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Math Help - sum...

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    sum...

    Is the next sum convergence?

    1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 + 1/7+ 1/8 - 1/9+...


    thanks!
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    yes, but conditionally, it's the alternating harmonic series.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The series can be written as...

    \sum_{n=0}^{\infty} (\frac{1}{3n+1}+ \frac{1}{3n+2} - \frac{1}{3n+3}) (1)

    ... where the general term for 'large n' can be approximated as...

    a_{n} \approx \frac{1}{3n} (2)

    ... so that the series (1) diverges...

    Kind regards

    \chi \sigma
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  4. #4
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     1 + \frac{1}{2} -  \frac{1}{3}  +  \frac{1}{4}  +  \frac{1}{5}  -  \frac{1}{6}  +  \frac{1}{7} +  \frac{1}{8}  -  \frac{1}{9}  + ...

     = 1 + \left( \frac{1}{2} -  \frac{1}{3}\right)  +  \frac{1}{4}  + \left( \frac{1}{5}  -  \frac{1}{6} \right) +  \frac{1}{7} + \left( \frac{1}{8}  -  \frac{1}{9} \right)  + ...

    We have

     \frac{1}{2} -  \frac{1}{3} > 0

      \frac{1}{5}  -  \frac{1}{6}   > 0  ...

    The series is thus greater than

     1 + \frac{1}{4} + \frac{1}{7} + ... = \sum_{k=0}^{\infty} \frac{1}{ 3k+1 }

    Divergent .
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  5. #5
    Senior Member roninpro's Avatar
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    Wait a second - if a series does not converge absolutely, is it acceptable to group terms like that?
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  6. #6
    Math Engineering Student
    Krizalid's Avatar
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    oh, i misread the problem, i thought the signs were alternated, my bad.
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