Square root n limit ( sum question )

[Vali]

Hi!

$$(x_{n})_{n\geq 2}\ \ x_{n}=\sqrt[n]{1+\sum_{k=2}^{n}(k-1)(k-1)!}$$
$$\lim_{n\rightarrow \infty }\frac{x_{n}}{n}=?$$
I know how to solve the limit but I don't know how to solve the sum $\sum_{k=2}^{n}(k-1)(k-1)!$ which should be $(n! - 1)$ The limit would become $\lim_{n\rightarrow \infty } \sqrt[n]{\frac{n!}{n^{n}}}$ which I know how to solve.
So, how to approach the sum such that the result to be $(n! - 1)$ ?

 

[Euge]

Hello again, Vali. (Wave)

Note $(k-1)\cdot (k-1)! = k(k-1)! - (k-1)! = k! - (k-1)!$, and thus the sum $\sum\limits_{k = 2}^n (k-1)\cdot (k-1)! = \sum\limits_{k = 2}^n [k! - (k-1)!]$ telescopes to $n! - 1$.

 

[Vali]

Thank you! :)