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Homework Help: Limit of a sequence n!/n^n

  1. Jun 1, 2012 #1
    The problem statement, all variables and given/known data

    Using the sandwich rule (which i understand) find the limit of n!/nn

    The attempt at a solution

    To my knowledge n! is the fastest growing function you can have, so I immediately thought the function did not have a limit, however, the answer states the limit to be 1

    I know that the bound on n! would be 1 =< n! < inf (?) but then dividing through by nn doesn't really help.

    I am stuck :/

    What am I doing wrong?
  2. jcsd
  3. Jun 1, 2012 #2

    After a bit of searching on the internet, the answer in the book is wrong. phew!

    Answer is 0 as i initially thought
  4. Jun 1, 2012 #3


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    What limits did you compare it to, for the Sandwich theorem?
  5. Jun 1, 2012 #4
    Just asking what the limit of f(x) is, gives no meaning. You need to specify to what value n approaches?
  6. Jun 1, 2012 #5
    apologies as n approaches infinity
  7. Jun 1, 2012 #6
    i said the lower limit for n! was 1 and the upper limit was infinity....
  8. Jun 1, 2012 #7
    Aye, 0 is correct, then. :smile:

    If, supposedly, n->0, then the limit would be 1 :wink:

    From this reasoning(incorrect), I don't see how you got 0 as your answer though....
  9. Jun 1, 2012 #8


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    And how did you end up with those limits? Just curious. :smile:
  10. Jun 1, 2012 #9
    Isn't it quite intuitive that 0 is correct?
  11. Jun 1, 2012 #10


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    Then your knowlege is incorrect. n! is the product of n terms each less than or equal to n. [itex]n^n[/itex] is the product of n terms, each equal to n. For all n, [itex]n!< n^n[/itex].

  12. Jun 1, 2012 #11

    Ray Vickson

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    Homework Helper

    n! is NOT the fastest growing function you can have; there are infinitely many functions that grow much faster than n! (for example, (n!)^2 or exp(n!) or n^(n!) to name a few).

    Anyway, do you know Stirling's formula? It says
    [tex] n! \sim \sqrt{2 \pi}\: n^{n + 1/2} e^{-n}, [/tex] for large n, where "~" means "is asymptotic to", which in turn means that the ratio of the two sides has limit 1 as n → ∞. Thus, [tex] \frac{n!}{n^n} \sim \sqrt{2 \pi} \: \sqrt{n} e^{-n} \rightarrow 0[/tex] as [itex] n \rightarrow \infty.[/itex].

    If you feel a bit suspicious about the use of "~" you can squeeze the ratio between two easily-analyzed expressions, because the original Stirling approximation can be turned into a rigorous pair of inequalities that are true for all n > 1:
    [tex] \sqrt{2 \pi}\: n^{n + 1/2} e^{-n} < n! < \sqrt{2 \pi} \: n^{n + 1/2} e^{-n + (1/(12 n))} \:\:\forall n > 1. [/tex] For a simple proof, see,. eg., W. Feller, "Introduction to Probability Theory and its Applications", Volume I, Wiley (1968).

    Last edited: Jun 1, 2012
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