How can I compute this limit: $\lim_{n \to \infty} \frac{n!}{n^n}$. I have tried taking the n-th root but then I have problem managing the factorial. Thanks to everyone who will help!
Asked
Active
Viewed 88 times
1
-
1$$ \frac{n!}{n^n} = \frac{1 \times 2 \times 3 \times 4 \dots \times n}{n \times n \times n \times n \dots \times n} $$ How does it look like to you? – Matti P. Jun 05 '18 at 08:22
-
@user502940 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Aug 03 '18 at 21:54
3 Answers
3
You don't have to involve a root or ratio test here. $$ \frac{n!}{n^n} = \frac{1 \times 2 \times \cdots \times n}{n \times n \times \cdots \times n} = \frac1n \times \frac2n \times \cdots \times \frac nn < \frac 1n. $$
Mees de Vries
- 26,947
2
Let $a_n=\frac{n!}{n^n}$
Prove that $\frac{a_{n+1}}{a_n} \to \frac{1}{e}<1$
Thus from ratio test for sequences we have that $a_n \to 0$
Marios Gretsas
- 21,906
1
HINT
By ratio test
$$\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!} =\frac{1}{\left(1+\frac1n\right)^n}$$
and recall that $\left(1+\frac1n\right)^n\to e$.
user
- 154,566