How to find $\lim_{n\to\infty}\frac{1!+2!+\cdots+n!}{n!}$?

Perhaps you might like the following argument:

Notice that $$\frac{\sum_{k=1}^{n} i!}{n!} = 1 + \frac{1}{n}\frac{\sum_{k=1}^{n-1} i!}{n-1!};$$ from this, we get the recurrence

$$\frac{\sum_{k=1}^{2} i!}{2!} = 1 + \frac{1}{2},$$ $$\frac{\sum_{k=1}^{3} i!}{3!} = 1 + \frac{1}{3}(1 + \frac{1}{2}),$$

and in general

$$\frac{\sum_{k=1}^{n} i!}{n!} = 1 + \frac{1}{n}(1 + \frac{1}{n-1}(...(1 + \frac{1}{3}(1+ \frac{1}{2}))...)) \\< 1 + \frac{1}{n}(1 + \frac{1}{2}(...(1 + \frac{1}{2}(1+\frac{1}{2}))...)) \\ < 1 + \frac{1}{n}(2)$$

and so your sequence is bounded above by one converging to $1$. Since it is also trivially bounded below by the constant sequence of $1$, your sequence thus converges to $1$.

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 $$ \lim_{n\to\infty}a_n=1 $$