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!

Stirling's approximation problem

  1. Mar 27, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\sum n^n/ n![/tex]


    This example is in the book, it concludes that the above series is :

    (1 +1/n)^n, which converges to e and n->infinity

    how is this so?

    If i take the root test then, is not answer 1 as n->infinity

    Can you explain?
     
  2. jcsd
  3. Mar 27, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Divergent/convergent

    How about using Stirling's approximation for the n!?
     
  4. Mar 27, 2009 #3
    Re: Divergent/convergent

    Haven't got there yet
     
  5. Mar 27, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Divergent/convergent

    That series clearly diverges. Every term is greater than 1. What's the real question? Does it have an x^n in it? Then use the ratio test.
     
    Last edited: Mar 27, 2009
  6. Mar 27, 2009 #5
    Re: Divergent/convergent

    Prove the following:

    lim (1 +1/n)^n = e
    n->infinity
     
  7. Mar 27, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Divergent/convergent

    That's not a series. That's a limit. What's your definition of e? Because that's one of them.
     
  8. Mar 27, 2009 #7
    Re: Divergent/convergent

    so how is it that the above limit goes to e = 2.7128282....
     
  9. Mar 27, 2009 #8
    Re: Divergent/convergent

    dont u just re-arrange to find n if e = 2.7?

    if u draw the graph on ur calculator u can see that the left side jumps up then stops at (-1.27, 7.15), then all of a sudden on the right side it shows up at about (0.3, 1.57) again, then moves up slowly
     
    Last edited: Mar 27, 2009
  10. Mar 27, 2009 #9
    Re: Divergent/convergent

    Alright forget about this minor problem.
     
  11. Mar 28, 2009 #10

    Mark44

    Staff: Mentor

    Re: Divergent/convergent

    Let y = (1 + 1/n)n (no limits just yet)
    ln y = ln (1 + 1/n)n = n ln (1 + 1/n)

    lim ln y = lim n ln (1 + 1/n) ( all limits as n --> [itex]\infty[/itex])

    = lim ln (1 + 1/n) / (1/n) (the limit on right is of the form 0/0, so L'Hopital's Rule applies)
    [tex]= lim \frac{\frac{-1/n^2}{1 + 1/n}}{\frac{-1}{n^2}}[/tex]
    = lim 1/(1 + 1/n) = 1

    So lim ln y = 1, or
    ln (lim y) = 1, or
    lim y = e1 = e

    Therefore, lim (1 +1/n)n = e
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Stirling's approximation problem
  1. Approximation problem (Replies: 12)

Loading...