Problems with limit of factorials: $\;\lim_{n\to\infty}\frac{1!+2!+\ldots+n!}{n!}\;$

HINT: The numerator is less than $(n-2)!+(n-2)!+...+(n-2)!+(n-1)!+n!$

Hint: Show that $S_{n+1} = S_n/(n+1) + 1$. If you really showed it is monotone decreasing, then the limit exists and you can take limits of the expression given here to find out what it is.

Hint 2: Fix $m$ and write $$\frac{1! + \cdots + (n-m)!}{n!} + \frac{(n-m+1)! + \cdots n!}{n!}$$ and realize $$\frac{1! + \cdots + (n-m)!}{n!} \leq \frac{(n-m)(n-m)!}{n!}$$

Since you have already shown that this limit exists. Now what you can do is the following

Let $x_n = \dfrac{1! + \cdots + n!}{n!}$. Look at $x_{n+1} - \dfrac{x_n}{n+1}$.

\begin{equation} x_{n+1} - \dfrac{x_n}{n+1} = \frac{(n+1)!}{(n+1)!} = 1. \end{equation} If we assume that $\lim_\limits{n \to \infty} x_n = a$ and we also observe that $\lim_\limits{n\to\infty}\dfrac{x_n}{n+1} = 0$ because $\lim_\limits{n \to \infty} x_n = a$. Thus taking limit as $n \to \infty$ in the above equation gives \begin{equation} \lim_\limits{n\to \infty}\left(x_{n+1} - \dfrac{x_n}{n+1}\right) =\lim_\limits{n \to \infty} x_n = a. \end{equation} And since $\lim_\limits{n\to \infty}\left(x_{n+1} - \dfrac{x_n}{n+1}\right) = 1$, it follows that $a = 1$.