用归纳法证明
n=2时,1+1/3>√5/2成立
设n=k时:
(1+1/3)(1+1/5)(1+1/7)…(1+1/(2k—1))>√(2k+1)/2成立
则n=k+1时
(1+1/3)(1+1/5)(1+1/7)…(1+1/(2k—1))(1+1/(2k+1))
>(1+1/(2k+1))*√(2k+1)/2
>1/2*[√(2k+1)+1/√(2k+1)]
=1/2*√[2k+1+2+1/(2k+1)]
>√(2k+3)/2
也成立,故对n>1,n属于正整数都成立
用归纳法证明
n=2时,1+1/3>√5/2成立
设n=k时:
(1+1/3)(1+1/5)(1+1/7)…(1+1/(2k—1))>√(2k+1)/2成立
则n=k+1时
(1+1/3)(1+1/5)(1+1/7)…(1+1/(2k—1))(1+1/(2k+1))
>(1+1/(2k+1))*√(2k+1)/2
>1/2*[√(2k+1)+1/√(2k+1)]
=1/2*√[2k+1+2+1/(2k+1)]
>√(2k+3)/2
也成立,故对n>1,n属于正整数都成立