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Attempt at solving a limit

  1. Feb 25, 2006 #1
    Show that
    [tex]\lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \infty \qquad a \in \mathbb{N}[/tex]

    Here is my try:

    Take [tex] a_n = a^{2n+1}/n![/tex] . Since
    \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{a^{2(n+1)+1}/\,(n+1)!}{a^{2n+1}/\,n!} = \lim_{n \to \infty} \frac{a^{2}}{n+1} = 0 < 1 [/tex](D'Alembert test),

    the series [tex]\sum_{n=0}^\infty a_n[/tex] is convergent, i.e.[tex]
    \lim_{n \to \infty} a_n = 0.


    \lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \lim_{n \to \infty} \frac{1}{a_n} = \infty.

    Correct? Is there another way to prove it?
    Last edited: Feb 25, 2006
  2. jcsd
  3. Feb 25, 2006 #2
    That works. You can also approach it from the standpoint that n! is eventually larger than any power of fixed a for n large enough. (Well, as long as we aren't talking about [tex]a^{n!}[/tex] or something crazy like that!)

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