设(1/2+1/3+···+1/2004)为a,则,
(1/2+1/3+···+1/2004)(1+1/2+1/3+···1/2003)-(1+1/2+1/3+···1/2004)(1/2+1/3+···+1/2003)
=a(1+a- 1/2004)-(1+a)(a- 1/2004)
=a+a^2- a/2004 -a +1/2004 -a^2+ a/2004
=1/2004
设(1/2+1/3+···+1/2004)为a,则,
(1/2+1/3+···+1/2004)(1+1/2+1/3+···1/2003)-(1+1/2+1/3+···1/2004)(1/2+1/3+···+1/2003)
=a(1+a- 1/2004)-(1+a)(a- 1/2004)
=a+a^2- a/2004 -a +1/2004 -a^2+ a/2004
=1/2004