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Homework Help: 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

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    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

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    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

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    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).
     
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