令a=1/2+1/3+...+1/2008
则原式=(a+1/2009)(1+a)-(1+a+1/2009)a
=a(1+a)+1/2009(1+a)-a(1-a)-1/2009a
=1/2009(1+a)-1/2009a
=1/2009+1/2009a-1/2009a
=1/2009
令a=1/2+1/3+...+1/2008
则原式=(a+1/2009)(1+a)-(1+a+1/2009)a
=a(1+a)+1/2009(1+a)-a(1-a)-1/2009a
=1/2009(1+a)-1/2009a
=1/2009+1/2009a-1/2009a
=1/2009