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)+(1/3n+3)+(1/3n+3)+(1/n+1)=0
已知f(n)是单调递增的
故f(n)＞f(2)=1/3+1/4+1/5+1/6=57/60＞5/6
当n等于1时带入检验成立
综上所述……
证明不等式 1\/n+1+1\/n+2……+1\/3n>5\/6
＞(1\/3n+3)+(1\/3n+3)+(1\/3n+3)+(1\/n+1)=0 已知f(n)是单调递增的 故f(n)＞f(2)=1\/3+1\/4+1\/5+1\/6=57\/60＞5\/6 当n等于1时带入检验成立 综上所述……
证明不等式 1\/n+1+1\/n+2……+1\/3n>5\/6
＞(1\/3n+3)+(1\/3n+3)+(1\/3n+3)+(1\/n+1)=0 已知f(n)是单调递增的 故f(n)＞f(2)=1\/3+1\/4+1\/5+1\/6=57\/60＞5\/6 当n等于1时带入检验成立 综上所述……
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，...
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\/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 ...
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
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...
(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 ...
1\/(n+1)+1\/(n+2)+1\/(n+3)+ ……+1\/3n 极限
用定积分解答如下:
求证:1\/(n+1) + 1\/(n+2) + 1\/(n+3) + ... + 1\/(3n)
1\/(n+1)+1\/(n+2)+...+1\/(2n)>1\/(2n)+1\/(2n)+...+1\/(2n)=1\/2 1\/(2n+1)+1\/(2n+2)+...+1\/(3n)>1\/(3n)+1\/(3n)+...+1\/(3n)=1\/3 因此1\/(n+1) + 1\/(n+2) + 1\/(n+3) + ... + 1\/(3n)>1\/2+1\/3=5\/6 希望可以帮到你！
对于n属于N,用数学归纳法证明:1\/n+1+1\/n+2+...+1\/3n+1大于1
注：从而n+1到3n,左边共有2n项．（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...
