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Math Help - Series problem

  1. #1
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    n =0

    I Hope this might help u to read better,

    assess the series

    infinity

    sum of ((-2)^n)/(3n+1)!

    n = 0
    Attached Thumbnails Attached Thumbnails Series problem-series.gif  
    Last edited by distance; June 22nd 2006 at 07:30 PM. Reason: i hope this will help
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  2. #2
    Grand Panjandrum
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    Don't post the same question multipe times

    RonL
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  3. #3
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    Sorry about that. Anyone can do this problem ( bouncing ball), the final is in next week, so i need to understand the problem
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  4. #4
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    Is your second thread titled "More series problems" something you would like deleted?
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  5. #5
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    \sum_{n=1}^{\infty}\frac{(-2)^n}{(3n+1)!}
    Use the generalized ratio test,
    \lim_{n\to\infty}\left| \frac{(-2)^{n+1}}{(3(n+1)+1)!}\cdot \frac{(3n+1)!}{(-2)^n}\right|
    Note that,
    (3(n+1)+1)!=(3n+4)!=(3n+4)(3n+3)(3n+2)(3n+1)!
    Thus,
    \left| \frac{-2}{(3n+4)(3n+3)(3n+2)} \right|= 0 as n\to\infty
    Which is less than 1 thus the series converges.
    ---
    Also, the sequence,
    \left\{ \frac{(-2)^n}{(3n+1)!} \right\}
    Fits the conditions for Leibniz's alternating series test, thus it absolutely convergent.
    Last edited by ThePerfectHacker; June 22nd 2006 at 08:04 PM.
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  6. #6
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    \sum_{n=0}^{\infty}\frac{n!}{e^n}

    Just analytically thinking about this, I can see that the factorial will very strongly outgrow the exponential as x gets larger. But more formally...

    R_{n}=\frac{(n+1)!}{e^n \times e} \times \frac{e^n}{n!}

    R_{n}=\frac{n}{e}

    \lim_{n \to \infty}R_{n}=\infty

    Thus this diverges.

    Another way is to just take the limit of the original series.

    \lim_{n \to \infty} \frac{n!}{e^n}=\infty

    If this limit is anything but zero, the series diverges. If it is zero, you must use another test to help you.
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  7. #7
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    sorry !

    <br />
\sum_{n=0}^{\infty}\frac{(-2)^n}{(3n+1)!}<br />
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  8. #8
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    ThePerfectHacker showed you how to show this converges. What's the problem? Don't worry about the lower bound of one versus zero. His work is still correct.
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  9. #9
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    thanks a lot

    Quote Originally Posted by Jameson
    ThePerfectHacker showed you how to show this converges. What's the problem? Don't worry about the lower bound of one versus zero. His work is still correct.
    <br />
\sum_{n=1}^{\infty}\frac{n-1}{(3n+1)}<br />
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  10. #10
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    \sum_{n=1}^{\infty}\frac{n-1}{(3n+1)}

    Always start by taking the limit of the series. If it's non-zero then the series diverges.

    \lim_{n \to \infty} \frac{n-1}{(3n+1)} = \frac{1}{3}

    Thus this diverges.
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  11. #11
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    Quote Originally Posted by Jameson
    If it is zero, you must use another test to help you.
    Correction, 'if it is one' then you should use a different test.
    This is my 15th Post!!!
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  12. #12
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    What about this series then... \sum_{n=1}^{\infty}\frac{1}{n}

    If the limit is one, this most certainly diverges.
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  13. #13
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    I do not understand you?
    The ratio test is inconlusive for L=1.
    In your other post you said for L=0,
    I was making a correction.
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  14. #14
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    Oh! Ok. I was talking about the nth term test. If the nth term does not approach 0, then the series diverges. Yes you are correct about the Ratio Test.
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