## anonymous one year ago Evaluate the limit

1. anonymous

$\lim_{n \rightarrow \infty}\frac{ n! }{ n^n }$

2. anonymous

If I expand the factorial the first n term cancels but I'm really sure where to go from there.

3. anonymous

|dw:1441847154717:dw|

4. anonymous

Not sure how to keep going after that...

5. zzr0ck3r

show that $$n!<n^n$$

6. anonymous

Induction. Ohh boy.

7. anonymous

I would begin by changing $\frac{n!}{n^n}$ into $n^{-n}n!$ then logarithms

8. anonymous

If we are going to use logs and then L'Hopitals rule I am not allowed to do that. I have to do this without L'hopital's rule.

9. anonymous

I mean, I could use induction to show that n! is less than n^n and then take the limit. Would that be sufficient?

10. Zarkon

$\frac{n!}{n^n}=\frac{n(n-1)\cdots 32}{nn\cdots nn}\frac{1}{n}\le1\cdot \frac{1}{n}=\frac{1}{n}$

11. anonymous

Where did you get a 32 from?

12. Zarkon

...3*2*1

13. anonymous

Ohh.

14. anonymous

Nope still not clicking.

15. anonymous

Wait I got it. it was just induction. Bah.

16. jim_thompson5910

Proving n! < n^n Here's an informal argument. It probably won't work as a proof, but it's the way I see it n! = n*(n-1)*(n-2)*...*3*2*1 n^n = n*n*n...*n*n There are n copies of factors being multiplied in each term. So we can pair up the factors between n! and n^n n pairs with n (first factors) n-1 pairs with n (second factors) n-2 pairs with n (third factors) and so on until... 3 pairs with n (third to last factors) 2 pairs with n (second to last factors) 1 pairs with n (last factors) On the left side we have n,n-1,n-2,... those factors are getting smaller. On the right side, the factors stay as n which don't change. So one side (n!) gets smaller factors while the other side (n^n) has the factors stay the same. So the product of all of this must have n! < n^n. There is no way to have n! > n^n since the factors of n! would have to be larger than the corresponding factors of n^n Again this is just an informal way to think of it and not a formal proof.

17. Zarkon

$\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\cdots\frac{3}{n}\frac{2}{n}\frac{1}{n}\le 1\cdot1\cdot1\cdots1\cdot1\cdot\frac{1}{n}=\frac{1}{n}$ then use squeeze theorem

18. anonymous

Aha! That's brilliant! I used induction but there seems to be 2 methods :) .