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Help prove the limit of x^n/n! = 0

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data
    lim x^n/n! = 0 for all x
    n->∞


    2. Relevant equations
    No equations, but I am currently in Calculus II in my first semester here at my University. Please help me prove why this limit is true!


    3. The attempt at a solution
    I've tried to use natural logarithms and L'Hospital's rule, but I don't know if I can derive (n!). Any ideas?
     
  2. jcsd
  3. Sep 26, 2012 #2

    Zondrina

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    Do you know that the factorial function grows faster than any x^n?
     
  4. Sep 26, 2012 #3
    Yeah I do know this, but I am not sure how to prove it.
     
  5. Sep 26, 2012 #4

    micromass

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    Have you seen series?? Can you show that

    [tex]\sum_{n=1}^{+\infty} \frac{x^n}{n!}[/tex]

    converges?
     
  6. Sep 26, 2012 #5

    Ray Vickson

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    Assume x > 0. We then have
    [tex] \log\left(\frac{x^n}{n!}\right) = n \log(x) - \log(1) - \log(2) - \cdots - \log(n).[/tex]
    Try to bound the sum of the logs. (Hint: compare log(j) with the integral of log(x) over [j-1,j] or [j,j+1], whichever one gives you what you need.)

    RGV
     
  7. Sep 26, 2012 #6
    To be honest I really don't know what that means. In high school our teacher didn't worry about that kind of sigma notation because it wasn't on the AP test. So I kind of understand it after reading into it myself, but it was never anything that got set into my mind.

    I think I know what you mean. I'm going to try this out tomorrow. But thanks guys.
     
  8. Sep 26, 2012 #7

    Zondrina

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    Fun fact :

    [tex]e^x = \sum_{n=1}^{\infty} \frac{x^n}{n!}[/tex]
     
  9. Sep 26, 2012 #8

    Ray Vickson

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    You can also use the fact that for 0 < x < N, and n > N we have
    [tex]0 < \frac{x^n}{n!} < \frac{N^n}{n!}
    = \frac{N^N}{N!} \frac{N}{N+1} \frac{N}{N+2} \cdots \frac{N}{n}. [/tex]
    For any r in (0,1) we have
    [tex] \frac{N}{N+1} \frac{N}{N+2} \cdots \frac{N}{n} < r^{n-N} [/tex] for all n sufficiently large (why?) so the desired result follows for any x.

    RGV
     
  10. Sep 27, 2012 #9

    jbunniii

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    Choose any fixed [itex]x[/itex]. If [itex]a_n = x^n / n![/itex], then
    [tex]\left|\frac{a_{n+1}}{a_n}\right| = \left| \frac{x^{n+1} n!}{x^n (n+1)!}\right| = \frac{|x|}{n+1}[/tex]
    which is arbitrarily small if [itex]n[/itex] is large enough. In particular, if I choose [itex]r[/itex] with [itex]0 < r < 1[/itex], then for sufficiently large [itex]n[/itex], say [itex]n \geq N[/itex], we have [itex]|a_{n+1}/a_{n}| < r[/itex]. What can you conclude?
     
  11. Sep 27, 2012 #10
    This explanation actually helped a lot. Thanks so much to you and the others that tried helping
     
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