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Math Help - Problem with limit/direct comparison for series

  1. #1
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    Problem with limit/direct comparison for series

    Hey guys, for some reason I am really struggling with the whole limit and direct comparison tests for series. Many times, I can't decide what to compare my series with.

    For example: (I am currently stuck on) -- the series from 2 to infinity of 2 / 7n ln(n)+4n
    I am assuming that I would use the limit comparison test (right??), which I am thinking would be ln(n)/n (a divergent series), and then divide An/Bn as limit goes to infinity? So then I am thinking the answer would be divergent?

    Appreciate any feedback, thanks
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  2. #2
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    Re: Problem with limit/direct comparison for series

    Quote Originally Posted by Beevo View Post
    Hey guys, for some reason I am really struggling with the whole limit and direct comparison tests for series. Many times, I can't decide what to compare my series with.

    For example: (I am currently stuck on) -- the series from 2 to infinity of 2 / 7n ln(n)+4n
    I am assuming that I would use the limit comparison test (right??), which I am thinking would be ln(n)/n (a divergent series), and then divide An/Bn as limit goes to infinity? So then I am thinking the answer would be divergent?

    Appreciate any feedback, thanks
    So that we know what you are talking about, please put some brackets where they're needed.
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  3. #3
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    Re: Problem with limit/direct comparison for series

    Quote Originally Posted by Prove It View Post
    So that we know what you are talking about, please put some brackets where they're needed.
    I have no idea how to type in the proper mathematical symbols on forums. Basically, the problem states: sigma (series) from n=2 to infinity, of 2 / 7n*ln(n) + 4n
    Which means 2 over 7n times ln(n) plus 4.

    Would I use the limit comparison test of say ln(n)/n? Then I am assuming it would be divergent.
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  4. #4
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    Re: Problem with limit/direct comparison for series

    Is this what you mean

    \sum_{n=2}^{\infty} \frac{2}{7n\log(n)+4n}?

    You type it like this \sum_{n=2}^{\infty} \frac{2}{7n\log(n)+4n} with tex /tex around it (with square brackets).
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  5. #5
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    Re: Problem with limit/direct comparison for series

    Quote Originally Posted by Vlasev View Post
    Is this what you mean

    \sum_{n=2}^{\infty} \frac{2}{7n\log(n)+4n}?

    You type it like this \sum_{n=2}^{\infty} \frac{2}{7n\log(n)+4n} with tex /tex around it (with square brackets).
    Yeah that's it, thanks dude. So basically:

    \sum_{n=2}^{\infty} \frac{2}{7n\ln(n)+4n}

    Would I take the limit comparison test with bn = \frac{ln(n)}{n} ?? Would that then be divergent??


    Any feedback appreciated.
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