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Math Help - Infinite diverging series

  1. #1
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    Infinite diverging series

    Having an issue with #17

    Do I divide the top and bottom by 1/2^n

    If so how does the bottom work since the exponet is n+1?Infinite diverging series-imageuploadedbytapatalk-21372816640.960175.jpg
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  2. #2
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    Re: Infinite diverging series

    \displaystyle \begin{align*} \frac{2^n + 1}{2^{n+1}} &= \frac{2^n}{2^{n+1}} + \frac{1}{2^{n+1}} \\ &= \frac{1}{2} + \frac{1}{2^{n+1}} \\ &= \frac{1}{2} + \left( \frac{1}{2} \right) ^{n+1} \end{align*}

    So the series must be divergent because constantly adding \displaystyle \begin{align*} \frac{1}{2} \end{align*} will eventually lead to \displaystyle \begin{align*} \infty \end{align*}.
    Last edited by Prove It; July 2nd 2013 at 07:16 PM.
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