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Proving lim (as n -> infinity) 2^n/n! = 0

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that lim n [tex]\rightarrow[/tex][tex]\infty[/tex] 2[tex]^{}n[/tex]/n! = 0

    2. Relevant equations
    This implies that 2[tex]^{}n[/tex]/n! is a null sequence and so therefore this must hold:
    ([tex]\forall[/tex] E >0)([tex]\exists[/tex]N E N[tex]^{}+[/tex])([tex]\forall[/tex]n E N[tex]^{}+[/tex])[(n > N) [tex]\Rightarrow[/tex] (|a[tex]_{}n[/tex]| < E)

    3. The attempt at a solution
    Whenever I have proved these before I have tried to eliminate n from the top and then use Archimedian Principle to finsih off the proof. However I don't know how to do this in this case. I have the idea that 2/k [tex]\leq[/tex]2/3 for any k[tex]\geq[/tex]3
  2. jcsd
  3. Jan 18, 2009 #2


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    It is very easy to see that
    [tex]\frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n}= \frac{2}{n+1}[/tex]
    so that each term is less than
    times the previous term.

    It is also very easy to see that, for n= 4,
    [tex]\frac{2^n}{n!}= \frac{16}{24}< 1[/itex]

    That is, if n> 4,
    [tex]\frac{2^n}{n!}< \left(\frac{2}{5}\right)^n\frac{2}{3}[/tex]
    and that last goes to 0.
  4. Jan 18, 2009 #3
    i see where you're going but where does that final 2/3 come from?
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