I'm having problems with following limit:


This is what I did now: I already proved this sequence is monotone descending, at least for $\;n\ge4\;$ , and since zero , or even one, is a low bound limit exists.

Also think limit is $\;1\;$ since $\;n!\;$ is $\;n\;$ times greater than $\;(n-1)!\;$ , so in numerator we can like omit all summands except last one, but of course I need something formalizing more.

I don't know of Stirling approximations or integrals, but someone told me that won't help in this.

Any help is appreciated.

  • 1
    $\begingroup$ @GregoryGrant $100n$ is much bigger than $n$, but it does not mean that the limit of $\frac{100n}{n}$ is infinity. $\endgroup$ – lisyarus Mar 30 '15 at 18:28
  • $\begingroup$ @GregoryGrant How do you figure that necessarily? $n!$ is the dominant term in the numerator. The catch is that there are a lot of terms in the numerator, which in principle could counterbalance that effect. $\endgroup$ – Ian Mar 30 '15 at 18:29
  • $\begingroup$ Wouldn't $\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$ work? $\endgroup$ – Miguelgondu Mar 30 '15 at 18:30

HINT: The numerator is less than $(n-2)!+(n-2)!+...+(n-2)!+(n-1)!+n!$

  • $\begingroup$ Wow, great hint! Thank you very much. +1 $\endgroup$ – user177692 Mar 30 '15 at 18:38

Hint: Show that $S_{n+1} = S_n/(n+1) + 1$. If you really showed it is monotone decreasing, then the limit exists and you can take limits of the expression given here to find out what it is.

Hint 2: Fix $m$ and write $$\frac{1! + \cdots + (n-m)!}{n!} + \frac{(n-m+1)! + \cdots n!}{n!}$$ and realize $$\frac{1! + \cdots + (n-m)!}{n!} \leq \frac{(n-m)(n-m)!}{n!}$$

  • $\begingroup$ Very nice indeed! Thank you very much. +1 $\endgroup$ – user177692 Mar 30 '15 at 18:40

Since you have already shown that this limit exists. Now what you can do is the following

Let $x_n = \dfrac{1! + \cdots + n!}{n!}$. Look at $x_{n+1} - \dfrac{x_n}{n+1}$.

\begin{equation} x_{n+1} - \dfrac{x_n}{n+1} = \frac{(n+1)!}{(n+1)!} = 1. \end{equation} If we assume that $\lim_\limits{n \to \infty} x_n = a$ and we also observe that $\lim_\limits{n\to\infty}\dfrac{x_n}{n+1} = 0$ because $\lim_\limits{n \to \infty} x_n = a$. Thus taking limit as $n \to \infty$ in the above equation gives \begin{equation} \lim_\limits{n\to \infty}\left(x_{n+1} - \dfrac{x_n}{n+1}\right) =\lim_\limits{n \to \infty} x_n = a. \end{equation} And since $\lim_\limits{n\to \infty}\left(x_{n+1} - \dfrac{x_n}{n+1}\right) = 1$, it follows that $a = 1$.

  • $\begingroup$ Thank you. This is basically nayrb's answer. $\endgroup$ – user177692 Mar 30 '15 at 18:49
  • $\begingroup$ I was little slow in typing I guess....lol... $\endgroup$ – Urban PENDU Mar 30 '15 at 18:50

Let $a_n=(1!+2!+\cdots+n!)/n!$. It's easy to see that

$$a_n=1+{a_{n-1}\over n}$$

From $a_1=1\lt2$, it follows by induction that $a_n\lt2$ for all $n$, since $a_{n-1}\lt2$ and $n\gt1$ together imply

$$a_n\lt1+{2\over n}\le1+{2\over2}=2$$

But $1\le a_n$ for all $n$ is also clear, so we now have

$$1\le a_n=1+{a_{n-1}\over n}\lt 1+{2\over n}$$

for all $n$, and so the Squeeze Theorem tells us $\lim_{n\to\infty}a_n=1$.


You can also solve this $\;\lim_{n\to\infty}\frac{1!+2!+\ldots+n!}{n!}\;$ by Cesaro - Stolz theorem



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