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anonymous
 one year ago
Evaluate the limit
anonymous
 one year ago
Evaluate the limit

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty}\frac{ n! }{ n^n }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I expand the factorial the first n term cancels but I'm really sure where to go from there.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441847154717:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not sure how to keep going after that...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would begin by changing \[\frac{n!}{n^n}\] into \[n^{n}n!\] then logarithms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean, I could use induction to show that n! is less than n^n and then take the limit. Would that be sufficient?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{n!}{n^n}=\frac{n(n1)\cdots 32}{nn\cdots nn}\frac{1}{n}\le1\cdot \frac{1}{n}=\frac{1}{n}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where did you get a 32 from?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope still not clicking.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait I got it. it was just induction. Bah.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0Proving 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*(n1)*(n2)*...*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) n1 pairs with n (second factors) n2 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,n1,n2,... 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.

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{n}{n}\frac{n1}{n}\frac{n2}{n}\cdots\frac{3}{n}\frac{2}{n}\frac{1}{n}\le 1\cdot1\cdot1\cdots1\cdot1\cdot\frac{1}{n}=\frac{1}{n}\] then use squeeze theorem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Aha! That's brilliant! I used induction but there seems to be 2 methods :) .
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