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Math Help - Absolute convergence

  1. #1
    Super Member Showcase_22's Avatar
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    Absolute convergence

    Does the sequence \sum_{n=1}^\infty \frac{sin(2n)}{1+n+n^2} converge? If it does, does it converge absolutely?
    I managed to get this series to converge but I have no idea whether it will converge absolutely.

    I gather that there will no longer be any positive terms but I can't prove whether it is absolutely convergent or not.

    Help would be appreciated!
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Showcase_22 View Post
    I managed to get this series to converge but I have no idea whether it will converge absolutely.

    I gather that there will no longer be any positive terms but I can't prove whether it is absolutely convergent or not.

    Help would be appreciated!
    I trust you for the convergence part.

    Now for the absolute convergence part, you have to prove that \sum \left|\frac{\sin(n)}{1+n+n^2}\right| converges.
    And we know that |\sin(n)| \leq 1, 1+n+n^2 \geq n^2

    Hence \left|\frac{\sin(n)}{1+n+n^2}\right|\leq \frac{1}{n^2}

    So \sum_{n \ge 1} \left|\frac{\sin(n)}{1+n+n^2}\right|\leq \sum_{n \ge 1} \frac{1}{n^2}
    And since the RHS is a convergence series, the LHS...
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  3. #3
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    I trust you for the convergence part.
    lol, well that's good! We are officially friends on here after all......

    Before I forget: Merry Christmas!!

    Now for the absolute convergence part, you have to prove that \sum \left|\frac{\sin(n)}{1+n+n^2}\right| converges.
    And we know that |\sin(n)| \leq 1, 1+n+n^2 \geq n^2

    Hence \left|\frac{\sin(n)}{1+n+n^2}\right|\leq \frac{1}{n^2}

    So \sum_{n \ge 1} \left|\frac{\sin(n)}{1+n+n^2}\right|\leq \sum_{n \ge 1} \frac{1}{n^2}
    And since the RHS is a convergence series, the LHS...
    hmm, you made that look so easy!

    I was getting confused that 1 \leq sin (n) \leq 1 and it would have been a simple step to |sin (n)| \leq 1. I'm not sure what I was thinking about.
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