設f(n) = 1/(n+1) + 1/(n+2) + ...+ 1/(3n+1)
則f(n+1) = 1/(n+2) + 1/(n+3) + ...+ 1/[3(n+1)+1]
= 1/(n+2) + 1/(n+3) + ...+ 1/(3n+4)
則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))……應用到下式】
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
因為f(n)-f(n+1)2a+5對所有自然數(shù)n成立
所以只要13/12>2a+5
解得a2a-5對一切正整數(shù)n成立.
解法同上,
數(shù)列f(n)的最小值等于f(1) = 1/2 + 1/3 + 1/4 = 13/12
因為f(n)>2a-5對所有自然數(shù)n成立
所以只要13/12>2a-5
解得a