1. Not finding help here? Sign up for a free 30min 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!

Does the series converge? n!/n^n

  1. Nov 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Σ(n!)/(n^n)

    2. Relevant equations
    Does the series converge or diverge?


    3. The attempt at a solution


    My prof. suggested we compare it to n^(n-2)/n^n.

    I found that n^(n-2)/n^n was convergent since it was like 1/n^2 which is a p series where p>1. Therefore by the comparison theorem, the original series is convergent.

    I'm hoping this is the right answer, but I thought it was too easy for what was deemed a tricky problem, even with the hint.
     
  2. jcsd
  3. Nov 2, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    All fine, but what about the comparison part? How did you show n!<n^(n-2)? It's not true for all n...
     
  4. Nov 2, 2007 #3

    Curious3141

    User Avatar
    Homework Helper

    Ratio test (where you prove the limit of the ratio of consecutive terms is less than one (in this case, it's 1/e) is MUCH more straightforward, given that you have to prove what Dick said if you use the comparison test.
     
  5. Nov 2, 2007 #4

    Curious3141

    User Avatar
    Homework Helper

    And in fact, this series folds very easily under the root test.

    General term is n!/(n^n) = a_n.

    (a_n)^(1/n) = [(n!)^(1/n)]/n = {[(n)(n-1)...1]^(1/n)}/n < [(n^n)^(1/n)]/n = 1

    Sorry that I didn't format with Tex.
     
  6. Nov 2, 2007 #5
    Dick: I have no idea. It was suggested and I just went through with it. I just took it as true.

    I'm still rather confused because the ratio test is what I should use, but we haven't actually covered that in class.
     
  7. Nov 2, 2007 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It is true for n>=5, which is enough to allow you to use the comparison test. The easiest proof of the inequality actually looks rather like the ratio test. But if your prof suggested you do that, then maybe a proof isn't needed.
     
  8. Nov 2, 2007 #7

    Curious3141

    User Avatar
    Homework Helper

    To "see" it is easy. To prove it, not so - it's messy (you have to consider the monotone increasing nature of the functions, sketch curves etc.) You can use induction, but it is inelegant.

    The easiest proof is the root test (as I've shown). The professor was only making a suggestion, I guess, and a better way can, and should be sought.
     
    Last edited: Nov 2, 2007
  9. Nov 3, 2007 #8

    Gib Z

    User Avatar
    Homework Helper

    Well, this series converges if [tex]\sum_{n=1}^{\infty} \frac{\sqrt{2\pi n} \left(\frac{e}{n}\right)^n}{n^n}[/tex] converges.
     
  10. Nov 3, 2007 #9
    i was supposed to do this without the ratio or root test. we just discussed ratio test today in class. I don't really know what the book/class was expecting.
     
  11. Nov 3, 2007 #10

    learningphysics

    User Avatar
    Homework Helper

    Think the idea was to use induction... as Dick pointed out for n = 5, n!<n^(n-2). so using induction... assume n>=5. assume n!<n^(n-2)... prove (n+1)!<(n+1)^[(n+1)-2] ie:

    (n+1)! < (n+1)^(n-1)

    so start with:

    (n+1)! = (n+1)n! < (n+1)[n^(n-2)] < (n+1)[(n+1)^(n-2)] = (n+1)^(n-1)

    so using the comparison test your series converges.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?