How to calculate $\lim_{n \to \infty} \sum_{k=1}^n\binom nk k!k\frac{1}{n^k}$? 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.
 A: 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,
$$ u_n = \sum_{k=1}^{n} k a_k = n \sum_{k=1}^{n} \left( 1 - \frac{n-k}{n} \right) a_k = n \sum_{k=1}^{n}(a_k - a_{k+1}) = n(a_1 - a_{n+1}) = n. $$
A: Hint: The number $u_n$ has a nice telescoping property. Let's for example consider the case $n=5$.

\begin{align*}
\color{blue}{u_5}&=1+2\cdot\frac{4}{5}+3\cdot\frac{4}{5}\cdot\frac{3}{5}+4\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}
+\color{blue}{5}\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}\cdot\color{blue}{\frac{1}{5}}\\
&=1+2\cdot\frac{4}{5}+3\cdot\frac{4}{5}\cdot\frac{3}{5}
+(4+1)\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}\\
&=1+2\cdot\frac{4}{5}+3\cdot\frac{4}{5}\cdot\frac{3}{5}
+\color{blue}{5}\cdot\frac{4}{5}\cdot\frac{3}{5}\cdot\color{blue}{\frac{2}{5}}\\
&=1+2\cdot\frac{4}{5}+(3+2)\cdot\frac{4}{5}\cdot\frac{3}{5}\\
&=1+2\cdot\frac{4}{5}+\color{blue}{5}\cdot\frac{4}{5}\cdot\color{blue}{\frac{3}{5}}\\
&=1+(2+3)\cdot\frac{4}{5}\\
&=1+\color{blue}{5}\cdot\color{blue}{\frac{4}{5}}\\
&=(1+4)\\
&\color{blue}{\,=5}
\end{align*}
indicating $\color{blue}{u_n=n}$ for $n\geq 1$.

We observe we can iteratively collect the two right-most summands whereby the factor $5$ and $\frac{1}{5}$ cancel in the right-most summand.
A: Note that $$\begin{aligned}
\sum_{k=1}^n\binom nk k!k\frac{1}{n^k} &= \sum_{k=1}^n\binom nk ((k+1)!-k!)\frac{1}{n^k}\\
&= \sum_{k=1}^n\binom nk (k+1)!\frac{1}{n^k}-\sum_{k=1}^n\binom nk k!\frac{1}{n^k}\\
&= \frac{n^2-n-1}{n+1}+\frac{e^n n^{-n} }{n+1}\Gamma (n+2,n) - \left(e^n n^{-n} \Gamma (n+1,n)-1 \right)\\
&= \frac{n^2-n-1}{n+1}+\frac{e^n n^{-n} }{n+1}\left((n+1)\Gamma (n+1,n) 
+ n^{n+1}e^{-n}\right) - \left(e^n n^{-n} \Gamma (n+1,n)-1 \right)\\
&= n
\end{aligned}$$
where $\Gamma$ is the incomplete Gamma function.
A: Since you asked " .. without using  Gamma function .. "
and if you just want to know whether the sum converges or not,
then you were on the right track: just carry on with a BigO
$$
\eqalign{
  & \sum\limits_{k = 1}^n {\left( \matrix{
  n \cr 
  k \cr}  \right)k!k{1 \over {n^{\,k} }}}  = \sum\limits_{k = 1}^n {k{{n^{\,\underline {\,k} } } \over {n^{\,k} }}}  = \sum\limits_{k = 1}^n {k{{n\left( {n - 1} \right)\quad \left( {n - k + 1} \right)} \over {n^{\,k} }}}  =   \cr 
  &  = \sum\limits_{k = 1}^n {k\left( 1 \right)\left( {1 - {1 \over n}} \right) \cdots \left( {1 - {{k - 1} \over n}} \right)}  = 1 + \sum\limits_{k = 2}^n {k\left( {1 + O\left( {{1 \over n}} \right)} \right)} \quad \left| {\;2 \le n} \right. \cr} 
$$
to show that it diverges.
