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Math Help - series

  1. #1
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    series

    Converge or Diverge?

    I'm not sure how to go about solving these...help?

    1. Sigma (Sqrt(2n+1) - Sqrt(2n))
    n = 1, and goes to infinity

    1. Sigma (1/(Sqrt(n)ln (n)))
    n= 2, and goes to infinity

    thanks.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by skabani View Post
    Converge or Diverge?

    I'm not sure how to go about solving these...help?

    1. Sigma (Sqrt(2n+1) - Sqrt(2n))
    n = 1, and goes to infinity

    1. Sigma (1/(Sqrt(n)ln (n)))
    n= 2, and goes to infinity

    thanks.
    1.Consider the sequence of partial sums for the first one. Do you notice a pattern?

    2. \sum_{n=2}^{\infty}\frac{1}{\sqrt{n}\ln(n)}>\sum_{  n=2}^{\infty}\frac{1}{n\ln(n)}

    Now consider that if a_n=\frac{1}{n\ln(n)}, then \forall{n}\in[2,\infty)~a_n>0\wedge{a_n\in\mathcal{C}}\wedge{a_n  \in\downarrow}

    So the integral test applies, and note that \int_2^{\infty}\frac{dx}{x\ln(x)} diverges.
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  3. #3
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    thanks.

    could we also use the limit comparison test and compare the series to 1/Sqrt(n), where c>0, and the series would diverge?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by skabani View Post
    thanks.

    could we also use the limit comparison test and compare the series to 1/Sqrt(n), where c>0, and the series would diverge?
    No, because \lim_{n\to\infty}\frac{\frac{1}{\sqrt{n}}}{\frac{1  }{\sqrt{n}\ln(n)}}=\infty. Thus one of the series dominates the other, so we can glean nothing of its convergence from the LCT.
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  5. #5
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    ok

    wait, so the original series converges then?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by skabani View Post
    ok

    wait, so the original series converges then?
    As I showed with the integral test the comparison series divergres, which implies that since the original series is greater than the comparison series it diverges.
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