∵|a-1|+

ab−2

=0，
∴a-1=0，ab-2=0，
解得a=1，b=2，

1

ab

+

1

(a+1)(b+1)

+

1

(a+2)(b+2)

+…+

1

(a+2009)(b+2009)

=

1

1×2

+

1

2×3

+…+

1

2010×2011

=1-

1

2

+

1

2

-

1

3

1

2010

-

1

2011

=1-

1

2011

=

2010

2011

．