如何证明:1\/n+1+1\/n+2+L+1\/3n>5\/6(n>=2 n属于正整数)
1\/(n+1)+1\/(n+2)+...+1\/3n=1\/3+1\/4+1\/5+1\/6+1\/(n+5)+...+1\/3n=19\/20+...+1\/3n≥19\/20＞5\/6
求证:1\/(n+1)+1\/(n+2)+...+1\/3n>5\/6(n大于等于2,且是整数!)
又因为f(2)=1\/3+1\/4+1\/5+1\/6>1\/3+1\/6+1\/6+1\/6=5\/6 所以f(n)>5\/6 解2：n=2易证，假设n=k成立,当n=k+1时 f(n+1)=1\/(n+2)+1\/(n+2)+1\/(n+3)+...+1\/(3n+3)=f(n)+1\/(3n+1)+1\/(3n+2)+1\/(3n+3)-1\/(n+1)>f(n)>5\/6 ...
求证:1\/n+1+1\/n+2+1\/n+3+...+1\/3n>5\/6(n≥2,n∈N*)
1\/(3n)> 5\/6 (n≥2,n∈N*)证明：(1)当 n=2 时，左边 = 1\/3 + 1\/4 + 1\/5 + 1\/6 > 2\/6 + 1\/6 + 1\/6 + 1\/6 = 5\/6 = 右边 原式成立。(2)假设当 n=k(k≥2,k∈N*)时原式成立，即 1\/(k+1)+ 1\/(k+2)+ 1\/(k+3)+ ...+ 1\/(3k)> 5\/6 那么，...
1\/(n+1)+1\/(n+2)+...+1\/3n>5\/6(n>=2,n属於N*) 用数学归纳法证明 题目没...
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
求证:1\/n+1+1\/n+2+...+1\/3n大于5\/6(n大于等于2)
n+1)>f(n)又因为f(2)=1\/3+1\/4+1\/5+1\/6>1\/3+1\/6+1\/6+1\/6=5\/6 所以f(n)>5\/6 解2：n=2易证，假设n=k成立,当n=k+1时 f(n+1)=1\/(n+2)+1\/(n+2)+1\/(n+3)+...+1\/(3n+3)=f(n)+1\/(3n+1)+1\/(3n+2)+1\/(3n+3)-1\/(n+1)>f(n)>5\/6 ...
证明不等式 1\/n+1+1\/n+2……+1\/3n>5\/6
1\/（n+1）+1\/（n+2）……+1\/3n＞5\/6吧 这个是典型的函数思想 当n大于等于2时有f(n)=(1\/n+1)+(1\/n+2)+...+(1\/3n)∴f(n+1)=(1\/n+2)+...+(1\/3n)+(1\/3n+1)+(1\/3n+2)+(1\/3n+3)∴f(n+1)-f(n)=(1\/3n+1)+(1\/3n+2)+(1\/3n+3)-(1\/n+1)＞(1\/...
证明不等式 1\/n+1+1\/n+2……+1\/3n>5\/6
1\/(n+1)+1\/(n+2)……+1\/3n＞5\/6吧 这个是典型的函数思想当n大于等于2时有f(n)=(1\/n+1)+(1\/n+2)+...+(1\/3n)∴f(n+1)=(1\/n+2)+...+(1\/3n)+(1\/3n+1)+(1\/3n+2)+(1\/3n+3)∴f(n+1)-f(n)=(1\/3n+1)+(1\/3n+2)+(1\/3n+3)-(1\/n+1)＞(1\/3n+3...
...n+3)+...+1\/(n+n)大于等于1\/2 (其中n属于正整数)
为方便起见，令原式=S(n)显然S(1)≥1\/2，下面证明如果S(n-1)≥1\/2，则S(n)也大于等于1\/2 S(n)-S(n-1)=1\/(2n)+1\/(2n-1)-1\/(n-1)=[1\/(2n)-1\/(2n-2)]+[1\/(2n-1)-1\/(2n-2)]>0 所以S(n)≥S(n-1)≥1\/2 得证 ...
...+1\/(n+2)+1\/(n+3)+……+1\/(3n+1)>1(N属于正整数)
令f(n)=1\/(n+1)+1\/(n+2)+1\/(n+3)+...+1\/(3n+1)f(n+1)-f(n)=1\/(3n+2)+1\/(3n+3)+1\/(3n+4)-1\/(n+1)=2\/(3n+2)(3n+3)(3n+4)>0 f(n)递增 所以f(n)最小值为f(1)=13\/12
若不等式1\/n+1+1\/n+2……+1\/3n+1>a\/24对一切正整数n都成立,求正整数a...
1\/3(n+1)- 1\/3(n+1)=[1\/(3n+2)-1\/(3n+3)]-[1\/(3n+3)-1\/(3n+4)]=1\/(3n+2)(3n+3)- 1\/(3n+3)(3n+4)>0 所以sn的最小值是当n=1时 s1=1\/2 + 1\/3 +1\/4 =13\/12 当a\/24<13\/12即有不等式对所有正整数n都成立 所以a<26 正整数a的最大值为25 欢迎追问!
