(k!+(k+1)!+(k+2)!)=k![1+(k+1)+(k+1)(k+2)]=k!(k+2)^2
∴(k+2)/(k!+(k+1)!+(k+2)!)=(k+2)/[k!(k+2)^2]=1/[k!(k+2)]=(k+1)/(k+2)!
∴Sn=∑(k+1)/(k+2)!
Sn+∑1/(k+2)!=∑(k+1)/(k+2)!+∑1/(k+2)!=∑1/(k+1)!(求和∑中k均从1到k)
∴Sn=∑1/(k+1)!-∑1/(k+2)!=1/2!-1/(k+2)!=1/2-1/(k+2)!