裂项
an=(n+2)/[n!+(n+1)!+(n+2)!]
=（n+2)[n!(1+n+1+(n+1)(n+2))]
=（n+2)/[n!(n+2)^2]
=1/[n!(n+2)]
=(n+1)/(n+2)!
=[(n+2)-1]/(n+2)!
=1/(n+1)!-1/(n+2)!
Sn=1/2!-1/3!+1/3!-1/4!+.+1/(n+1)!-1/(n+2)!
=1/2!-1/(n+2)!
=1/2-1/(n+2)!
数列{(n+2)/[n!+(n+1)!+(n+2)!]}的前n项和为--------