请问怎么计算 3/(1!+ + )+ 4/(2!+ + ) +…… +2013/(2011!+ 2012!+ 2013!

(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)!
原式=1／2-1／2013!