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.
 A: Your sequence can be rewritten recursively as:
$$u_0 = 1$$
$$u_n = 1 + \frac{u_{n-1}}{n}$$
Assuming that this sequence does approach a limit $L$, then for sufficiently large $n$, $L_n$ and $L_{n-1}$ both get arbitrarily close to $L$, so:
$$L \approx 1 + \frac{L}{n}$$
As $n \rightarrow \infty$, we get $L = 1$.
A: Write it as
$$\frac{1}{n!} \sum_{k=0}^n k! = \frac{n!}{n!} + \frac{(n-1)!}{n!}+...+\frac{1!}{n!} = 1+\frac{1}{n}+\frac{1}{n(n-1)}+...+\frac{1}{n!}$$
Now this is clearly greater or equal to $1$. On the other hand the last $(n-1)$ terms (all except the first two) are smaller than $\frac{1}{n(n-1)}$ so we have an upper bound of
$$1+\frac{1}{n} + \frac{1}{n} = 1+\frac{2}{n}$$
which converges to $1$.
Use the Sandwich theorem.
A: 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 $x_n \downarrow 1$.
A: By the Stolz–Cesàro theorem,
$$
\lim_{n\to\infty} u_n = \lim_{n\to\infty}\frac{n!}{n! - (n-1)!} = \lim_{n\to\infty}\frac{1}{1-1/n} = 1.
$$
A: 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.
