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Ratio test for infinite series

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Investigate the convergence of the following series.

    (2n)!/(n!n!)

    Screenshot2012-02-20at122944AM.png


    3. The attempt at a solution

    Number one, I don't see how they get that
    if an = (2n)!, then an+1 = ((2n+2))!, it should be (2n+1)!

    Number two, I don't see how they go to the second step, why is the second step not:

    (n!n!(2n+2)!)/((n+1)!(n+1)!(2n)!)? why do they have that extra (2n)! in the numerator?

    Number three, I don't see why n!n! cancels in the numerator in step 3.
     
  2. jcsd
  3. Feb 20, 2012 #2

    SammyS

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    Number one:
    an+1 = ((2n+2))! results from basic substitution.

    If [itex]\displaystyle a_{n}= (2n)!\,,\ \text{ then }a_{n+1}=(2(n+1))!\,,\text{ so that }a_{n+1}=(2n+2)![/itex]

    Of course, this is not the same an as in the image you posted.
     
  4. Feb 20, 2012 #3

    SammyS

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    [itex]\displaystyle \frac{n!n!(2n+2)!}{(n+1)!(n+1)!(2n)!}=\frac{n!n!(2n+2)(2n+1)(2n)!}{(n+1)!(n+1)!(2n)!}[/itex]

    This is because
    [itex](2n+2)!=(2n+2)((2n+2)-1)((2n+2)-2)((2n+2)-3)((2n+2)-4)\dots(3)(2)(1)[/itex]
    [itex]=(2n+2)(2n+1)(2n)(2n-1)(2n-2)\dots(3)(2)(1)[/itex]

    [itex]=(2n+2)(2n+1)(2n)![/itex]​
     
  5. Feb 20, 2012 #4

    SammyS

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    [itex](n+1)!=(n+1)(n!)[/itex]
     
  6. Feb 20, 2012 #5
    Great advice I really appreciate it.
     
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