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Math Help - Divergence of a series

  1. #1
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    Divergence of a series

    Hi, I want to show that \sum_{n=1}^{\infty}b_n, where b_n = \ln(1 - \frac{1}{2^n}), is divergent (approaches negative infinity).

    How would I go about approach this? I was thinking of using direct comparison or limit form of comparison test. However, I am not sure of what to compare b_n with.

    Any help will be kindly appreciated.
    Last edited by shinn; March 15th 2010 at 05:49 AM.
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    Quote Originally Posted by shinn View Post
    Hi, I want to show that \sum_{n=0}^{\infty}b_n, where b_n = \ln(1 - \frac{1}{2^n}), is divergent (approaches negative infinity).

    How would I go about approach this? I was thinking of using direct comparison or limit form of comparison test. However, I am not sure of what to compare b_n with.

    Any help will be kindly appreciated.
    It should be obvious that the very first term \to \ln{0} \to -\infty.


    If you can show the rest of the terms create a convergent series, when you add them, the sum is still going to \to -\infty.
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    sorry I made a mistake when typing up the question, n is suppose to start at 1. thanks for pointing that out.
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    \sum_{n = 1}^{\infty}\ln{\left(1 - \frac{1}{2^n}\right)} = \ln{\left(1 - \frac{1}{2}\right)} + \ln{\left(1 - \frac{1}{4}\right)} + \ln{\left(1 - \frac{1}{8}\right)} + \ln{\left(1 - \frac{1}{16}\right)} + \dots

     = \ln{\frac{1}{2}} + \ln{\frac{3}{4}} + \ln{\frac{7}{8}} + \ln{\frac{15}{16}} + \dots

     = \ln{\left(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{15}{16} \cdot \dots\right)}

     \to \ln{0}

     \to -\infty.
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  5. #5
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    the series converges.
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    Quote Originally Posted by Krizalid View Post
    the series converges.
    Sorry, but I've just shown that it diverges...
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    Quote Originally Posted by Prove It View Post
     = \ln{\left(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{15}{16} \cdot \dots\right)}

     \to \ln{0}

     \to -\infty.
    i don't get this step when you say \to\ln0.

    mind if you explain it?
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  8. #8
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    No. It converges.
    Here is a hint:
    Make a common denominator.
    And use one of the logarithmic functions' properties.
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    Quote Originally Posted by Krizalid View Post
    i don't get this step when you say \to\ln0.

    mind if you explain it?
    When you multiply any number by a number <1, the product is smaller than the original number.

    Here I have continually multiplied by numbers <1, so the product \to 0.
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  10. #10
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    Quote Originally Posted by Miss View Post
    No. It converges.
    Here is a hint:
    Make a common denominator.
    And use one of the logarithmic functions' properties.
    Sorry but I don't see how

    \ln{\left(1 - \frac{1}{2^n}\right)} = \ln{\left(\frac{2^n - 1}{2^n}\right)} = \ln{(2^n - 1)} - \ln{(2^n)}

    helps us in this case.


    When you take the sum you just end up with

    \ln{1} - \ln{2} + \ln{3} - \ln{4} + \ln{7} - \ln{8} + \dots.

    This is not a telescopic series so I don't see how you get that the series is convergent...
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  11. #11
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    Quote Originally Posted by Prove It View Post
    Sorry but I don't see how

    \ln{\left(1 - \frac{1}{2^n}\right)} = \ln{\left(\frac{2^n - 1}{2^n}\right)} = \ln{(2^n - 1)} - \ln{(2^n)}

    helps us in this case.


    When you take the sum you just end up with

    \ln{1} - \ln{2} + \ln{3} - \ln{4} + \ln{7} - \ln{8} + \dots.

    This is not a telescopic series so I don't see how you get that the series is convergent...
    But I did not say it will be a telescoping series.
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  12. #12
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    alright, so suppose i have the series \sum{\ln \left( 1+\frac{1}{n^{2}} \right)}=\ln 2+\ln \frac{5}{4}+\ln +\frac{10}{9}+\cdots and i'm seeing here that the numerator grows faster than the denominator, thus the series diverges.

    that's just wrong, doesn't provide a solid proof of the convergence of the series.

    my series converges by limit comparison test with b_n=\frac1{n^2} which is a solid proof of the convergence of the series.
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  13. #13
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    Considering that the OP has stated that he/she needs help getting to the answer of -\infty he/she has been given, I have shown correctly how to get that answer, as the logarithm laws and logic I have used is not flawed.
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  14. #14
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    Quote Originally Posted by Krizalid View Post
    alright, so suppose i have the series \sum{\ln \left( 1+\frac{1}{n^{2}} \right)}=\ln 2+\ln \frac{5}{4}+\ln +\frac{10}{9}+\cdots and i'm seeing here that the numerator grows faster than the denominator, thus the series diverges.

    that's just wrong, doesn't provide a solid proof of the convergence of the series.

    my series converges by limit comparison test with b_n=\frac1{n^2} which is a solid proof of the convergence of the series.
    Actually it does work in this case.

    Simple application of the same logarithm rule \ln{a} + \ln{b} = \ln{(ab)}.

    So \ln{2} + \ln{\frac{5}{4}} + \ln{\frac{10}{9}} + \dots = \ln{\left(2 \cdot \frac{5}{4} \cdot \frac{10}{9} \cdot \dots \right)}

    and as stated before, since you're multiplying each term in the product by a number >1, the product grows without bound (albeit slowly), as does the logarithm. So this series diverges.
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  15. #15
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    a_n=\ln\bigg(1-\frac1{2^n}\bigg) is strictly negative, so there's no difference on studying the series \sum\ln\bigg(1-\frac1{2^n}\bigg) or -\sum \ln\bigg(1-\frac1{2^n}\bigg) where the first one is negative and the second one positive, so limit comparison test with \frac1{2^n} does apply, and the series converges.
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