若n为大于1的自然数，求证：1⼀n+1+1⼀n+2+…+1⼀2n＞13⼀24

n为大于1的自然数
可以用数学归纳法来证：
（1）当n=2时
1/(2+1)+1/(2+2)=1/3+1/4=7/12=14/24>13/24成立
（2）假设当n=k时成立
即：1/(k+1)+1/(k+2)+1/(k+1)+---+1/(k+k)>13/24
那么当n=k+1时
1/(k+2)+1/(k+1)+---+1/(k+k)+1/(2k+1)+1/(2k+2)
=1/(k+1)+1/(k+2)+1/(k+1)+---+1/(k+k)+1/(2k+1)+1/(2k+2)-1/(k+1)
>13/24+1/(2k+1)+1/(2k+2)-1/(k+1)
>13/24+1/(2k+2)+1/(2k+2)-2/(2k+2)=13/24
说明当n=k+1时也成立
由（1）（2）可知不等式对于大于1的自然数都成立

数学归纳法证明，当n=1,左边=1/2>13/24成立，
假设n=k时也成立，即1/k+1+1/k+2+…+1/2k＞13/24

当n=k+1时，左边=1/(k+1)+1+1/k+1+2+…+1/2(k+1)=1/k+2+1/k+3+…+1/2(k+1)>13/24-1/(k+1)+1/2(k+1)+1/(2k+1)
现在只要证明-1/(k+1)+1/2(k+1)+1/(2k+1)>0就好了。-1/(k+1)+1/2(k+1)+1/(2k+1)=-1/2(k+1)+1/(2k+1)》0是成立的，所以由归纳法知道对所有自然数都成立。