Limit of $u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k !$ I encountered a sequence $(u_n)_{n \in \mathbb{N}} $ defined as
$$ u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k ! $$
And I wonder what is the limit. It seems to be 1 but even Wolfram Alpha cannot figure it out.
My first idea was to write $$ \frac{k!}{n!} = \frac{1}{(k+1)...(n-1)n}$$ and
$$  (k+1)^{n-k} \leq (k+1)...(n-1)n \leq n^{n-k} $$
EDIT : I found a solution but Yanko found something more elementary.
My solution :
\begin{align}
  u_n &= 1 + \frac{1}{n} + \frac{1}{n(n-1)} + \frac{1}{n(n-1)(n-2)} + ... + \frac{1}{n!}   \\ 
&= 1 + \frac{1}{n} + (n-1) \times o(\frac{1}{n^2})  \\
&= 1 + \frac{1}{n}  + o(\frac{1}{n}) \\
\end{align}
Thus all terms converge towards 0 except the first one, and the limit is 1.

To go beyond the limit itself, using the same approach as @Yanko, let
$$a_p=\frac 1{\prod_{i=0}^p (n-i)}$$ and consider
$$1+\sum_{p=0}^\infty a_p=1+\sum_{k=0}^\infty \frac {B_k}{n^{k+1}}=1+\frac{1}{n}+\frac{1}{n^2}+\frac{2}{n^3}+\frac{5}{n^4}+O\left(\frac{1}{n^5}
   \right)$$ where appear Bell numbers.
