1/(n+1)+1/(n+2)+.+1/3n>5/6(n>=2,n属於N*)用数学归纳法证明

Question

1/(n+1)+1/(n+2)+.+1/3n>5/6(n>=2,n属於N*) 用数学归纳法证明

Answer 1

证明：n=2时,有1/3+1/4+1/5+1/6=0.95>5/6
假设n=k时,有1/(k+1)+1/(k+2)+.+1/3k>5/6成立
则当n=k+1时,
左边=1/(k+2)+1/(k+3)+.+1/3(k+1)
=1/(k+2)+1/(k+3)+.+1/3k+1/(3k+1)+1/(3k+2)+1/3(k+1)
下面比较n=k+1比n=k时,表达式多出的后三项之和与缺少的第一项两者的大小
设1/(3k+1)+1/(3k+2)+1/3(k+1)=(27k²+36k+11)/(3k+1)(3k+2)(3k+3)=a
1/(k+1)=(27k²+27k+2)/(3k+1)(3k+2)(3k+3)=b
k＞0,显然a＞b
所以左边=1/(k+2)+1/(k+3)+.+1/3k+1/(3k+1)+1/(3k+2)+1/3(k+1)
＞1(k+1)+1/(k+2)+1/(k+3)+.+1/3k>5/6
即n=k成立时,n=k+1也成立
所以对于任意n>=2,n∈N*,都有1/(n+1)+1/(n+2)+.+1/3n>5/6