We can show that $\sum_{k=3}^n \frac{n!}{(n-k)! n^k k} = \Theta(\ln(n))$
using the fact that $\sum_{k=3}^n \frac{1}{k} = \Theta(\ln(n))$
and $\frac{n!}{(n-k)! n^k} = \Theta(1)$ when $k=[\sqrt{n}]$.
But is it possible to find the exact constant $c$ such that $\sum_{k=3}^n \frac{n!}{(n-k)! n^k k} \sim c \ln{n}$?
