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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.

Martin Sleziak
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Lely
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    Note that $n! \leq n^{n-1}$. – Ivan Loh Nov 24 '13 at 22:36
  • It would converge to $0$ as for a very large $n$, $n!$ is puny compared to $n^n$ – Ali Caglayan Nov 24 '13 at 22:39
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    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 $n\to\infty$... – TheVal Nov 24 '13 at 22:40
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    See also http://math.stackexchange.com/questions/61713/whats-the-limit-of-the-sequence-lim-n-rightarrow-infty-fracnnn – Martin Sleziak May 24 '15 at 07:30
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    Related: http://math.stackexchange.com/questions/1904113/limit-cn-n-nn-as-n-goes-to-infinity – Watson Dec 14 '16 at 12:40

3 Answers3

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

Leif Sabellek
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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 $$\lim_{n \to \infty} \frac{n!}{n^n}=0. $$

user1337
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Hint:

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

Haha
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