已知在数列an中,a1=1/2,an+1=3an/an+3,已知bn的前n项和为sn
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已知在数列an中,a1=1/2,an+1=3an/an+3,已知bn的前n项和为sn
,且对任意正整数N,都有bn·n(3-4an)/an=1成立,求证,1/2≤sn<1
,且对任意正整数N,都有bn·n(3-4an)/an=1成立,求证,1/2≤sn<1
![已知在数列an中,a1=1/2,an+1=3an/an+3,已知bn的前n项和为sn](/uploads/image/z/4466480-32-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%A8%E6%95%B0%E5%88%97an%E4%B8%AD%2Ca1%3D1%2F2%2Can%2B1%3D3an%2Fan%2B3%2C%E5%B7%B2%E7%9F%A5bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn)
a(n+1)=3a(n)/[a(n)+3],
若 a(n+1)=0,则, a(n)=0, ..., a(1)=0,与a(1)=1/2矛盾.
因此,a(n)不为0.
1/a(n+1) = (1/3)[a(n)+3]/a(n) = 1/a(n) + 1/3,
{1/a(n)}是首项为1/a(1)=2,公差为1/3的等差数列.
1/a(n) = 2 + (n-1)/3 = (n+5)/3,
a(n) = 3/(n+5).
3-4a(n)=3-12/(n+5)=(3n+15-12)/(n+5)=3(n+1)/(n+5),
[3-4a(n)]/a(n) = 3(n+1)/(n+5) * (n+5)/3 = n+1,
1 = b(n)*n[3-4a(n)]/a(n) = b(n)*n(n+1),
b(n) = 1/[n(n+1)] = 1/n - 1/(n+1).
s(n) = b(1)+b(2)+...+b(n-1)+b(n)
=1/1-1/2 + 1/2-1/3 + ... + 1/(n-1)-1/n + 1/n-1/(n+1)
=1 - 1/(n+1)
=n/(n+1),
=1时,
n/(n+1)-1/2 = (2n-n-1)/[2n+2]=(n-1)/(2n+2)>=0,
所以,n/(n+1)>=1/2.
因此,
1/2
若 a(n+1)=0,则, a(n)=0, ..., a(1)=0,与a(1)=1/2矛盾.
因此,a(n)不为0.
1/a(n+1) = (1/3)[a(n)+3]/a(n) = 1/a(n) + 1/3,
{1/a(n)}是首项为1/a(1)=2,公差为1/3的等差数列.
1/a(n) = 2 + (n-1)/3 = (n+5)/3,
a(n) = 3/(n+5).
3-4a(n)=3-12/(n+5)=(3n+15-12)/(n+5)=3(n+1)/(n+5),
[3-4a(n)]/a(n) = 3(n+1)/(n+5) * (n+5)/3 = n+1,
1 = b(n)*n[3-4a(n)]/a(n) = b(n)*n(n+1),
b(n) = 1/[n(n+1)] = 1/n - 1/(n+1).
s(n) = b(1)+b(2)+...+b(n-1)+b(n)
=1/1-1/2 + 1/2-1/3 + ... + 1/(n-1)-1/n + 1/n-1/(n+1)
=1 - 1/(n+1)
=n/(n+1),
=1时,
n/(n+1)-1/2 = (2n-n-1)/[2n+2]=(n-1)/(2n+2)>=0,
所以,n/(n+1)>=1/2.
因此,
1/2
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