Question

用放缩法证明: 1/2-1/(n+1)<1/(2^2)+1/(3^3)+````+1/(n^2)<(n-1)/n (n=2,3,````)

用放缩法证明: 1/2-1/(n+1)<1/(2^2)+1/(3^3)+````+1/(n^2)<(n-1)/n (n=2,3,````)

Answer 1

1/(2*2)+1/(3*3)+````+1/(n^2)<1/(1*2)+1/(3*3)+````+1/(n*(n-1)=(1-1/2)+(1/2-1/3) +...(1/(n-1)-1/n)
=1-1/n=(n-1)/n, (n=2,3,````)
1/(2*2)+1/(3*3)+````+1/(n^2)＞1/(2*3)+1/(3*4)+````+1/n*（n+1）=(1/2-1/3) +...(1/n-1/（n+1）)
=1/2-1/（n+1）=(n-1)/n, (n=2,3,````)
1/2-1/(n+1)<1/(2*2)+1/(3*3)+````+1/(n^2)<(n-1)/n (n=2,3,````)
不清楚请追问，晚上好。。。

追问

请问为啥放大的那一步 最后加的是(1/n-1/（n+1）)啊

追答

因为最后一项是1/n*（n+1）=1/n-1/（n+1） 相对应的

Answer 2

1/(2^2)+1/(3^3)+````+1/(n^2)>1/2*3+1/3*4+.....+1/n(n+1)=1/2-1/3+1/3-1/4+....+1/n-1/(n+1)
=1/2-1/(n+1)
1/(2^2)+1/(3^3)+````+1/(n^2)<1/2*1+1/3*2+....1/n(n-1)=1/2+1/2-1/3+1/3-1/4+.....+1/(n-1)-1/n
=1/2+1/2-1/n
=1-1/n
=(n-1)/n

Answer 3

1/(2^2)+1/(3^2)+````+1/(n^2)>1/(2*3)+1/(3*4)+````+1/[n*(n+1)]=1/2-1/3+1/3-1/4+...+1/n-1/(n+1)=1/2-1/(n+1);
1/(2^2)+1/(3^2)+````+1/(n^2)<1/(1*2)+1/(2*3)+````+1/[(n-1)*n]=1-1/2+1/2-1/3+...+1/(n-1)-1/n=1-1/n.
n=2,3,...

Answer 4

用数学归纳法啊。。。明摆着的