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!

Infinite series convergence question:

  1. Jun 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Does [tex]\sum _{ n=1 }^{ \infty }{ \frac { { \alpha }^{ n }{ n }! }{ { n }^{ n } } } [/tex] converge [tex]\forall |\alpha |<e[/tex]
    and if so, how can I prove it?



    2. Relevant equations
    [tex]{ e }^{ x }=\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n } }{ n! } } [/tex]


    3. The attempt at a solution
    In the case of e I get the strange and probably wrong [tex]\sum _{ n=1 }^{ \infty }{ \frac { { (\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } } ) }^{ n }{ n }! }{ { n }^{ n } } } [/tex]
    which seems to be the boundary between convergence and divergence for the series, but I have no idea how to actually make anything of this.
     
  2. jcsd
  3. Jun 18, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Have you tried the ratio test for convergence?
     
  4. Jun 18, 2012 #3
    Yup.
    I tried it on [tex]\frac { { \alpha }^{ n }{ n }! }{ { n }^{ n } } [/tex]
    and ended up with: [tex]\lim _{ n\rightarrow \infty }{ \frac { { \alpha }^{ n }{ (n+1){ n }^{ n } } }{ { (n+1) }^{ n+1 } } } [/tex] but I don't see anything relating alpha to e.. I also don't know how to calculate that limit, i'm guessing it's zero but that would mean it's independent of alpha(≠0)..

    EDIT: typo correction:
    [tex]\lim _{ n\rightarrow \infty }{ \frac { { \alpha }{ (n+1){ n }^{ n } } }{ { (n+1) }^{ n+1 } } } [/tex]
     
    Last edited: Jun 18, 2012
  5. Jun 18, 2012 #4
    I'll try to manipulate the expression to get e out of the form of (1+1/n)^n
     
  6. Jun 18, 2012 #5
    ah, I did the above and my result is that for any alpha>e the ratio test yields r>1 and for any alpha<e the ratio test yields r<1. I'll think about r=1
     
  7. Jun 18, 2012 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You need absolute values around that, and I don't think you get ##\alpha^n## after you simplify the ratio. Check that.
     
  8. Jun 18, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Use Stirling's formula for the asymptotic form of n!. You can even improve it to a rigorous inequality that holds for all n > 1 (even small values of n):
    [tex] \sqrt{2\pi n} n^n e^{-n} < n! < \sqrt{2\pi n} n^n e^{\left(-n + \frac{1}{12n}\right)}.[/tex]
    Note: the left-hand quantity above is Stirling's asymptotic approximation to n!; the right-hand quantity is an improvement, and it is surprising just how good it is: for n = 2 we get 2! = 2, but RHS = 2.000652048, 5! = 120, RHS = 120.0026371, etc. Various "simple" proofs of these inequalities are available; my favourite one is in Feller, "Introduction to Probability Theory", Vol. 1, Wiley 1968.

    RGV
     
  9. Jun 18, 2012 #8
    @MLCKurtz
    Yes, I obtained alpha without an exponent, made a typo. Didn't forget the absolute value. Thanks
    @Ray Vickson
    Thanks for that reading material! I think i'll stick to my simpler idea, considering I don't have justifications to apply Stirling's formula (haven't gotten around to it yet).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Infinite series convergence question:
Loading...