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Math Help - Convergene of series

  1. #1
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    Convergene of series

    I think I copied this down right. It's a problem our professor gave us at the end of class, and his handwriting is a little sketchy. Also, I think I'm suppose to test to see if it converges.

    problem:

    (1/3)+(1*4)/ (3*5)+(1*4*7)/ (3*5*7)+..+[1*4*7*..*(3n-2)] /[3*5*7*..*(2n+1)]

    Can I get a hint? Just a shot in the dark here, but it looks recursive.. kinda like the fibonacci sequence (the future term depends on the previous one); but this is a series. So.. not too sure what to do.

    [edit] Sorry about the typo in the title! Totally meant "convergence".
    Last edited by Hikari; July 23rd 2009 at 09:58 PM.
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  2. #2
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    Quote Originally Posted by Hikari View Post
    I think I copied this down right. It's a problem our professor gave us at the end of class, and his handwriting is a little sketchy. Also, I think I'm suppose to test to see if it converges.

    problem:

    (1/3)+(1*4)/ (3*5)+(1*4*7)/ (3*5*7)+..+[1*4*7*..*(3n-2)] /[3*5*7*..*(2n+1)]

    Can I just get a hint?
    Just a hint.

    For n > 3, (3n-2) > (2n+1)
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  3. #3
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    Are you suggesting a comparison test?

    Could I compare that whole crazy series like this:

    crazy series > [SUM] (3n-2)/(2n+1)

    And then use a test for divergence on the smaller one? And then by comparison test to say that because the smaller one diverges, so does the bigger one? Sorry, I'm rather tired, so if that made absolutely no sense, I apologize.
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  4. #4
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    was I wrong?
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  5. #5
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     <br />
s_n = \sum_{n=0}^{\infty} \frac{3n-2}{2n+1}<br />

    The divergence test says: if the \lim_{n-\infty} a_n does not equal 0 or it does not exist, then the series s_n diverges.


    \lim_{n-\infty} \frac{3n-2}{2n+1} = \frac{3}{2} , using L'Hospital rule (sp?). Therefore, the series diverges.

    Edit: sorry, I only read your last post. Didn't read the OP!
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