Σ(k=1~n) 1/[k(k+1)(k+2)]
= Σ(k=1~n) { 1/[k(k+1)] - 1/[(k+1)(k+2)] }/2
= [1/(1 * 2) - 1/(2 * 3)]/2 + [1/(2 * 3) - 1/(3 * 4]/2 + ...+ { 1/[n(n + 1)] - 1/[(n + 1)(n + 2)] }/2
= 1/4 - 1/[2(n + 1)(n + 2)],裂项法,中间的项全部互相抵消了