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How to show that lim 2^n/n! = 0 with n--> infinity

  1. Sep 21, 2006 #1

    kreil

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    Gold Member

    Using the definition of a limit, show that

    [tex]\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0[/tex]

    If someone could get me started that would be great.

    thanks
    josh
     
    Last edited: Sep 21, 2006
  2. jcsd
  3. Sep 21, 2006 #2
    [tex]\frac{2^n}{n!}=\frac{2}{n}\frac{2}{n-1}.....\frac{2}{2}
    \frac{2}{1}[/tex]

    so when [tex]n\rightarrow\infty[/tex]

    we get [tex]\frac{2^n}{n!}<\frac{4}{n}[/tex]

    to any [tex]\varepsilon > 0[/tex]

    [tex]\exists N=[\frac{4}{\varepsilon}]+1[/tex]

    when n>N

    [tex]\frac{2^n}{n!}<\frac{4}{n}
    <\frac{4}{[\frac{4}{\varepsilon}]+1}<\varepsilon[/tex]

    now we get the conclusion:

    [tex]\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0[/tex]
     
    Last edited: Sep 21, 2006
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