6
$\begingroup$

This question already has an answer here:

I am trying to calculate the following limit without Stirling's relation. \begin{equation} \lim_{n \to \infty} \dfrac{n!}{n^n} \end{equation} I tried every trick I know but nothing works. Thank you very much.

$\endgroup$

marked as duplicate by Watson, Dirk, KittyL, Eric Wofsey, Stefan Mesken Dec 21 '16 at 20:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 4
    $\begingroup$ Note that $n! \leq n^{n-1}$. $\endgroup$ – Ivan Loh Nov 24 '13 at 22:36
  • $\begingroup$ It would converge to $0$ as for a very large $n$, $n!$ is puny compared to $n^n$ $\endgroup$ – Ali Caglayan Nov 24 '13 at 22:39
  • 1
    $\begingroup$ Hint:$$\frac{n!}{n^n}=\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\frac{n-3}{n}\cdot\ldots\cdot\frac{3}{n}\cdot\frac{2}{n}\cdot\frac{1}{n}$$From here you can "see" that one is certainly _losing_ as $n\to\infty$... $\endgroup$ – alandella Nov 24 '13 at 22:40
  • 6
    $\begingroup$ See also math.stackexchange.com/questions/61713/… $\endgroup$ – Martin Sleziak May 24 '15 at 7:30
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/1904113/… $\endgroup$ – Watson Dec 14 '16 at 12:40
16
$\begingroup$

By estimating all the factors in $n!$ except the first one, we get: $$0 \leq \lim_{n \rightarrow \infty} \frac{n!}{n^n} \leq \lim_{n \rightarrow \infty} \frac{n^{n-1}}{n^n} = \lim_{n \rightarrow \infty} \frac{1}{n} = 0$$

$\endgroup$
3
$\begingroup$

Consider the series $$\sum_{n=1}^\infty \frac{n!}{n^n} $$ of positive terms. The ratio of two consecutive terms is $$\frac{a_{n+1}}{a_n}=\frac{(n+1)!/(n+1)^{n+1}}{n!/n^n}= \left( \frac{n}{n+1} \right)^n=\left[ \left(1+\frac{1}{n} \right)^n \right]^{-1} $$ which tends to $e^{-1}<1$. It follows from the ratio test that the series converges, and by the necessary condition for convergence of series the limit is obtained. We have $$\lim_{n \to \infty} \frac{n!}{n^n}=0. $$

$\endgroup$
0
$\begingroup$

Hint:

We have that $n!=1\cdot 2\cdot ...\cdot n<n\cdot n \cdot ...\cdot n$ ,$n-1$ times.

$\endgroup$
  • 1
    $\begingroup$ You probably want to have $n-1$ times in there not $n$ because that doesn't tell you anything. $\endgroup$ – Deven Ware Nov 24 '13 at 22:43
  • $\begingroup$ Haha,correct,wrote it wrong.thnx $\endgroup$ – Haha Nov 24 '13 at 23:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.