Question

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的最大值，并证明结论。

Answer 1

设f(n)=1/(n+1)+1/(n+2)+…+1/(3n+1)
则f(n+1)-f(n)=1/(3n+2)+1/(3n+3)+1/(3n+4)-1/(n+1)>0
故f(n)单调增加，要使f(n)>a/24对一切正整数n都成立,
只要 f(1)=1/2+1/3+1/4>a/24 即a<26 故正整数a的最大值为25

Answer 2

令sn=1/n+1+1/n+2……+1/3n+1则s(n+1)=1/(n+2) +1/(n+3)+.....1/(3n+4)s(n+1)-sn=1/(3n+2) + 1/(3n+3) + 1/(3n+4) - 1/(n+1) =1/(3n+2) + 1/(3n+3) + 1/(3n+4)- 1/3(n+1)- 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!

Answer 3

向楼上学习了。不过楼主的题目让我理解错了