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:
$$1=\dfrac{n!}{n!}<\dfrac{1!+2!+\cdots+n!}{n!}<\dfrac{(n-2)(n-2)!+(n-1)!+n!}{n!}=\dfrac{n-2}{n(n-1)}+\dfrac{1}{n}+1$$

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 $1$.
