Question

用数学归纳法证明1/n+1/(n+1)+1/(n+2)+……1/n²>1(n∈N*,n>1)

Answer 1

n=2略

n=k时有1/k+1/(k+1)+……+1/k²>1
k≥2
令a=1/k+1/(k+1)+……+1/k²>1
则n=k+1
1/(k+1)+1/(k+2)+……+1/(k+1)²
=a-1/k+1/(k²+1)+……+1/(k+1)²
因为1/(k²+1)>1/(k+1)²
1/(k²+2)>1/(k+1)²
……
所以a-1/k+1/(k²+1)+……+1/(k+1)²>a-1/k+1/(k+1)²+……+1/(k+1)²
=a-1/k+(2k+1)*1/(k+1)²
=a+(2k²+k-k²-2k-1)/k(k+1)²
=a+[(k-1/2)²-5/4]k(k+1)²
k≥2
所以a+[(k-1/2)²-5/4]k(k+1)²>a>1
所以n=k+1
1/(k+1)+1/(k+2)+……+1/(k+1)²>1

综上，……

Answer 2

n=2

1/2+1/3+1/4=(6+4+3)/12=13/12>1

成立

设n=k时，关系成立

1/k+1/（k+1）+...+1/k²>1

n=k+1时

左边=1/（k+1）+1/（k+2）+...+1/k²+1/（k²+1）+...+1/（k+1）²

=1/k+1/（k+1）+1/（k+2）+...+1/k²+1/（k²+1）+...+1/（k+1）²-1/k

>1+1/（k²+1）+...+1/（k+1）²-1/k

要证明：

1/（k²+1）+...+1/（k+1）²-1/k≥0

(k+1）²-k²=2k+1项，

1/（k²+1）+...+1/（k+1）²>（2k+1）[1/（k²+1）(k²+2）...（k+1）²]^(2k+1)

>（2k+1）/(k+1)²

1/（k²+1）+...+1/（k+1）²-1/k>（2k+1）/(k+1)²-1/k

=[k(2k+1)-(k+1)²]/k(k+1)²

=(2k²+k-k²-2k-1）/k(k+1)²

=（k²-k-1）/k(k+1)²

=[(k-1/2)²-5/4]/k(k+1)²

k≥2，

k-1/2≥2-1/2=3/2

(k-1)²≥6/4

(k-1)²-5/4≥6/4-5/4=1/4>0

∴1/（k²+1）+...+1/（k+1）²-1/k≥0成立。

得证.

Answer 3

证：
n=2时，1/2 +1/3 +1/4=13/12>1，不等式成立。
假设当n=k(k∈N*)时，不等式成立，即1/k +1/(k+1)+...+1/k²>1
则当n=k+1时
1/(k+1)+1/(k+2)+...+1/k²+1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1)
=[1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1) -1/k] +[1/k +1/(k+1)+...+1/k²]
>[1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1) -1/k] +1
1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1) -1/k
>1/(k+1)²+ 1/(k+1)²+...+1/(k+1)² -1/k
=(2k+1)/(k+1)² -1/k
=[k(2k+1)-(k+1)²]/[k(k+1)²]
=(k²-k-1)/[k(k+1)²]
=(k²-k-2+1)/[k(k+1)²]
=[(k+1)(k-2)+1]/[k(k+1)²]
k≥2，(k+1)(k-2)≥0，1>0，(k+1)²>0
[(k+1)(k-2)+1]/[k(k+1)²]>0
1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1) -1/k
>1/(k+1)²+ 1/(k+1)²+...+1/(k+1)² -1/k
>0
1/(k+1)+1/(k+2)+...+1/k²+1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1)
>[1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1) -1/k] +1
>1
不等式同样成立。
k为任意正整数，因此对于任意正整数n，不等式恒成立。
即：1/n +1/(n+1)+ 1/(n+2)+...+1/n²>1

解题思路：
1、运用数学归纳法，先解得n=2时不等式成立；再设n=k，进而证明n=k+1时不等式成立。
2、在证明过程中，用到了放缩法。