Question

高等数学常数项？

如图，求下这个等式计算过程，没看懂，谢谢啦

Answer 1

lim<n→∞>[(n+1)!/(n+1)^(n+1)] (n^n/n!)
= lim<n→∞>[(n+1)/(n+1)^(n+1)] (n^n)
= lim<n→∞>[1/(n+1)^n] (n^n)
= lim<n→∞>[n/(n+1)]^n = lim<n→∞>[(n+1-1)/(n+1)]^n
= lim<n→∞>[1-1/(n+1)]^n
= lim<n→∞>{[1-1/(n+1)]^[-(n+1)]}^[-n/(n+1)]
= e^lim<n→∞>[-n/(n+1)] = e^lim<n→∞>[-1/(1+1/n)] = e^(-1) = 1/e