-1.
1/[a(n+1)+1] = (1/4)[a(n)+1+1]/[a(n)+1] = (1/4)/[a(n)+1] + 1/4,
b(n+1)=1/[a(n+1)+1] = (1/4)/[a(n)+1] + 1/4 = b(n)/4 + 1/4,
b(n+1)-1/3 = b(n)/4 + 1/4 - 1/3 = b(n)/4 - 1/12 = [b(n)-1/3]/4
{b(n)-1/3}是首項為b(1)-1/3=1/[a(1)+1]-1/3=1/[1/2+1] - 1/3=1/3,公比為(1/4)的等比數(shù)列.
b(n)-1/3=(1/3)(1/4)^(n-1),
b(n) = 1/3 + (1/3)(1/4)^(n-1)