设bn=an比Sn平方,求证b1+b2+.bn<1

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设bn=an比Sn平方,求证b1+b2+.bn<1
设数列{an}的前n项和为Sn,且满足S1=2，Sn+1=3Sn+2(n=1,2,3)
设bn=2,Sn+1=3Sn+2(n=1,2,3.....)
设bn=an比Sn平方，求证b1+b2+b3.....bn

S(n+1) + 1 = 3(Sn + 1),{Sn + 1}是等比数列.
S1 + 1 = 3,故Sn + 1 = 3^n.故Sn = 3^n - 1.
a1 = S1 = 2,an = Sn - S(n-1) = 3^n - 3^(n-1) = 2 * 3^(n-1).
bn = an / (Sn)^2 = 2 * 3^(n-1) / (3^n - 1)^2.
证bn < 2/3 * (5/2)^n即可.

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