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Lightning question before the test : )

  1. Nov 6, 2007 #1
    How can I show that the sequence { 4^(n)/n! }

    converges to zero (by squeeze theorem)??

    Thank you! : )
  2. jcsd
  3. Nov 6, 2007 #2


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    If the Squeeze Theorem is required as part of the proof, how about comparing

    [tex]\frac{4^{n}}{n!}[/tex] < [tex]\frac{4^{n}}{n^{n}}[/tex] (for n > 1)

    and showing that the latter goes to zero as n--> infinity?
  4. Nov 7, 2007 #3

    Gib Z

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    Minor correction, but the "for n > 1" part isn't correct. You can find the correct value, or just say sufficiently large n.
  5. Nov 7, 2007 #4


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    as n->infinity n^n>n! so [tex]\frac{4^n}{n!} > \frac{4^{n}}{n^{n}}[/tex].

    I think probably the idea being expressed was to use a constant in the denominator instead of n. [tex]\frac{4^n}{8^n}[/tex]. (or any constant^n in the denominator where the constant>4)

    [tex]\frac{4^n}{n!}< \frac{4^n}{8^n}[/tex] for large enough n.
    Last edited: Nov 7, 2007
  6. Nov 7, 2007 #5


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    Whoops! Quite so -- I was only looking at the denominators. That needs to be n > 2 , doesn't it? (n=3 works...)
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