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]... 展开
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 收起