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Math Help - Help With Series

  1. #1
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    Help With Series

    Determine whether the series ∑_(n=1)^(infinity) ((squareroot(n)+ln(n))/((n^2)+1)) converges or diverges. clearly identigy any test(s) you are using and label all relevant data and/or properties.

    I have no idead where to start or what tests to use.
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  2. #2
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    Re: Help With Series

    It can be shown that \ln(n) \leq \sqrt{n}

    Can you then find a bound on each term of your series?


    Hint - think comparison test and p-test to arrive at your final answer!
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  3. #3
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    Re: Help With Series

    Quote Originally Posted by benb89 View Post
    It can be shown that \ln(n) \leq \sqrt{n}

    Can you then find a bound on each term of your series?


    Hint - think comparison test and p-test to arrive at your final answer!
    Are you saying compare it to squareroot(n)/n?
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  4. #4
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    Re: Help With Series

    So,

    \frac{\ln n + \sqrt{n}}{n^2 + 1}\leq \frac{\sqrt{n} + \sqrt{n}}{n^2 + 1}

    Can you go forward from here?
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  5. #5
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    Re: Help With Series

    NO!
    \frac{\sqrt{n}+\ln(n)}{n^2+1}\le\frac{2\sqrt{n}}{n  ^2+1}<\frac{2}{n^{3/2}}.
    The last is a p-series that converges.
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  6. #6
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    Re: Help With Series

    Quote Originally Posted by benb89 View Post
    So,

    \frac{\ln n + \sqrt{n}}{n^2 + 1}\leq \frac{\sqrt{n} + \sqrt{n}}{n^2 + 1}

    Can you go forward from here?
    I'm not sure. I'm using the comparison test right? Or could I use the limit comparison test?
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  7. #7
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    Re: Help With Series

    Quote Originally Posted by Plato View Post
    NO!
    \frac{\sqrt{n}+\ln(n)}{n^2+1}\le\frac{2\sqrt{n}}{n  ^2+1}<\frac{2}{n^{3/2}}.
    The last is a p-series that converges.
    Oh! Okay. So am I going to have to do multiple comparison tests?
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  8. #8
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    Re: Help With Series

    Quote Originally Posted by Preston019 View Post
    Oh! Okay. So am I going to have to do multiple comparison tests?
    No you don't.
    Just say \sum\limits_{k = 1}^\infty  {\frac{2}{{\sqrt[2]{{n^3 }}}}} converges.
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  9. #9
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    Re: Help With Series

    Quote Originally Posted by Plato View Post
    No you don't.
    Just say \sum\limits_{k = 1}^\infty  {\frac{2}{{\sqrt[2]{{n^3 }}}}} converges.
    Okay. And where did you get ln(n) is less than or equal to the square root of n?
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