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Thread: Does the series converge?

  1. November 4th 2008, 10:26 AM #1
    veronicak5678
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    Does the series converge?

    Summation from n=1 ti infinity of (sin(1))^n.

    How can I break this up? I think it will be a geometric series...
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  2. November 4th 2008, 10:34 AM #2
    Plato
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    \left| r \right| < 1 \Rightarrow \quad \sum\limits_{n = 1}^\infty {r^n } = \frac{r}{{1 - r}}
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  3. November 4th 2008, 10:36 AM #3
    veronicak5678
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    I thought it was r ^ (n-1)? And what is my r and my a?
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  4. November 4th 2008, 01:12 PM #4
    Mathstud28
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    Quote Originally Posted by veronicak5678 View Post
    Summation from n=1 ti infinity of (sin(1))^n.

    How can I break this up? I think it will be a geometric series...
    |\sin(1)|=|r|<1\Rightarrow\sum)_{n=1}^{\infty}\sin ^n(1)=\frac{\sin(1)}{1-\sin(1)}
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  5. November 4th 2008, 01:22 PM #5
    veronicak5678
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    Got it. Thank you!
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