<s>[INST] 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.
 [/INST] Even more elementary:
Note that $x_{n+1} = {1 \over n+1} x_n +1$. Note that
$x_{n+1} \le x_n$ iff $x_n \ge {n+1 \over n}$. Also note that $x_1 = 2$.
If $x_n \ge {n+1 \over n}$ then $x_{n+1} \ge {n+1 \over n} \ge {n+2 \over n+1}$. Hence $x_n$ is non increasing and bounded below by $0$. Hence it has a limit $L$ and from continuity we have $L = 1$. Hence $x_n \downarrow 1$.
</s>