如果ab-2的绝对值+(b-1)的2次方=0,试着求出1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+..

绝对值和平方不能为负,两者均为0,得b=1,a=2.又1/(2*3)=1/2-1/3,原式=1/2+1/2-1/3+1/3-1/4+…+1/2008-1/2009+1/2009-1/2010=1/2+1/2-1/2010=2009/2010