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Math Help - series with n!

  1. #1
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    series with n!

    i'm supposed to test this series for convergence/divergence

    \sum_{n=1}^{\infty} \frac{(n!)^n}{n^{4n}}

    i guess my real hang-up is with the factorial..but i'm just kind of lost with this one.. please help!
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  2. #2
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    Quote Originally Posted by buttonbear View Post
    i'm supposed to test this series for convergence/divergence

    \sum_{n=1}^{\infty} \frac{(n!)^n}{n^{4n}}

    i guess my real hang-up is with the factorial..but i'm just kind of lost with this one.. please help!
    it's divergent because for sufficiently large values of n we have: n! \geq n(n-1)(n-2)(n-3)(n-4) \geq n^4. thus: \frac{(n!)^n}{n^{4n}} \geq 1.

    this method also proves that for any real number c the series \sum_{n=1}^{\infty} \frac{(n!)^n}{n^{cn}} is divergent.
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  3. #3
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    is this any particular test? or is it just some sort of comparison.. i'm sorry but i need to be able to justify everything i say about these problems
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  4. #4
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    Quote Originally Posted by buttonbear View Post
    i'm supposed to test this series for convergence/divergence

    \sum_{n=1}^{\infty} \frac{(n!)^n}{n^{4n}}

    i guess my real hang-up is with the factorial..but i'm just kind of lost with this one.. please help!

    you can use the root test and make it

    \frac{(n!)^n}{(n^4)^n} so that it is
    limit (x is approaching infinity) of the nth root of \frac{(n!)^n}{(n^4)^n} and it becomes
    limit of \frac{n!}{n^4}
    and it is \frac{fast}{slow}
    diverging
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