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
1个回答
f(n)=1/（n+1) + 1/(n+2) +1/(n+3) +……+1/（3n+1）
f(n+1)=1/（n+2) + 1/(n+3) +1/(n+4) +……+1/[3(n+1)+1]
f(n+1)-f(n)=1/（n+1) - 1/(3n+2)-1/(3n+3)-1/(3n+4)>0
所以函数f(n)对于n为正整数时为单调增函数
所以原不等式等效于a/2425/24
当k=n+1时
由于
9(n+1)^2=9n^2+18n+9>9n^2+18n+8=(3n+2)(3n+4)
即
9(n+1)^2/[(3n+2)(3n+4)]-1>0
左侧为
1/[(n+1)+1]+1/[(n+1)+2]+1/[(n+1)+3]+...+1/[3(n+1)+1]
=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+{1/(3n+2)+1/(3n+3)+1/(3n+4)-1/(n+1)}
=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+{6(n+1)/[(3n+2)(3n+4)]-2/(3n+3)}
=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+2/(3n+3)*{9(n+1)^2/[(3n+2)(3n+4)]-1}
>1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24.
结论成立.
这样可以么?
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