Results 1 to 5 of 5

Math Help - [SOLVED] Don't understand why this sequence is divergent.

  1. #1
    Newbie
    Joined
    Jan 2008
    Posts
    12

    [SOLVED] Don't understand why this sequence is divergent.

     a_n = \frac{5n!}{2^n}

    I don't understand why this sequence is divergent. Any help is much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by miniwheats View Post
     a_n = \frac{5n!}{2^n}

    I don't understand why this sequence is divergent. Any help is much appreciated.
    Becasue \lim_{n \to {\infty}}a_n=\lim_{n \to {\infty}}\frac{5n!}{2^n}=\infty this can be either shown by knowledge of rate of increase of n! vesrsus 2^n...or using a Gamma function substitution for n! and applying L'hopital's rule
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2008
    Posts
    12
    Sorry, I should've rephrased my question. I don't understand how to prove that the sequence is divergent. My solutions manual shows this:

     a_n = \frac{5n!}{2^n} = \frac{5}{2} \times \frac{2}{2} \times \frac{3}{2} \times \frac{4}{2} ... \frac{n-1}{2} \times \frac{n}{2} \geq \frac{5}{2} \times \frac{n}{2} \ \ for\ n > 2 = \frac{5n}{4} \rightarrow \infty, so\ {a_n}\ diverges.

    But I don't understand what they did.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by miniwheats View Post
    Sorry, I should've rephrased my question. I don't understand how to prove that the sequence is divergent. My solutions manual shows this:

     a_n = \frac{5n!}{2^n} = \frac{5}{2} \times \frac{2}{2} \times \frac{3}{2} \times \frac{4}{2} ... \frac{n-1}{2} \times \frac{n}{2} \geq \frac{5}{2} \times \frac{n}{2} \ \ for\ n > 2 = \frac{5n}{4} \rightarrow \infty, so\ {a_n}\ diverges.

    But I don't understand what they did.
    Saying \lim_{n \to {\infty}}a_n\ne{0} the definition of a divergent sequence isn't enough?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2008
    Posts
    12
    That works too. Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 26th 2011, 03:33 AM
  2. Divergent Sequence 61.17
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 27th 2009, 07:15 AM
  3. Replies: 2
    Last Post: October 9th 2009, 08:23 AM
  4. Convergent or divergent sequence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 12th 2009, 11:05 AM
  5. Replies: 3
    Last Post: November 30th 2008, 06:57 AM

Search Tags


/mathhelpforum @mathhelpforum