<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] 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.
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