f（n）=1/(n+1)+1/(n+2)+1/(n+3)+.+1/2n
f(n+1)=1/(n+2)+1/(n+3)+.+1/(2n+1)+1/(2n+2)
f(n+1)-f(n)=[1/(n+2)+1/(n+3)+.+1/(2n+1)]-[1/(n+1)+1/(n+2)+1/(n+3)+.+1/2n]
=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)-1/(2n+2)
=1/(2n+1)(2n+2)