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Math Help - Test this Series for Convergence or Divergence

  1. #1
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    Test this Series for Convergence or Divergence

    Test this Series for Convergence or Divergence.

    Summation from 1 to infinity of (-1)^n * sin(8Pi/n)

    It's from the Alternating Series Test section. I wasn't sure how to do this. I compared b(sub n)= sin(8Pi/n) to b(sub n+1)=sin(8Pi/n+1), b(sub n+1) is always less than b(sub n), so I just take the limit of b(sub n), right? If it's 0, it's convergent?

    I don't have my book with me so I'm not quite sure.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Here you go

    [quote=thegame189;126859]Test this Series for Convergence or Divergence.

    Summation from 1 to infinity of (-1)^n * sin(8Pi/n)

    It's from the Alternating Series Test section. I wasn't sure how to do this. I compared b(sub n)= sin(8Pi/n) to b(sub n+1)=sin(8Pi/n+1), b(sub n+1) is always less than b(sub n), so I just take the limit of b(sub n), right? If it's 0, it's convergent?

    you have \sum_{n=1}^{\infty}\bigg[sin\bigg(\frac{8\pi}{n}\bigg)\cdot{cos(\pi{n})}\bi  gg]...does that make it clearer?
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  3. #3
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    Not at all

    Where did the cos(nPi) come from?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by thegame189 View Post
    Not at all

    Where did the cos(nPi) come from?
    It came from the fact that for integer values for x... (-1)^{x}=cos{\pi{x}}..think about it cos(\pi{\cdot{1}})=-1,cos(\pi\cdot{2})=-1,cos(\pi\cdot{3})=-1... ad infinitum...so is it easier to work with now?
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  5. #5
    Super Member PaulRS's Avatar
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    Since it's a alternating series, \lim_{n\rightarrow{+\infty}}{\sin\left(\frac{8\cdo  t{\pi}}{n}\right)}=0 *, and \sin\left(\frac{8\cdot{\pi}}{n}\right) is decreasing, it converges

    Alternating series test - Wikipedia, the free encyclopedia

    * \sin(x)\approx{x} as x\rightarrow{0}
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