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Math Help - Series convergence

  1. #1
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    Series convergence

    How can I show that the series [(-1)^n]/n^(1/3) converges?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by jules027 View Post
    How can I show that the series [(-1)^n]/n^(1/3) converges?
    \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}

    By the alternating series test since \forall{n}>0,\frac{1}{\sqrt[3]{n}}>\frac{1}{\sqrt[3]{n+1}}

    so for an arbitrary value we have showed that for every value greater than that value that

    a_{n+1}<a_n

    Now we must show that \lim_{n\to\infty}\bigg|\frac{(-1)^n}{\sqrt[3]{n}}\bigg|=\lim_{n\to\infty}\frac{1}{\sqrt[3]{n}}=0

    So we have also established that \lim_{n\to\infty}|a_n|=0

    So this series conditionally converges

    but it does not absolutely converge since

    \sum_{n=1}^{\infty}\bigg|\frac{(-1)^n}{\sqrt[3]{n}}\bigg|=\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}}

    diverges by the integral test

    since it meets all its requirements and

    \int_1^{\infty}\frac{1}{\sqrt[3]{n}}dn=\infty
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  3. #3
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    That makes sense. This was from a bigger problem: "Give an example of two convergent series an and bn such that the series(an*bn) diverges. So would an being [(-1)^n]/n^(1/3) and bn being [(-1)^n]/n^(1/2) work for this problem?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by jules027 View Post
    That makes sense. This was from a bigger problem: "Give an example of two convergent series an and bn such that the series(an*bn) diverges. So would an being [(-1)^n]/n^(1/3) and bn being [(-1)^n]/n^(1/2) work for this problem?
    Assuming you mean \sum_{n=1}^{\infty}\bigg[\frac{(-1)^n}{\sqrt{n}}\cdot\frac{(-1)^n}{\sqrt[3]{n}}\bigg] and not \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}\cdot\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}

    then yes you are correct the multiplication of both would yield \frac{1}{n^{\frac{5}{6}}}

    which diverges by the integral test
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