数列{An}中,A1=1 An=n^2[1+1/2*2+1/3*3+1/(n-1)*(n-1)] n>=2

数列{An}中,A1=1 An=n^2[1+1/2*2+1/3*3+1/(n-1)*(n-1)] n>=2
求证(An +1)/An+1=n^2/(n+1)^2
求证(1+1/A1)(1+1/A2)....(1+1/An)<4 n>=1

证明:
An +1=n^2[1+1/2^2+1/3^3+...+1/(n-1)^2]+1
=n^2[1+1/2^2+1/3^3+...+1/(n-1)^2+1/n^2]
A(n+1)=(n+1)^2[1+1/2^2+1/3^2+...+1/n^2]
所以(An +1)/A(n+1)=n^2/(n+1)^2

(1+/A1)(1+/A2)...(1+1/An)=[(A1+1)/A1][(A2+1)/A2]...[(An+1)/An]
=(1/A1)[(A1+1)/A2][(A2+1)/A3]...[(A(n-1)+1)/An](An+1)
=[(An+1)/A1][(1^2/2^2)*(2^2/3^2)...[(n-1)^2/n^2]
=[(An+1)/A1][1/n^2]
=(An+1)/n^2
=1+1/2^2+1/3^3+...+1/(n-1)^2+1/n^2
<1+[1/(1*2)]+[1/(2*3)]+...+[(1/((n-1)n)]
=1+[1-(1/2)]+[(1/2)-(1/3)]+...+[1/(n-1)-1/n]
=2-1/n<2
所以(1+1/A1)(1+1/A2)....(1+1/An)<4
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第1个回答  2008-09-08
所以(1+1/A1)(1+1/A2)....(1+1/An)<4
第2个回答  2008-09-08
fg
第3个回答  2008-09-10
2
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