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Math Help - Convergence Divergence ?

  1. #1
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    Convergence Divergence ?

    To find Convergence/Divergence of
     u(n) = \sum \frac{1}{n!}

    and of this one

     u(n) =\sum \frac{1.2.3.4.5.6.....}{1.3.5.7.9......}

    Thank you.
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  2. #2
    Super Member girdav's Avatar
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    Re: Convergence Divergence ?

    You should write u(n)=\frac 1{n!} for the first, not for the whole sum since it won't depend on n. Use the fact that n!\geq n(n-1). For the second, could you write the terms with a formula on which n appears?
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  3. #3
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    Re: Convergence Divergence ?

    I do not understand how the fact that n! >= n(n-1) can be used for to solve this .
    The second one can be simplified to get :-
     \frac{2(n!)(n!)}{(2n-1)!}

    Enlighten me.
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  4. #4
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    Re: Convergence Divergence ?

    It is e itself actually, so it converges !
    e=  \sum \frac{1}{n!}
    But how do you prove it ?
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  5. #5
    Super Member girdav's Avatar
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    Re: Convergence Divergence ?

    We have \frac 1{n!}\leq \frac 1{n(n-1)} and the sum of the terms of the RHS is telescopic.

    For the second series, first try ratio test. If you can't conclude, try Raabe-Duhamel test.
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  6. #6
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    Re: Convergence Divergence ?

    thanks girdav. I shall try it using the tests you mentioned. However this problem was presented before discussion on ratio test or raabe-duhamel test. So i guess we are expected to solve it using comparison test.
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