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Math Help - Series

  1. #1
    Member Nacho's Avatar
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    Series

    Converge?

    <br />
\sum\limits_{n = 1}^\infty  {\frac{{\sin (n)}}<br />
{n}} <br />

    thanks
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  2. #2
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    Quote Originally Posted by Nacho View Post
    Converge?

    <br />
\sum\limits_{n = 1}^\infty  {\frac{{\sin (n)}}<br />
{n}} <br />

    thanks
    Yes. It can be proven by Dirichlet's test that \sum_{n=1}^{\infty}\frac{\sin n}{n} and \sum_{n=1}^{\infty} \frac{\cos n}{n} converge.
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  3. #3
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    Quote Originally Posted by Nacho View Post
    <br />
\sum\limits_{n = 1}^\infty  {\frac{{\sin (n)}}<br />
{n}} <br />
    It is more interesting to sum this series.

    Remember that 2i\sin (x) = \sinh (ix).

    Therefore,
    \sum_{n=1}^{\infty} \frac{\sin (n)}{n} = \frac{1}{2i} \left( \sum_{n=1}^{\infty} \frac{e^{in}}{n} - \sum_{n=1}^{\infty} \frac{e^{-in}}{n} \right)

    Now use the formula,
    \sum_{n=1}^{\infty} \frac{z^n}{n} = - \log (1-z) for |z| < 1

    Therefore, we get,
    \frac{1}{2i}\left(  \log (1 - e^{-i}) - \log (1 - e^{i}) \right) = \sin 1

    I hope I did not make any mistakes.
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  4. #4
    Member Nacho's Avatar
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    Thanks!

    but I don`t know about complex number and haven`t demostration of Dirichlet`s test

    again, than you ThePerfectHacker
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  5. #5
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    Quote Originally Posted by Nacho View Post
    Thanks!

    but I don`t know about complex number and haven`t demostration of Dirichlet`s test

    again, than you ThePerfectHacker
    How can I help you?
    What do you need explained?
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  6. #6
    Member Nacho's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    How can I help you?
    What do you need explained?
    Maybe we can solve delimiting?
    I always find other serie minor, but this coverge or a major, but this diverge...

    Too I try whit quotien rule o the integral, but never obey the hypothesis

    Moreover I am in first year of university, so I dont know powerful tool...

    anyways...thank you ThePerfectHacker
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  7. #7
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    But you asked if the series does converge or not and that was answered. Now, wanna know how to find the value of that series? Its value is \frac{\pi-1}{2}.
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  8. #8
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    Quote Originally Posted by Krizalid View Post
    Now, wanna know how to find the value of that series? Its value is \frac{\pi-1}{2}.
    Yes, thank you Krizalid. I thought I made a mistake. But my method should work if fixed.
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  9. #9
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    \frac{1}{2i}\ln \frac{1-e^{-i}}{1-e^{i}}=\frac{1}{2i}\ln \left( -e^{-i} \right)=\frac{1}{2i}\ln \left( e^{(\pi -1)i} \right)=\frac{\pi -1}{2}.

    Quote Originally Posted by ThePerfectHacker View Post

    But my method should work if fixed.
    Yes.
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