MHB Series using Geometric series argument

[karush]

$\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*}

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

$\tiny{206.10.3.75}$

 

[MarkFL]

I would write:

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

Now you're ready to plug and play. :D

 

[karush]

$\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*}
$\tiny{206.10.3.75}$