证明：
设f(n)=1/（n+1）+1/(n+2)+1/(n+3).1/(2n),
则f(n+1)=1/(n+2)+1/(n+3).1/2(n+1),
f(n+1)-f(n)
=[1/(n+2)+1/(n+3).1/2(n+1)]-
[1/（n+1）+1/(n+2)+1/(n+3).1/(2n)]
=1/(2n+1)+1/2(n+1)-1/(n+1)
=1/2(n+1)(2n+1)>0
所以f(n)为单调递增函数,又因为n≥2,
所以f(n)≥f(2)=1/3+1/4=7/12>13/24
原式得证
PS：f(n)取不到13/24