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Math Help - Convergence / Divergence

  1. #1
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    Convergence / Divergence

    Question: Determine whether the series is convergent or divergent, and if it is possible, find the sum.

    \sum_{n = 1}\ln(\frac{n}{2n + 5})

    At this point in my class, we only know geometric series, the divergence test, and the p-series. So we cant use integral test, comparison test etc. My initial attempt was to use the divergence test. But I am having a hard time evaluating the limit. Thanks in advance.

    p.s the sigma goes from n=1 to infinity.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zellster87 View Post
    Question: Determine whether the series is convergent or divergent, and if it is possible, find the sum.

    \sum_{n = 1}\ln(\frac{n}{2n + 5})

    At this point in my class, we only know geometric series, the divergence test, and the p-series. So we cant use integral test, comparison test etc. My initial attempt was to use the divergence test. But I am having a hard time evaluating the limit. Thanks in advance.

    p.s the sigma goes from n=1 to infinity.
    the natural log is a "nice" continuous function, therefore we can pass the limit "through" it. so \lim_{n \to \infty} \ln \left( \frac n{2n + 5} \right) = \ln \left( \lim_{n \to \infty} \frac n{2n + 5}\right)

    can you continue?
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  3. #3
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    I think so.

    <br />
\lim_{n \to \infty} \ln \left( \frac n{2n + 5} \right) = \ln \left( \lim_{n \to \infty} \frac n{2n + 5}\right)<br />


    is equal to:

    <br />
\ln \left( \lim_{n \to \infty}\frac{1}{2 + \frac{5}{n}}\right)<br />

    Then from that, i assume the limit is equal to 1/2?

    Would I then have to take the natural log on 1/2, or is it automatically divergent since the limit is not equal to zero.

    Thanks for you reply.
    Last edited by zellster87; February 18th 2008 at 10:09 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zellster87 View Post
    I think so.

    <br />
\lim_{n \to \infty} \ln \left( \frac n{2n + 5} \right) = \ln \left( \lim_{n \to \infty} \frac n{2n + 5}\right)<br />


    is equal to:

    <br />
\ln \left( \lim_{n \to \infty}\frac{1}{2 + \frac{5}{n}}\right)<br />

    Then from that, i assume the limit is equal to 1/2?

    Would I then have to take the natural log on 1/2, or is it automatically divergent since the limit is not equal to zero.

    Thanks for you reply.
    yes, you are correct. ln(1/2) is not zero, we can stop there. the series diverges by the test for divergence
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  5. #5
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    Got it. Thanks for your help.
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  6. #6
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    Quote Originally Posted by zellster87 View Post
    Question: Determine whether the series is convergent or divergent, and if it is possible, find the sum.

    \sum_{n = 1}\ln(\frac{n}{2n + 5})

    At this point in my class, we only know geometric series, the divergence test, and the p-series. So we cant use integral test, comparison test etc. My initial attempt was to use the divergence test. But I am having a hard time evaluating the limit. Thanks in advance.

    p.s the sigma goes from n=1 to infinity.
    As n \to \infty,\ \ln\left( \frac{n}{2n + 5} \right) \to \ln(1/2) \ne 0 so the series diverges.

    For a series to converge the limit of the sequence of terms must be 0.

    RonL
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