点p(an,a(n+1))(n属于N)在直线x-y+1=0上,
an+1-an=1
数列{an}中,a1=1.d=1
所以an=a1+(n-1)*1=n
∴f(n)=1/(n+a1)+1/(n+a2)+1/(n+a3)+…+1/（n+an）（n∈N,且n≥2）
==>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)
则当n≥2时,f(n)＞f(n-1)＞f(n-2)＞...＞f(2)
∴n=2时,
f(n)取最小值f(2)=1/(2+1)+1/(2+2)=7/12