square root n limit $$(x_{n})_{n\geq 2}\ \ x_{n}=\sqrt[n]{1+\sum_{k=2}^{n}(k-1)(k-1)!} $$
$$\lim_{n\rightarrow \infty }\frac{x_{n}}{n}=?$$
I'm trying to solve the sum I but I don't know how to rewrite it.Can you give me some hints?
edit: 
The sum $\sum_{k=2}^{n}(k-1)(k-1)!$ should be $(n! - 1)$ so $x_{n}=\sqrt[n]{n!}$.The limit would be $\lim_{n\rightarrow \infty } \sqrt[n]{\frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion and make some simplifications I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
 A: Very roughly, the sum will be dominated by the last term.  You can then apply Stiriling's approximation to the last term
$$(n-1)(n-1)!\approx \frac {(n-1)^n}{e^{n-1}}\sqrt {2\pi (n-1)}$$
The $n^{th}$ root of this goes to $\frac {n-1}e$ so the limit goes to $\frac 1e$
You need to justify the approximations made, particularly the one about considering only the last term of the sum.
A: Well, using Stirling's formula $m!=(m/e)^m\sqrt{2\pi m}(1+o(1))$ as $m\to\infty$, we find that the radicand for large $k$ is $((n-1)/e)^{n-1} A_n$, where $1\le A_n\le Cn^{5/2}$ with some positive constant $C$ independent of $n$. Hence $\lim_{n\to\infty}\sqrt[n]{A_n}=1$, and we obtain
$$
\lim_{n\to\infty} \frac{x_n}n=\lim_{n\to\infty}\frac{(n-1)^{(n-1)/n}}{ne^{(n-1)/n}}=\frac1e.
$$
A: Note that $(k-1)(k-1)!$ is very nearly $k(k-1)!$, which is $k!$. We'll use that:
$$
\sum_{k=2}^n (k-1)(k-1)! + \sum_{k=2}^n (k-1)! = \sum_{k=2}^n (k-1+1)(k-1)! = \sum_{k=2}^n k!
$$
Then (careful with the limits):
$$
\sum_{k=2}^n (k-1)(k-1)! = \sum_{k=2}^n k! - \sum_{k=2}^n (k-1)! = \sum_{k=2}^n k! - \sum_{k=1}^{n-1} k! = n! - 1! \tag{*}
$$
because almost all the terms cancel out.

Edit: neater train of thought, but technically the same:
Observe that $(k-1)(k-1)! = k! - (k-1)!$, and then follow with $(*)$.
