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设1/2006+1/2007+1/2008=x
1/2006+1/2007+1/2008+1/2009=y
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=(1+x)*y-(1+y)*x
=y+xy-x-xy
=y-x
=1/2009

设1/2006+1/2007+1/2008=a,1/2009=b
原式=(1+a)(a+b)-a(1+a+b)
=a+a*a+b+ab-a-a*a-ab抵消
=b
=1/2009（二零零九分之一）