sn=1+1/2+1/3+...1/n,证明:当n≥2时,sn^2≥2(s2/2+s3/3+...sn/n)
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sn=1+1/2+1/3+...1/n,证明:当n≥2时,sn^2≥2(s2/2+s3/3+...sn/n)
![sn=1+1/2+1/3+...1/n,证明:当n≥2时,sn^2≥2(s2/2+s3/3+...sn/n)](/uploads/image/z/16540884-36-4.jpg?t=sn%3D1%2B1%2F2%2B1%2F3%2B...1%2Fn%2C%E8%AF%81%E6%98%8E%EF%BC%9A%E5%BD%93n%E2%89%A52%E6%97%B6%2Csn%5E2%E2%89%A52%EF%BC%88s2%2F2%2Bs3%2F3%2B...sn%2Fn%29)
Sn^2=(S(n-1)+1/n)^2=[S(n-1)]^2+2S(n-1)/n+1/n^2
=[S(n-1)]^2+[2S(n-1)+2/n]/n-1/n^2
=[S(n-1)]^2+2Sn/n-1/n^2
以此类推可得
Sn^2=S1^2-1/2^2-1/3^2-……-1/n^2+2(s2/2+s3/3+...sn/n)
只要证明n>=2时,
S1^2-1/2^2-1/3^2-……-1/n^2>=0即可
因为,S1=1
且对任意n>=1均有,1/n-1/(n+1)^2
=[(n+1)^2-n]/n(n+1)^2=(n^2+n+1)/n(n+1)^2>(n^2+n)/n(n+1)^2=1/(n+1)
故S1^2-1/2^2-1/3^2-……-1/n^2
=(1/1^2-1/2^2)-1/3^2-……-1/n^2
>1/2-1/3^2-……-1/n^2
>1/3-……-1/n^2
>1/(n-1)-1/n^2>0
原式得证.
=[S(n-1)]^2+[2S(n-1)+2/n]/n-1/n^2
=[S(n-1)]^2+2Sn/n-1/n^2
以此类推可得
Sn^2=S1^2-1/2^2-1/3^2-……-1/n^2+2(s2/2+s3/3+...sn/n)
只要证明n>=2时,
S1^2-1/2^2-1/3^2-……-1/n^2>=0即可
因为,S1=1
且对任意n>=1均有,1/n-1/(n+1)^2
=[(n+1)^2-n]/n(n+1)^2=(n^2+n+1)/n(n+1)^2>(n^2+n)/n(n+1)^2=1/(n+1)
故S1^2-1/2^2-1/3^2-……-1/n^2
=(1/1^2-1/2^2)-1/3^2-……-1/n^2
>1/2-1/3^2-……-1/n^2
>1/3-……-1/n^2
>1/(n-1)-1/n^2>0
原式得证.
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