设数列an满足a1=0且1/(1-an+1)-1/(1-an)=1,设bn=(1-根号an+1)/根号n,记Sn为bn的前n项和,证明Sn
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![设数列an满足a1=0且1/(1-an+1)-1/(1-an)=1,设bn=(1-根号an+1)/根号n,记Sn为bn的前n项和,证明Sn](/uploads/image/z/634611-3-1.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D0%E4%B8%941%2F%281-an%2B1%29-1%2F%281-an%29%3D1%2C%E8%AE%BEbn%3D%281-%E6%A0%B9%E5%8F%B7an%2B1%29%2F%E6%A0%B9%E5%8F%B7n%2C%E8%AE%B0Sn%E4%B8%BAbn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E8%AF%81%E6%98%8ESn)
设数列an满足a1=0且1/(1-an+1)-1/(1-an)=1,设bn=(1-根号an+1)/根号n,记Sn为bn的前n项和,证明Sn
设数列an满足a1=0且1/(1-an+1)-1/(1-an)=1,设bn=(1-根号an+1)/根号n,记Sn为bn的前n项和,证明Sn<1
设数列an满足a1=0且1/(1-an+1)-1/(1-an)=1,设bn=(1-根号an+1)/根号n,记Sn为bn的前n项和,证明Sn
证明:
令cn=1/(1-an),则c1=1/(1-a1)=1,所以:
c(n+1)-cn=1,是等差数列,即:
cn=c1+(n-1)=n,则:
an=(n-1)/n
bn=[1-√a(n+1)]/n
={1-√[n/(n+1)]} / n
=1/√n - 1/ √(n+1)
Sn=b1+...+bn=1-1/√2 +.+ 1/√n - 1/ √(n+1)=1- 1/ √(n+1)
n为正整数,所以上式中- 1/ √(n+1)<0,即:
Sn<1
h