求自然数a的最大值,使得不等式1/(n+1)+1/(n+2)+……+1/(3n+1)>2a+5对一切正整数n

设f(n) = 1/(n+1) + 1/(n+2) + ...+ 1/(3n+1)
则f(n+1) = 1/(n+2) + 1/(n+3) + ...+ 1/[3(n+1)+1]
= 1/(n+2) + 1/(n+3) + ...+ 1/(3n+4)
则f(n)-f(n+1) = 1/(n+1) - [1/(3n+2) + 1/(3n+3) + 1/(3n+4)]
= 1/(n+1) - [1/(3n+3)+(3n+2+3n+4)/((3n+2)(3n+4))]
= 1/(n+1) - [1/(3n+3) + (6n+6)/((3n+2)(3n+4))]
【(3n+2)(3n+4)=9n^2+18n+81 /((3n+3)(3n+3))……应用到下式】
f(n)-f(n+1)< 1/(n+1) - [1/(3n+3) + (6n+6)/((3n+3)(3n+3))]
= 1/(n+1) - [1/(3n+3) + 2/(3n+3)]
= 1/(n+1) - 3/(3n+3)
= 0
因为f(n)-f(n+1)2a+5对所有自然数n成立
所以只要13/12>2a+5
解得a2a-5对一切正整数n成立.
解法同上,
数列f(n)的最小值等于f(1) = 1/2 + 1/3 + 1/4 = 13/12
因为f(n)>2a-5对所有自然数n成立
所以只要13/12>2a-5
解得a