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  1. MHB Master
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    #1
    $\displaystyle\text{if} \left| r \right|< 1 \text{ the geometric series }
    a+ar+ar^2+\cdots ar^{n-1}+\cdots \text{converges} $
    $\displaystyle\text{to} \frac{a}{(1-r)}.$
    $$\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{(1-r)}, \ \ \left| r \right|< 1$$
    $\text{if} \left| r \right|\ge 1 \text{, the series diverges}$

    $\text{Evaluate using geometric series argument}$
    \begin{align*}
    \displaystyle
    S&=\sum_{k=1}^{\infty}\left[\frac{8}{5^k}-\frac{8}{5^{k+1}}\right] \\
    \left[\frac{8}{5^k}-\frac{8}{5^{k+1}}\right]&=\frac{32}{5\cdot5^k} \\
    &=
    \end{align*}

    $\text{ok wasn't sure how to plug this in?}$

    $\tiny{206.10.3.75}$
    Last edited by karush; October 19th, 2016 at 20:53. Reason: Correct LaTeX Code

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    #2
    I would write:

    $ \displaystyle \frac{32}{5\cdot5^k}=\frac{32}{25}\left(\frac{1}{5}\right)^{k-1}$

    Now you're ready to plug and play.

  4. MHB Master
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    #3 Thread Author
    $\displaystyle\text{if} \left| r \right|< 1 \text{ the geometric series }
    a+ar+ar^2+\cdots ar^{n-1}+\cdots \text{converges} $
    $\displaystyle\text{to} \frac{a}{(1-r)}.$
    $$\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{(1-r)}, \ \ \left| r \right|< 1$$
    $\text{if} \left| r \right|\ge 1 \text{, the series diverges}$

    $\text{Evaluate using geometric series argument}$
    \begin{align*}
    \displaystyle
    S&=\sum_{k=1}^{\infty}\left[\frac{8}{5^k}-\frac{8}{5^{k+1}}\right] \\
    \left[\frac{8}{5^k}-\frac{8}{5^{k+1}}\right]&=\frac{32}{5\cdot5^k} =\frac{32}{25}\left(\frac{1}{5}\right)^{k-1 }\\
    \text{so then } a&=\frac{32}{25}; r=\frac{1}{5} \\
    \sum_{n=1}^{\infty}\frac{32}{25}
    \left(\frac{1}{5}\right)^{n-1}
    &=\frac{\frac{32}{25}}{(1-\frac{1}{5})}=\frac{8}{5}
    \end{align*}
    $\tiny{206.10.3.75}$

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