1/[n(n+1)]=(1/n)-[1/(n+1)].如1/(1×2)=1-(1/2).1/(2×3)=(1/2)-(1/3).1/(3×4)=(1/3)-(1/4).1/[n(n+1)]=(1/n)-[1/(n+1)].将这些等式累加得：Sn=1/1×2+1/2×3+1/3×4+1/4×5+...+1/n×(n+1)=1-[1/(n+1)].当n=10时,S10=1-(1/11)==10/11.