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Math Help - Series/Sequences Questions

  1. #1
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    Series/Sequences Questions

    The first question is to find \Sigma^{b}_{n = 1}\frac{\sin n}{2^n} and the second being to find \Sigma^{\infty}_{n = 1}\frac{\sin n}{2^n}.

    The thing is I'm not sure how to even begin these, so if someone could help me out that would be great. Thanks.
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  2. #2
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    Quote Originally Posted by carnage View Post
    The first question is to find \Sigma^{b}_{n = 1}\frac{\sin n}{2^n} and the second being to find \Sigma^{\infty}_{n = 1}\frac{\sin n}{2^n}.

    The thing is I'm not sure how to even begin these, so if someone could help me out that would be great. Thanks.
    Are you asked to evaluate them or to show that they are convergent?

    Showing convergence is easy, just use the comparison test with

    \sum_{n = 1}^{\infty}\frac{1}{2^n} (this is a geometric series).
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Are you asked to evaluate them or to show that they are convergent?

    Showing convergence is easy, just use the comparison test with

    \sum_{n = 1}^{\infty}\frac{1}{2^n} (this is a geometric series).
    I believe we're supposed to evaluate them.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Is...

     \sin n = \frac{e^{in} - e^{-in}}{2i} (1)

    ... so that...

    \sum_{n=1}^{\infty} \frac{\sin n}{2^{n}} = \frac{1}{2i}\cdot \{\sum_{n=1}^{\infty} (\frac{e^{i}}{2})^{n} - \sum_{n=1}^{\infty} (\frac{e^{-i}}{2})^{n}\}=

    = \frac{1}{2i}\cdot \{\frac{e^{i}}{2-e^{i}} - \frac{e^{-i}}{2-e^{-i}} \} (2)

    The numerical evaluation of (2) is tedious but not difficult and it is left to the reader ...

    Kind regards

    \chi \sigma
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