mathURL
http://mathurl.com/cgq32bp
<img src="http://mathurl.com/cgq32bp.png" />
![\[
-8\cdot\sum_{n=1}^\infty \left(-\frac{1}{3}\right)^n.
\]
Now since $\left|-1/3\right|<1$ we have
\[
\sum_{k=0}^\infty x^n = \frac{1}{1-r}
\]](http://mathurl.com/cgq32bp.png)