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:
$$1=\dfrac{n!}{n!}<\dfrac{1!+2!+\cdots+n!}{n!}<\dfrac{(n-2)(n-2)!+(n-1)!+n!}{n!}=\dfrac{n-2}{n(n-1)}+\dfrac{1}{n}+1$$
 A: 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 $1$.
A: 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
$$
\lim_{n\to\infty}a_n=1
$$
A: You can write this as a kind of 'added fraction', or fraction of continued numerator.  Such fractions were used, for example, by Fibonacci in Liber Aceri
Thus  $1 \frac {a+}A \frac{b+}B \dots = 1 \frac{a+\frac{b+ \dots}B}A$  For example, one might regard decimals, as a series of added tenths, as $1m \frac{dm+}{10} \frac{cm+}{10} \frac{mm}{10}$  The $+$ serves to show it's the numerator continued. 
It gives  $A = 1 \frac {1+}n \frac {1+}{n-1} \frac {1+}{n-2} \dots$.  This is identical to writing it in a base, where the size of the base gets smaller as one goes along.  So, for example, when $x=10$ it gives $1 \frac {1+}{10} \frac {1+}9 \frac {1+}8$
As n goes large, one sees that the limiting factor (which is less than the sum), is $1 \frac {1+}n \frac {1+}n \frac {1+}n \dots$, which is $A_2 = \frac n{n-1} = 1 \frac 1{n-1}$.  In fact, the first two fractions of the number add to this.  
The sum of the first three fractions, then goes $A_3 = 1 \frac {1+}{n} \frac{1}{n-2}$.  This is $1 \frac{n-1}{n^2-2n}$.
The value between $A-A_3$ is much less than between $A-A_2$, to the extent that it is admissable to suppose an upper limit of $1 \frac{n-1}{n^2-2n-1}$ be larger than $A$ by the same order which $A_2$ is less, and that this difference is nearly $n (A-A_3)$.
The limit as n goes large is $1$, since the fractional part approaches zero, but for those of us who track the disappearence direction, it is in the order of $1 \frac 1n$.
A: Let 
$$b_n={1!+2!+\cdots(n-1)!\over n!}$$  
It suffices to show that $\lim_{n\rightarrow\infty}b_n=0$.
Note that $0\lt b_n\lt1$ for all $n\gt1$.  (There are fewer than $n$ terms in the numerator, none larger than $(n-1)!$.)  This implies
$$0\lt b_n={1\over n}\left({1!+\cdots+(n-2)!+(n-1)!\over(n-1)! }\right)={1\over n}(b_{n-1}+1)\lt{2\over n}$$
so the limit $0$ follows.
