问题标题：
【数学归纳法证明1/n+1+1/n+2+1/n+3+...+1/3n>9/10n>=21）当n=2时,左=1/3+1/4+1/5+1/6=57/60>54/60=9/10,成立．（2）假设n=k时,有1/(k+1)+1/(k+2)+...+1/3k>9/10那么1/(k+2)+1/(k+3)+...+1/3(k+1)=[1/(k+1)+1/(k+2)+...+1/3k]+1/(3k+1)】
问题描述：
数学归纳法证明1/n+1+1/n+2+1/n+3+...+1/3n>9/10n>=2
1）当n=2时,左=1/3+1/4+1/5+1/6=57/60>54/60=9/10,成立．
（2）假设n=k时,有1/(k+1)+1/(k+2)+...+1/3k>9/10
那么 1/(k+2)+1/(k+3)+...+1/3(k+1)
=[1/(k+1)+1/(k+2)+...+1/3k]+1/(3k+1)+1/(3k+2)+1/(3k+3)-1/(k+1)
>9/10+1/(3k+3)+1/(3k+3)+1/(3k+3)-1/(k+1)
=9/10
即n=k+1时命题也成立,
从而 原不等式对n∈N,且n>1成立．
第二步中为什么是
>9/10+1/(3k+3)+1/(3k+3)+1/(3k+3)-1/(k+1)
不应该是
>9/10+1/(3k+1)+1/(3k+2)+1/(3k+3)-1/(k+1)的么
苏显斌回答：
1）当n=2时,左=1/3+1/4+1/5+1/6=57/60>54/60=9/10,成立．（2）假设n=k时,有1/(k+1)+1/(k+2)+...+1/3k>9/10那么 1/(k+2)+1/(k+3)+...+1/3(k+1)=[1/(k+1)+1/(k+2)+...+1/3k]+1/(3k+1)+1/(3k+2)+1/(3k+3)-1/(...
刘立钧回答：
1/(3k+3)+1/(3k+3)+1/(3k+3)=1/(3k+1)+1/(3k+2)+1/(3k+3)??
苏显斌回答：
当然不等于啦~~~~~~~~~~~~~放说法你们没学过吗？？1/(3k+3)+1/(3k+3)+1/(3k+3)[1/(k+1)+1/(k+2)+...+1/3k]+1/(3k+3)+1/(3k+3)+1/(3k+3)-1/(k+1)，而[1/(k+1)+1/(k+2)+...+1/3k]+1/(3k+3)+1/(3k+3)+1/(3k+3)-1/(k+1)的值是9/10从而[1/(k+1)+1/(k+2)+...+1/3k]+1/(3k+1)+1/(3k+2)+1/(3k+3)-1/(k+1)>9/10~~~~~~算了做给你看吧~~~~~~~~~~~~~~~~1）n=2,时，1/3+1/41/5+1/6=19/20>9/102）假设n=k时，1/k+1+1/k+2+1/k+3+...+1/3k-1>9/10-1/3k那么当n=k+1时，1/k+2+1/k+3+...+1/3k-1+1/3k+1/(3k+1)+1/(3k+2)>9/10+1/3k+1/(3k+1)+1/(3k+2)-1/k+1那么只需要证明1/3k+1/(3k+1)+1/(3k+2)-1/k+1>-1/(3k+3)即1/3k+1/(3k+1)+1/(3k+2）>2/(3k+3)上式显然成立，那么当n=k+1时，假设也成立综合1），2）可知道不等式1/n+1+1/n+2+1/n+3+...+1/3n>9/10对于任意n>=2都成立。
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