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Sequence limit (factorial derivative?)

  1. Nov 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the limit of the sequence given by [itex]S_{n}=\frac{n^{n}}{n!}[/itex]

    2. Relevant equations

    [itex]lim_{n->∞}\frac{n^{n}}{n!}[/itex]

    3. The attempt at a solution

    I know the sequence diverges, but that doesn't mean the limit is also ∞, right?
     
  2. jcsd
  3. Nov 15, 2012 #2

    Ray Vickson

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    Either the function f(n) = n^n / n! converges, or else f(n) → +∞ or f(n) → -∞ or else f(n) "oscillates" as n → ∞ in such a way that f(n) does not approach a definite value---not even ± ∞. You need to decide which applies here.

    RGV
     
  4. Nov 15, 2012 #3

    Zondrina

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    Ask yourself, which grows faster, the numerator or the denominator.
     
  5. Nov 15, 2012 #4
    Thank you both. As the numerator grows faster, and it's a divergence sequence, then the limit is +∞. Now, where may I start to formally prove it?
     
  6. Nov 15, 2012 #5

    Zondrina

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    Can you prove that [itex] n^n > n! [/itex] think about intervals.
     
  7. Nov 15, 2012 #6
    Yes, I can. For instance, I just evaluated [itex]lim_{n->∞}\frac{3^{n}}{(n+3)!}[/itex] as being 0 by showing that [itex]\forall x \geq 0, (n+3)!>3^{n}[/itex]. In a similar way, I may show that [itex]n^{n}>n![/itex] in the same interval.

    It just seems to me that this method isn't rigorous enough, you know? As an example, 3>1 is true, but that does't mean that [itex]lim_{x->∞}\frac{x}{3x}=0[/itex]

    Thank you
     
  8. Nov 15, 2012 #7

    dextercioby

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    You don't see that for arbitrary n

    n times n > 1 times 2 times ... times n ?
     
  9. Nov 15, 2012 #8
    Yes, I do, but this seems as an intuitive approach, to me. Isn't it?

    Just to be sure I got the concepts correctly: Let [itex]s_{n}=\frac{3^{n}}{(n+3)!}[/itex]. Then, [itex]s_{1}, s_{2}, ..., s_{n}[/itex] is a sequence, and the partial sum is [itex]S_{x}=s_{1}+s_{2}+...+s_{x}[/itex]. That being said, when I say I want to know the limit of the sequence [itex](lim_{n->∞}\frac{3^{n}}{(n+3)!})[/itex], I'm evaluating the "last" term, [itex]s_{n}[/itex], not the sum to the "last" term, [itex]S_{n}[/itex] right?

    Many thanks
     
  10. Nov 15, 2012 #9

    Ray Vickson

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    This is not quite enough: you need [itex] n^n / n! [/itex] to be unbounded, not just > 1.

    RGV
     
  11. Nov 15, 2012 #10

    dextercioby

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    Can you prove that, for n>3

    [tex] n^n > \frac{n^n}{n!} > n +1 [/tex] ?
     
  12. Nov 18, 2012 #11
    May it be done by induction? It clearly holds for n=3, then I assume it also holds for n=j, and show it's also valid for n=j+1. (Sorry, no paper and pen around right now, I'll try it as soon as I can)

    Thanks
     
  13. Nov 19, 2012 #12

    dextercioby

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    I don't think my method for the second inequality (the 1st is obvious) is induction.
     
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