TITLE: How to calculate $\lim_{n \to \infty} \sum_{k=1}^n\binom nk k!k\frac{1}{n^k}$?
QUESTION [6 upvotes]: How to calculate this limit :
$$\lim_{n\rightarrow +\infty}u_n$$
with :
$$u_n=\sum_{k=1}^n\binom nk k!k\frac{1}{n^k}$$
We can write this :
$$u_n=\sum_{k=1}^n\frac{n}{n}\times\frac{n-1}{n}\times\cdots\times\frac{n-k+1}{n}\times k$$
But I can't find the solution.

REPLY [4 votes]: Here is a simplification of @Markus Scheuer's answer. Let
$$ a_k = \prod_{j=1}^{k} \frac{n-j+1}{n}. $$
Then by OP's observation,
$$ u_n = \sum_{k=1}^{n} k a_k = n \sum_{k=1}^{n} \left( 1 - \frac{n-k}{n} \right) a_k = n \sum_{k=1}^{n}(a_k - a_{k+1}) = n(a_1 - a_{n+1}) = n. $$