1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    FALSE ALARM

    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

    sharks

    User Avatar
    Gold Member

    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

    sharks

    User Avatar
    Gold Member

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Science Advisor
    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).

    RGV
     
    Last edited: Jun 1, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook