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Math Help - Infinite Series

  1. #1
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    Infinite Series

    Determine convergence or divergence:

    a_n = \frac {1+(-1)^n}{n^2}

    Does the this function converge and is the limit equal to 0? That is what I found it to be.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by saiyanmx89 View Post
    Determine convergence or divergence:

    a_n = \frac {1+(-1)^n}{n^2}

    Does the this function converge and is the limit equal to 0? That is what I found it to be.
    yes. one way to see this is to use the squeeze theorem
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  3. #3
    Senior Member Twig's Avatar
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    hi

    I guess you could do

     \frac{1 +(-1)^{n}}{n^{2}} \, \leq \, \frac{(-1)^{n}}{n^{2}}

    And the series to the right is converging by Leibniz convergence theorem, because the series is alternating and decreasing.
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  4. #4
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    Quote Originally Posted by Twig View Post
    hi

    I guess you could do

     \frac{1 +(-1)^{n}}{n^{2}} \, \leq \, \frac{(-1)^{n}}{n^{2}}

    [snip]
    This inequality is false for even values of n.
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  5. #5
    Math Engineering Student
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    (-1)^n is bounded and \frac1{n^2}\to0 as n\to\infty, thus \frac{(-1)^n}{n^2}\to0 as n\to\infty, and the conclusion follows.
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