原式=∑(k+2)/[k!(1+k+1+(k+1)(k+2)]=∑1/（k!(k+2))
令S(x）=∑1/k!(k+2)*x^(k+2) ,顯然S(0)=0
S'(x)=∑1/k!x^(k+1)=x∑1/k!*x^k
=x(e^x-1)
S(X)=∫x(e^x-1)dx=∫xe^xdx-∫xdx
=xe^x-e^x-x^2/2+c
S(0)=-1+C=0
∴C=1
S(X)=xe^x-e^x-x^2/2+1
令x=1得
原級數(shù)=e-e-1/2+1=1/2