根据两个连续自然数A,B.1/A-1/B=1/AB得原式=1-1/2+(1/2-1/3)+(1/4-1/3)+(1/5-1/4)+...+(1/2006-1/2005)=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5+...+1/2005-1/2006)=1-1/2006=2005/2006.