17/10/24 23:42:12.62 AZXk3eOu.net
>>661
a(n+1)=[x^(n+1)e^x][0,1]-∫[0,1](n+1)x^ne^xdx=e-(n+1)an
a(n+1)/(n+1)!=e/(n+1)!-an/n!
？
bn=∫[0,1]x^ne^(-x)dx=[-x^ne^(-x)][0,1]+n∫x^(n-1)e^(-x)dx=-(1/e)+nb(n-1)
bn/n!=-(1/e)/n!+b(n-1)/(n-1)!=-(1/e)(1/n!+…+1/1!)+b0/0!=-(1/e)(1/1!+…+1/n!)+[-e^(-x)][0,1]=1-(1/e)(1/0!+1/1!+…+1/n!)
0<e^(-x)<1 (0<x<1)
0<bn<∫[0,1]x^ndx=1/(n+1)→0
1=(1/e)(1/0!+…1/n!+…)
1/0!+…1/n!+…=e