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[Series] Divergence Test

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that the following series diverges

    [tex]\sum_{n=1}^{\infty}\frac{n!}{2^{n}} [/tex]

    2. Relevant equations

    The Divergence Test: In order for a series to be divergent, the following must be true

    [tex]\lim_{n\rightarrow \infty} a_n \neq 0 [/tex], or

    [tex]\lim_{n\rightarrow \infty} a_n \nexists [/tex]

    3. The attempt at a solution

    Alright, I know how to work it out with the denominator, as it is a geometric series and therefore as [tex]n \rightarrow \infty,\ 2^{n} \rightarrow 1 [/tex]

    But how do I do whenever I find a factorial? How do I work it out? I don't know what to do with this factorial, can I assume the following in this case

    As [tex]n \rightarrow \infty[/tex],

    [tex]n! \approx n[/tex]

    Then as [tex]n \rightarrow \infty[/tex] it would summarize to

    [tex]a_n = \frac{n}{2^{n}} [/tex], so by using L'hôpital's

    [tex]\frac{\frac{d}{dn}n}{\frac{d}{dn} 2^{n}} [/tex]

    [tex]\frac{1}{2^{n}}[/tex], and then as [tex]n \rightarrow \infty[/tex],

    [tex]\frac{1}{1} = 1 \neq 0 [/tex]


    Is this it?
     
    Last edited: Sep 6, 2010
  2. jcsd
  3. Sep 6, 2010 #2
    No.
    n!=n(n-1)(n-2).... *3*(2)*(1) [ n terms]
    [tex]2^{n}= 2*2*.... *2[/tex] n times.

    I have no idea what you are doing there.

    Please write the expansion of n! and [tex]2^{n}[/tex] as the numerator and denominator then try to see why the limit does not exist.
     
  4. Sep 6, 2010 #3
    Nevermind what I wrote before.

    I was trying to show that the series was divergent by [tex]\lim_{n\rightarrow \infty} a_n \neq 0[/tex], but that's not possible because the limit doesn't exist, that's why I messed up. But by using [tex]\lim_{n\rightarrow \infty} a_n \nexists [/tex] it can be shown that the series diverge. I had forgotten this second case and that's why I did some mess before on working out the first case heh.
     
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