裂项相消{1/n(n+1)}=1/n-1/(n+1)a1=1/1-1/(1+1)=1-1/2a2=1/2-1/(2+1)=1/2-1/3a3=1/3-1/(3+1)=1/3-1/4a4=1/4-1/(4+1)=1/4-1/5a5=1/5-1/(5+1)=1/5-1/6...a99=1/99-1/(99+1)=1/99-1/100S99=a1+a2+a3+a4+a5+……+a99最后...