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Math Help - Prove Convergence to (3/2)ln2

  1. #1
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    Prove Convergence to (3/2)ln2

    Hello,
    I have to prove that
    1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + 1/11 - 1/6 + ...
    converges to (3/2)ln2 (which is also equal to ln8/2 if it is easier to work with).
    I have no idea what to do with this, as the alternating harmonic series has conditional and not absolute convergence. How would I go about actually proving this convergence?
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  2. #2
    Super Member girdav's Avatar
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    Try to write the sum with the form \sum_k a_k. I think we will have to use the result \displaystyle \sum_{k=1}^{+\infty}\frac{(-1)^k}k=\ln 2.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Let's indicate s_{j k} the sum...

    \displaystyle 1 + \frac{1}{2} + \frac{1}{3} +... + \frac{1}{j k} (1)

    ... t_{2 j k} the sum...

    \displaystyle 1 - \frac{1}{2} + \frac{1}{3} +... + \frac{1}{2 j k-1} - \frac{1}{2 j k} (2)

    ... and u_{3k} the sum...

    \displaystyle 1 + \frac{1}{3} - \frac{1}{2} +\frac{1}{5}+ \frac{1}{7} - \frac{1}{4}... + \frac{1}{4k-3} + \frac{1}{4k-1} - \frac{1}{2k} (3)

    With a little of patience You can verify that is...

    \displaystyle u_{3n} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{4n} - \frac{1}{2}\ (1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n}) -

    \displaystyle - \frac{1}{2}\ (1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}) = s_{4n}- s_{2n} + \frac{1}{2}\ (s_{2n} - s_{n}) (4)

    But is also...

    \displaystyle t_{2n} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n} - 2\ (1 + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{2n}) =

    \displaystyle = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n} - (1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}) = s_{2n} - s_{n} (5)

    Comparing (4) and (5) You obtain that is...

    \displaystyle u_{3n}= t_{4n}+ \frac{1}{2}\ t_{2n} (6)

    ... and now You have to find the \displaystyle \lim_{n \rightarrow \infty} of (6)...

    Kind regards

    \chi \sigma
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