Ia-1I+(ab-2)的2次方=0
Ia-1I≥0
(ab-2)的2次方≥0
和＝0
所以分别等于0
a-1=ab-2=0
a=1 b=2
1/ab+1/(a+1)(b+1)+.1/（a+2004)(b+2004)
= 1-1/2+1/2-1/3……+1/2005-1/2006【根据1/n(n+1)=1/n-1/(n+1)】
=2005/2006