How to find $\lim_{n\to\infty}\frac{1!+2!+\cdots+n!}{n!}$? How to evaluate the following limit?
$$\lim_{n\to\infty}\dfrac{1!+2!+\cdots+n!}{n!}$$
For this problem I have two methods. But I'd like to know if there are better methods.
My solution 1:
Using Stolz-Cesaro Theorem, we have
$$\lim_{n\to\infty}\dfrac{1!+2!+\cdots+n!}{n!}=\lim_{n\to\infty}\dfrac{n!}{n!-(n-1)!}=\lim_{n\to\infty}\dfrac{n}{n-1}=1$$
My solution 2:
$$1=\dfrac{n!}{n!}<\dfrac{1!+2!+\cdots+n!}{n!}<\dfrac{(n-2)(n-2)!+(n-1)!+n!}{n!}=\dfrac{n-2}{n(n-1)}+\dfrac{1}{n}+1$$

If we let
$$
a_n=\frac1{n!}\sum_{k=1}^nk!
$$
then obviously, $a_n\ge1$. Furthermore, we get that
$$
a_{n+1}=1+\frac{a_n}{n+1}
$$
Suppose that for some $n\ge1$, $a_n\le2$, then
$$
\begin{align}
a_{n+1}
&=1+\frac{a_n}{n+1}\\
&\le1+\frac{2}{n+1}\\
&\le2
\end{align}
$$
Since $a_1=1$, we have that $a_n\le2$ for all $n\ge1$. Now finally,
$$
\begin{align}
1\le a_{n+1}=1+\frac{a_n}{n+1}\le1+\frac2{n+1}
\end{align}
$$
By the Squeeze Theorem, we get that
$$
\lim_{n\to\infty}a_n=1
$$
