这里 可能是楼主学生抄题目的失误.
应该是f(n)=1/(n+a1)+1/(n+a2)+1/(n+a3)+…+1/(n+an),肯定不是f(x),因为整个题目不存在自变量x.只有n!
显然an=n,
f(n)=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(n+n)
=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(2n)
f(n+1)=1/(n+1+1)+1/(n+1+2)+1/(n+1+3)+...+1/(n+1+n+1)
=1/(n+2)+1/(n+3)+...+1/(2n)+1/(2n+1)+1/(2n+2)
∴当n≥2时,有
f(n+1)-f(n)=1/(2n+1)+1/(2n+2)-1/(n+1)＞1/(2n+2)+1/(2n+2)-1/(n+1)=0
即f(n+1)＞f(n)
于是当n≥2时,f(n)＞f(n-1)＞f(n-2)＞...＞f(2)
即当n=2时,f(n)达到最小值f(2)=1/(2+1)+1/(2+2)=7/12