I am given the sequence $(x_n)_{n \ge 2}$ such that:
$$x_n = \sqrt[n]{1 + \sum\limits_{k=2}^n(k-1)(k-1)!}$$
And I am asked to find $\lim\limits_{n \to \infty} \dfrac{x_n}{n}$ . While trying to solve this, I first made the following observation:
$$\begin{align}\sum\limits_{k = 2}^n (k-1)(k-1)! &= \\ &=\sum\limits_{k = 2}^n(k-1+1)(k-1)! - (k-1)! \\ &=\sum\limits_{k=2}^nk(k-1)! - (k-1)! \\ &=\sum\limits_{k=2}^nk!-(k-1)! \\ &=2!-1!+3!-2!+4!-3!+5!-4!+...n!-(n-1)! \\ &= \boxed{\color{#D04}{-1 + n!}} \end{align}$$
So, $x_n$ becomes:
$$ \begin{align}x_n &= \sqrt[n]{1-1+n!} \\ x_n &= \color{#a90}{\sqrt[n]{n!}} \end{align}$$
So the limit I have to find is:
$$\lim\limits_{n \to \infty} \dfrac{x_n}{n} = \lim\limits_{n \to \infty} \dfrac{\sqrt[n]{n!}}{n} = \lim\limits_{n \to \infty} \sqrt[n]{\dfrac{n!}{n^n}}$$
And this is where I got stuck, kinda. I could find the limit by using the Stirling Approximation:
$$n! \approx \sqrt{2 \pi n} \bigg (\dfrac{n}{e} \bigg )^n$$
By replacing the $n!$ term in my limit with this approximation, I eventually found that
$$\lim\limits_{n \to \infty} \dfrac{x_n}{n} = \dfrac{1}{e}$$
Which seems to be the right result, according to my textbook.
What I don't like about this is that use of the Stirling Approximation. It always feels a bit off using an approximation to get an exact result. My question is this: is there any other way of arriving at this answer without using the Stirling Approximation? In other words, can I find the following limit:
$$\lim\limits_{n \to \infty} \sqrt[n]{\dfrac{n!}{n^n}}$$
Without using an approximation?
