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Math Help - Integral Test for Infinite Series - Convergence or Divergence

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    Integral Test for Infinite Series - Convergence or Divergence

    Problem: In each of the following, use the integral test (if possible) to determine whether the given series converges. (NOTE: Before applying the I.T., you must be sure your function f(x) is continuous, decreasing, and non-negative on the interval [1, \infty)

    \displaystyle \Sigma\frac{n}{ln(n)}

    I'm not sure exactly how to approach this problem, but I started out with the supposition that the I.T. can't be applied since the function isn't continuous on the interval [1, \infty). Is this correct? If so, I was thinking of using the Limit Comparison Test next, but I wasn't sure what similiar function to pick...I tried \frac{1}{ln(x)} but that limit wasnt finite, so now I'm stuck. Can someone help me get on the right track? Thanks is advance.
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    Super Member TheChaz's Avatar
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    This is certainly curious, since the first term in your series isn't even defined!
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    Quote Originally Posted by dbakeg00 View Post
    Problem: In each of the following, use the integral test (if possible) to determine whether the given series converges. (NOTE: Before applying the I.T., you must be sure your function f(x) is continuous, decreasing, and non-negative on the interval [1, \infty)
    \displaystyle \Sigma\frac{n}{ln(n)}
    A simple comparison test is so much easier.
    Suppose that n\ge 2. Then \ln(n)\le 2\sqrt{n}.

    This \dfrac{n}{\ln{n}}\ge \dfrac{\sqrt{n}}{2}.
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    Quote Originally Posted by Plato View Post
    A simple comparison test is so much easier.
    Suppose that n\ge 2. Then \ln(n)\le 2\sqrt{n}.

    This \dfrac{n}{\ln{n}}\ge \dfrac{\sqrt{n}}{2}.
    Would you mind explaing that solution a little more? I'm guessing that you chose 2 bc the function doesn't exist at 1, but where does the square root of n come from?
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    Quote Originally Posted by dbakeg00 View Post
    Would you mind explaing that solution a little more? I'm guessing that you chose 2 bc the function doesn't exist at 1, but where does the square root of n come from?
    See reply #2. Because \ln(1)=0 we cannot start with n=1.

    If you mean the inequality here is the way that works.
    For all x>0 we have \ln(x)<x.
    Then \begin{gathered}<br />
  \ln \left( {\sqrt x } \right) \leqslant \sqrt x  \hfill \\<br />
  \frac{1}<br />
{2}\ln (x) \leqslant \sqrt x  \hfill \\<br />
  \ln (x) \leqslant 2\sqrt x  \hfill \\ <br />
\end{gathered}
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    Super Member TheChaz's Avatar
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    ln(n) \leq 2\sqrt{n} is what might referred to as a known result. "Known results" are sort of your toolbox from which you draw comparisons. You know that (...) 1/n diverges by p-series test, so you can use it to show convergence (divergence) of "smaller" ("bigger") sums.
    If you "know" the result that Plato cites, then you can use it to show what he showed.
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    Everytime I come across of those "known results," its one that I don't already know...sure is frustrating! Thanks for the help guys
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    Super Member TheChaz's Avatar
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    Just wait till they start using "obviously/clearly..." to justify statements! I have said and heard many-a-time
    What's obvious (clear) to you might not be obvious (clear) to me!
    Anyway, you're welcome. Stop by anytime!
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    Member Miss's Avatar
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    What about using the test for divergence?

    \displaystyle \lim_{n\to\infty} \frac{n}{ln(n)} \neq 0 .. so ..
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    Quote Originally Posted by Miss View Post
    What about using the test for divergence?

    \displaystyle \lim_{n\to\infty} \frac{n}{ln(n)} \neq 0 .. so ..
    ye olde "nth" term test ... very nice!
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    Super Member TheChaz's Avatar
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    That's true and useful, but what "known result" (to use my own nomenclature...) would we employ if not Plato's?
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