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Series and convergence

  1. Nov 13, 2005 #1
    to solve this question i need to know whether the following series converges. however both the root test and ratio test give 1.

    sum from n=1 to infinity:
    [tex]\sum\frac{-1(2n-2)!}{n!(n-1)!2^{2n-1}}[/tex]
     
    Last edited: Nov 13, 2005
  2. jcsd
  3. Nov 13, 2005 #2

    benorin

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    Do you mean [tex]\sum\frac{(-1)^{n}(2n-2)!}{n!(n-1)!2^{2n-1}}[/tex] or [tex]-\sum\frac{(2n-2)!}{n!(n-1)!2^{2n-1}}[/tex] ?
     
  4. Nov 13, 2005 #3
    yes, the second one as you have written it, with the -1 out front
     
  5. Nov 13, 2005 #4

    benorin

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  6. Nov 13, 2005 #5
    this doesnt seem to help me in any way..
     
  7. Nov 13, 2005 #6

    benorin

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    Does [tex]-\sum_{n=1}^{\infty}\frac{(2n-2)!}{n!(n-1)!2^{2n-1}}=-1[/tex] help?
     
  8. Nov 13, 2005 #7

    benorin

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    Try partial sums

    Try partial sums:

    Prove that:

    [tex]S_{N}:=-\sum_{n=1}^{N}\frac{(2n-2)!}{n!(n-1)!2^{2n-1}}= \frac{\left( 2N\right)!}{2^{2N}\left( N!\right)^2} -1 = \frac{1}{2^{2N}} \left(\begin{array}{cc}2N\\N\end{array}\right)-1[/tex]


    by induction (I hope induction will work, anyway) where [tex]\left(\begin{array}{cc}n\\k\end{array}\right)[/tex] is a binomial coefficient, and then show that the partial sum [tex]S_{N} \rightarrow 1\mbox{ as } N\rightarrow \infty[/tex].

    But I won't kid you, this is NOT the easy way to do this problem.
     
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