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Math Help - convergent series

  1. #1
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    convergent series

    For which positive integers k is the following series convergent?

    [(n!)^2]/[(kn)!] as n goes from 1 to infinity. Please show the steps using the ratio test.
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  2. #2
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    Quote Originally Posted by Taurus3 View Post
    For which positive integers k is the following series convergent?
    [(n!)^2]/[(kn)!] as n goes from 1 to infinity. Please show the steps using the ratio test.
    \dfrac{{\left[ {\left( {n + 1} \right)!} \right]^2 }}<br />
{{\left( {kn + k} \right)!}}\dfrac{{\left( {kn} \right)!}}<br />
{{\left( {n!} \right)^2 }} = \dfrac{{\left( {n + 1} \right)^2 \left( {n!} \right)^2 }}<br />
{{\left( {n!} \right)^2 }}\dfrac{{\left( {kn} \right)!}}<br />
{{\left( {kn + k} \right)!}}
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  3. #3
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    I don't think that's right. Plus what are the possible values for k?
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  4. #4
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    Why don't you think it is correct?
    Do you understand the ratio test.
    You finish the algebra.
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  5. #5
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    doesn't (kn)! get cancelled since (kn + k)! = (kn)! (kn + k) (kn + (k-1)) (kn + (k-2)) + ....so on? So doesn't that just leave us with (n+2)^2 on the numerator?
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  6. #6
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    If k=1 we get \dfrac{1}{n+1}

    If k=2 we get \dfrac{(2n)!}{(2n+2)!}=\dfrac{1}{(2n+2)(2n+1)}

    If k=3 we get \dfrac{(3n)!}{(3n+3)!}=\dfrac{1}{(3n+3)(3n+2)(3n+1  }

    Get it?
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  7. #7
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    but what happened to the (n + 1)^2 in the numerator? It's still a bit confusing. So what are all the possible values for k that makes this convergent?
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  8. #8
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    Look, I am not going to do this problem for you.
    So the answer to your question is: I left that factor out.
    You must do somethings for yourself.
    I am now done with this thread.
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  9. #9
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    oh, thanks. Sorry.
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