裂项相消法
根据 1/n*(n+1)= 1/n -1/(n+1)
1/2+1/(2*3)+1/(3*4)+.+1/(2004*2005)+1/(2005*2006)
=1/2 +1/2-1/3+1/3-1/4+...+1/2005-1/2006
=2005/2006