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Math Help - Convergent or divergent? Find sum if convergent. (problem 2)

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    Convergent or divergent? Find sum if convergent. (problem 2)

    \sum\limits_{i=1}^{infty} \frac{(-2)^{n+4}}{\pi^n}
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    Re: Convergent or divergent? Find sum if convergent. (problem 2)

    Hello, superduper1!

    \sum\limits_{n=1}^{\infty} \frac{(\text{-}2)^{n+4}}{\pi^n}

    This is a geometric series.
    . . The first term is: . a \;=\;\frac{(\text{-}2)^5}{\pi^1} \;=\;\text{-}\frac{32}{\pi}
    . . The common ratio is: . r \;=\;\text{-}\frac{2}{\pi}

    The sum is: . S \;=\;\frac{a}{1-r}

    . . S \;=\;\dfrac{\text{-}\frac{32}{\pi}}{1-(\text{-}\frac{2}{\pi})} \;=\; \dfrac{\text{-}\frac{32}{\pi}}{\frac{\pi+2}{\pi}} \;=\;  \frac{\text{-}32}{\pi+2}
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    Re: Convergent or divergent? Find sum if convergent. (problem 2)

    How did you get the common ratio?
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    Re: Convergent or divergent? Find sum if convergent. (problem 2)

    Quote Originally Posted by superduper1 View Post
    How did you get the common ratio?
    r is the ratio of the n+1th term by the nth term. In this case:
    r = \frac{ \frac{(-2)^{(n + 4) + 1}}{\pi ^ {n + 1}} }{\frac{(-2)^{n + 4}}{\pi ^ n} }
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  5. #5
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    Re: Convergent or divergent? Find sum if convergent. (problem 2)

    \sum_{i=1}^{\infty} \frac{(-2)^{n+4}}{\pi^n} = 16 \sum_{i=1}^{\infty} \left(\frac{-2}{\pi}\right)^n = 16 \cdot \frac{\frac{-2}{\pi}}{1 + \frac{2}{\pi}} = \frac{-32}{\pi+2}
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