数学归纳法证明1/n+1+1/n+2+1/n+3+...+1/3n>9/10 n>=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/(...1/(3k+3) +1/(3k+3)+1/(3k+3)=1/(3k+1) +1/(3k+2)+1/(3k+3) ??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/(3k+1) +1/(3k+2)+1/(3k+3) [1/(k+1) +1/(k+2)+...+1/3k] +1/(3k+1) +1/(3k+2)+1/(3k+3) -1/(k+1)>[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。