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Math Help - Series convergence question

  1. #1
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    Series convergence question

    The question:

    Determine if the following alternating series converges. Is it absolutely convergent?

    \sum\limits_{k = 3}^\infty \frac{(-1)^k}{ln(k)}


    My attempt:
    Using the alternating series test, we need:
    a) a_k \ge 0
    b) a_{k} \ge a_{k + 1}
    c)  \lim\limits_{k \to \infty} a_k = 0

    a) \frac{1}{ln(k)} > 0 for all k > 1
    b) \frac{1}{ln(k)} \ge \frac{1}{ln(k+1)} for all k > 1
    c) Limit is obviously 0 as k \to \infty

    Thus the series converges.

    Now we consider \sum\limits_{k = 3}^\infty |\frac{1}{ln(k)}|

    Using ratio test:

    r = \frac{a_{k + 1}}{a_k} = |\frac{1}{ln(k + 1)}}| |\frac{ln(k)}{1}}|
    = \frac{ln(k)}{ln(k+1)} < 1

    Therefore converges. So this series should be absolutely convergent. However the answer is actually 'conditionally convergent'.

    Where have I gone wrong? Thanks.
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  2. #2
    Ted
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    First, in ratio test you'll calculate the limit of the fraction \dfrac{a_{k+1}}{a_k} as k goes to infinity not just the fraction!
    Second, the ratio test will be failed.
    Third, I'll use the comparison test for the absolute series.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Glitch View Post
    Where have I gone wrong? Thanks.

    \displaystyle\lim_{k \to{+}\infty}{\dfrac{\log k}{\log (k+1)}}=1


    Fernando Revilla
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Ted View Post
    First, in ratio test you'll calculate the limit of the fraction \dfrac{a_{k+1}}{a_k} as k goes to infinity not just the fraction!
    There is a ratio test without calculating the limit:

    If the series of general positive term a_k satisfies a_{k+1}/a_k\leq r for k sufficiently large and 0\leq r <1 fixed then , the series is convergent. In our case does not exist such r .


    Fernando Revilla
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  5. #5
    Ted
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    Well, am talking about the standard ratio test .
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  6. #6
    MHF Contributor chisigma's Avatar
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    Dirichlet's Test -- from Wolfram MathWorld

    Kind regards

    \chi \sigma
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  7. #7
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    Quote Originally Posted by Glitch View Post
    Determine if the following alternating series converges. Is it absolutely convergent?
    \sum\limits_{k = 3}^\infty \frac{(-1)^k}{ln(k)}

    Using the alternating series test, we need:
    a) a_k \ge 0
    b) a_{k} \ge a_{k + 1}
    c)  \lim\limits_{k \to \infty} a_k = 0

    a) \frac{1}{ln(k)} > 0 for all k > 1
    b) \frac{1}{ln(k)} \ge \frac{1}{ln(k+1)} for all k > 1
    c) Limit is obviously 0 as k \to \infty
    Thus the series converges.
    The above is correct. The given series converges.

    BUT \sum\limits_{k = 3}^\infty \left|\dfrac{(-1)^k}{ln(k)}\right|=\sum\limits_{k = 3}^\infty \dfrac{1}{ln(k)}.

    It easy to prove that if \alpha>0 then \sum\limits_{k = 3}^\infty \dfrac{1}{[ln(k)]^{\alpha}} diverges.
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  8. #8
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    This is what I've tried:

    Let a_k = \abs{\frac{1}{ln(k)}}
    Let b_k = \abs{\frac{1}{kln(k)}}

    b_k \le a_k

    \int\limits_{3}^{n} \frac{1}{kln(k)} \le \sum\limits_{k = 3}^\infty \frac{1}{ln(k)}

    = ln(ln(n + 1)) - ln(ln(3)) \to \infty as n \to \infty, therefore a_k must diverge.
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  9. #9
    MHF Contributor FernandoRevilla's Avatar
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    Better:

    \displaystyle\lim_{k \to{+}\infty}{\dfrac{1/\log k}{1/k}}=+\infty

    and

    \displaystyle\sum_{k=3}^{+\infty}\dfrac{1}{k}

    is divergent.


    Fernando Revilla
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