∵a(n+1)=an(an+1)
∴1/[a(n+1)]=1/an×1/（an+1)=1/an-1/（an+1)
∴1/（an+1)=1/an-1/[a(n+1)]
∴s10=1/a1-1/a2+1/a2-1/a3+……+1/a10-1/a11=1/a1-1/a11
∵a1=1/3 a2=4/9 ……a4＞1 a(n+1)＞an
∴a11＞1 ∴0＜1/a11＜1
∵1/a1=3 ∴s10的整数部分是2
数列an满足a1=1/3,an+1=an^2+an,记sn=1/(a1+1)+1/(a2+1)+.+1/(1+an),求
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