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Thread: convergence of piecewise function

  1. #1
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    convergence of piecewise function

    Hello,

    I have the following sequence:

    a_{n}=\frac{1}{\ln(x)} when n is even
    a_{n}=-\frac{1}{\ln(x)} when n is uneven

    I want to determine whether the sequence is absolute convergent, divergent or conditional convergence.

    I have tried showing that the sequence is absolute covergent by showing that \sum_{2}^{\infty} \vert a_{n} \vert is convergent. I tried using the integral test but I ended up with a strange expression (which contains li(x)).
    I have also tried using the ration test which leads to no conclusion.

    Suggestions would be appreciated.

    Thanks
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  2. #2
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    Re: convergence of piecewise function

    I presume you mean "log(n)", not "log(x)"? The problem is that this is not absolutely convergent.

    (If you do mean " \frac{1}{log(x)} for constant "x" then if it very easy!)
    Thanks from surjective
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  3. #3
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    Re: convergence of piecewise function

    Sorry but I presented the problem in a wrong way.

    I have the sequence a_{n} given in the previous post.

    How do I determine if the series \sum_{2}^{\infty} a_{n} is absolutely convergent, conditionally convergent or divergent.

    I say that the series is divergent since \lim_{n \rightarrow \infty} a_{n} \neq 0.

    What say you?
    Last edited by surjective; May 20th 2017 at 03:58 AM.
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  4. #4
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    Re: convergence of piecewise function

    $\displaystyle \lim_{n \to \infty} \dfrac{1}{\log{n}} = 0$

    $\displaystyle \sum_{n=2}^\infty \dfrac{(-1)^n}{\log{n}}$ converges by the alternating series test

    $\displaystyle \sum_{n=2}^\infty \dfrac{1}{\log{n}}$ diverges by direct comparison with $\displaystyle \sum_{n=2}^\infty \dfrac{1}{n}$, a known divergent series.
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