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Math Help - Convergence and divergence, two questions

  1. #1
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    Convergence and divergence, two questions

    hi y'all, here is a two part question that i just cant figure out. I've been staring at them for a while no, and i think that the firs tone is not convergent and the second is convergent, but i dont know how to go about proving that (or confirming if i'm right/wrong) Please let me know what you think.

    do the following series converge or diverge? Whatever you put, prove you answer.

    a.) [epsilon]infinity,n=1 (e^(1/n))/(n^2)




    b.) [epsilon]infinity,n=1 (n!)/(2*4*6. . . . . .(2n))
    Last edited by enormousface; May 15th 2009 at 08:11 AM. Reason: addition of a though
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  2. #2
    MHF Contributor Calculus26's Avatar
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    They both converge--See attachment
    Attached Files Attached Files
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  3. #3
    MHF Contributor chisigma's Avatar
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    a)  \forall n is e^{\frac{1}{n}}<3 so that is

    \frac{e^{\frac{1}{n}}}{n^{2}} < \frac{3}{n^{2}}

    The series...

    \sum_{n=1}^{\infty} \frac{3}{n^{2}}

    ... converges so that also converges the series...

    \sum_{n=1}^{\infty} \frac{e^{\frac{1}{n}}}{n^{2}}

    b) is...

    \frac{1\cdot 2 \cdot 3 \dots (n-1)\cdot n}{2\cdot 4\cdot 6 \dots (2n-2)\cdot 2n} = \frac{1}{2^{n}}

    ... so that...

    \sum_{n=1}^{\infty} \frac{1\cdot 2 \cdot 3 \dots (n-1)\cdot n}{2\cdot 4\cdot 6 \dots (2n-2)\cdot 2n} = \sum_{n=1}^{\infty} \frac {1}{2^{n}} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1

    Kind regards

    \chi \sigma
    Last edited by chisigma; May 15th 2009 at 10:44 PM. Reason: first term of series b) was erroneous... sorry!...
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