Results 1 to 6 of 6

Math Help - Convergence/Divergence check

  1. #1
    Junior Member cinder's Avatar
    Joined
    Feb 2006
    Posts
    60

    Convergence/Divergence check

    Not sure how to test this one:

    \sum_{n=0}^{\infty} \frac{(-1)^{n-1}4}{n!}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by cinder View Post
    Not sure how to test this one:

    \sum_{n=0}^{\infty} \frac{(-1)^{n-1}4}{n!}
    \sum_{n=0}^{\infty} \frac{(-1)^{n-1}4}{n!}=-4~ e^{-1}

    so I guess its convergent.

    Alternativly as it is an alternating series of tems whose absolute values are
    decreasing and whose limit is 0, then it is convergent. Also it is absolutely
    convergent as:

    1/n! < n^2,~ \forall~ n>3

    and

    \sum_1^{\infty}1/n^2

    converges


    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member cinder's Avatar
    Joined
    Feb 2006
    Posts
    60
    I think it can be tested with the alternating series test, but I'm not sure how to go about it. The book says it converges if 0 < a_{n+1} \leq a_n and \lim_{n\rightarrow\infty} a_n=0, but I don't understand that.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member cinder's Avatar
    Joined
    Feb 2006
    Posts
    60
    And is \sum_{n=0}^{\infty} e^{-n} divergent?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by cinder View Post
    And is \sum_{n=0}^{\infty} e^{-n} divergent?
    Its a geometric series with common ratio less than 1, so yes it does converge.

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,662
    Thanks
    298
    Awards
    1
    Quote Originally Posted by cinder View Post
    Not sure how to test this one:

    \sum_{n=0}^{\infty} \frac{(-1)^{n-1}4}{n!}
    Quote Originally Posted by cinder View Post
    I think it can be tested with the alternating series test, but I'm not sure how to go about it. The book says it converges if 0 < a_{n+1} \leq a_n and \lim_{n\rightarrow\infty} a_n=0, but I don't understand that.
    Either my definition of a_n from the series is wrong or the theorem should be:
    0 < |a_{n + 1}| \leq |a_n|

    So:
    |a_n| = \frac{4}{n!}

    and

    |a_{n + 1}| = \frac{4}{(n + 1)!}

    We obviously have that
    0 < |a_{n + 1}| \leq |a_n|
    so that part of the theorem is satisfied.

    Now for
    \lim_{n \to \infty} \frac{(-1)^{n-1}4}{n!}

    The constant term, of course, is immaterial here. The (-1) term simply oscillates between -1 and 1, so if the other part of the limit is zero then we're okay. And, in fact \lim_{n \to \infty}\frac{1}{n!} = 0, so
    \lim_{n \to \infty} \frac{(-1)^{n-1}4}{n!} = 0
    as required.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convergence/divergence
    Posted in the Calculus Forum
    Replies: 9
    Last Post: March 30th 2011, 07:37 AM
  2. convergence or divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 23rd 2009, 05:42 PM
  3. Divergence answer check.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 31st 2009, 03:21 PM
  4. Convergence or Divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 28th 2009, 11:53 AM
  5. Convergence / Divergence
    Posted in the Calculus Forum
    Replies: 8
    Last Post: April 6th 2008, 07:51 AM

Search Tags


/mathhelpforum @mathhelpforum