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Math Help - Integral Test and comparison test help!

  1. #1
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    Integral Test and comparison test help!

    Hello, I have been looking around for how to solve these to series by the integral test.

    Problem 1 Evaluate


    Σ (n^(1/n))/n^3
    n=1

    decide if the series converges or diverges?

    And,

    Problem 2 Evaluate


    Σ (3^n*n!)/(3k)!
    n=1

    decide if the series converges or diverges?
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  2. #2
    MHF Contributor Amer's Avatar
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    \sum_{n=1}^{\infty}\frac{n^{\frac{1}{n}}}{n^3}

    \sum_{n=1}^{\infty}\frac{3^n(n!)}{(3n)!}


    the second one by root test

    lim_{n\rightarrow\infty}\left(\frac{3^n(n!)}{(3n)!  }\right)^{\frac{1}{n}}

    lim_{n\rightarrow\infty}\frac{3(n!)^{\frac{1}{n}}}  {((3n)!)^{\frac{1}{n}}}=3

    limit is 3>1 so the series diverge the first one sorry I do not know how to decide yet ..
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  3. #3
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    Hey thanks for the help but in the denominator there's a k instead of an n. That's the trick question!
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  4. #4
    MHF Contributor Amer's Avatar
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    k is constant or what ???
    I am sorry
    it is higher level
    I waste your time
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  5. #5
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    Quote Originally Posted by tapiaghector View Post
    Hello, I have been looking around for how to solve these to series by the integral test.

    Problem 1 Evaluate


    Σ (n^(1/n))/n^3
    n=1

    decide if the series converges or diverges?

    And,

    Problem 2 Evaluate


    Σ (3^n*n!)/(3k)!
    n=1

    decide if the series converges or diverges?
    First question - why on earth would you want to use the integral test (remember, you will need to integrate).

    I would personally use a limit comparison test for the first. Compare with

     <br />
\sum_{n=1}^{\infty} \frac{1}{n^3}<br />

    for the second (it looks a little fishy) but

    \frac{1}{(3k)!} \sum_{n=1}^{\infty} 3^n n! which clearly diverges - the terms don't go to zero.
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  6. #6
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    So, for the second one: how are you sure that the 3^n*n! diverges since the you dont know 1/3k!. Also, if you plug numbers in the series it goes to larger and larger...

    there isnt another method to compared this serie?
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  7. #7
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    Quote Originally Posted by tapiaghector View Post
    So, for the second one: how are you sure that the 3^n*n! diverges since the you dont know 1/3k!. Also, if you plug numbers in the series it goes to larger and larger...

    there isnt another method to compared this serie?
    k does not depend on n and so is treated as a constant. So when dealing with the summation you only have to worry about the stuff that depends on n.
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