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

Question

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:

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

Answer 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 $1$.

Answer 2

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 $1 \frac 1n$.

Answer 3

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

Answer 4

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.