Results 1 to 6 of 6

Math Help - Find the closed-form of the series.

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    54

    Find the closed-form of the series.

    Prove that

    \sum_{n=0}^{\infty}{{2n}\choose n} x^n diverges for |x| = 1/4.

    Also find the closed-form of the series for x in the radius of convergence.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    As for |x| < \frac{1}{4} the closed form can be obtained from applying the binomial series to (1-4x)^{-1/2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by mathwizard View Post
    Prove that

    \sum_{n=0}^{\infty}{{2n}\choose n} x^n diverges for |x| = 1/4.

    Also find the closed-form of the series for x in the radius of convergence.
    \lim_{n\to\infty}\frac{\sqrt{4\pi{n}}(n^n)^2(e^{-n})^24^n}{2\pi{n}(n^n)^2(e^{-n})^24^n}=0

    Therefore when x=\frac{-1}{4}

    The series converges
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2008
    Posts
    54
    Quote Originally Posted by Mathstud28 View Post
    \lim_{n\to\infty}\frac{\sqrt{4\pi{n}}(n^n)^2(e^{-n})^24^n}{2\pi{n}(n^n)^2(e^{-n})^24^n}=0

    Therefore when x=\frac{-1}{4}

    The series converges
    Huh? What test are you using to determine the convergence (are you using Stirling's approximation for n!?)?

    The ratio test seems to yield an inconclusive result (i.e., the limit is 1).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    For x=1/4 I think you can use Abel's theorem. It would mean the series is continous at x=1/4 which is a problem because it is equal to (1-4x)^{-1/2} when |x|<1/4 and therefore not defined at x=1/4.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by mathwizard View Post
    Huh? What test are you using to determine the convergence (are you using Stirling's approximation for n!?)?

    The ratio test seems to yield an inconclusive result (i.e., the limit is 1).
    Correct me if I am wrong but,

    Let a_n=\frac{(2n)!}{4^n(n!)^2}

    Well

    \exists{N}\backepsilon\forall{n>N}\text{ }a_{n+1}<a_n

    and \lim_{n\to\infty}a_n=0

    Thus the series converges by the alternating series test
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Closed form solution of matrix series?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 19th 2011, 01:41 PM
  2. Is there a closed form solution to this infinite series?
    Posted in the Advanced Math Topics Forum
    Replies: 3
    Last Post: November 26th 2010, 09:35 AM
  3. Closed-form of a power series solution to an ODE
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: February 22nd 2010, 08:09 AM
  4. Find a closed form
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: September 28th 2009, 06:31 AM
  5. power series in closed form.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 4th 2008, 02:17 PM

Search Tags


/mathhelpforum @mathhelpforum