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anonymous

  • one year ago

Evaluate the limit

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  1. anonymous
    • one year ago
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    \[\lim_{n \rightarrow \infty}\frac{ n! }{ n^n }\]

  2. anonymous
    • one year ago
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    If I expand the factorial the first n term cancels but I'm really sure where to go from there.

  3. anonymous
    • one year ago
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    |dw:1441847154717:dw|

  4. anonymous
    • one year ago
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    Not sure how to keep going after that...

  5. zzr0ck3r
    • one year ago
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    show that \(n!<n^n\)

  6. anonymous
    • one year ago
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    Induction. Ohh boy.

  7. Rizags
    • one year ago
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    I would begin by changing \[\frac{n!}{n^n}\] into \[n^{-n}n!\] then logarithms

  8. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \[\frac{n!}{n^n}=\frac{n(n-1)\cdots 32}{nn\cdots nn}\frac{1}{n}\le1\cdot \frac{1}{n}=\frac{1}{n}\]

  11. anonymous
    • one year ago
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    Where did you get a 32 from?

  12. Zarkon
    • one year ago
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    ...3*2*1

  13. anonymous
    • one year ago
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    Ohh.

  14. anonymous
    • one year ago
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    Nope still not clicking.

  15. anonymous
    • one year ago
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    Wait I got it. it was just induction. Bah.

  16. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    \[\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\cdots\frac{3}{n}\frac{2}{n}\frac{1}{n}\le 1\cdot1\cdot1\cdots1\cdot1\cdot\frac{1}{n}=\frac{1}{n}\] then use squeeze theorem

  18. anonymous
    • one year ago
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    Aha! That's brilliant! I used induction but there seems to be 2 methods :) .

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