若Ia-1I+Iab-2I=0,求1/ab+1/(a+1)(b+1) +.+1/（a+2005)(b+2005)=

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因为Ia-1I与Iab-2I都大于等于0,而Ia-1I+Iab-2I=0,
所以Ia-1I=0,Iab-2I=0,
解得a=1,b=2
1/ab+1/(a+1)(b+1) +.+1/（a+2005)(b+2005)
=1/1*2+1/2*3+1/3*4+……+1/2006*2007
=1-1/2+1/2-1/3+1/3-1/4+……+1/2006-1/2007
=1-1/2007
=2006/2007

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