I'm asked to find the sum for $s_n=10-8+6.4-5.12+...$ as $n\rightarrow \infty$. I discovered that the sum can be written as $$10\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{8}{10}\right)^{n-1}$$
I know from the ratio/roots test the series indeed converges. My problem is figuring out what it converges to. I don't see how I can use the geometric formula $\frac{a}{1-r}$.
$$\sum_{n=1}^{\infty}10(-1)^{n-1}\biggr(\frac{8}{10}\biggr)^{n-1} = \sum_{n=0}^{\infty}10\left(\frac{-8}{10}\right)^{n}=10\frac{1}{1+\frac{8}{10}}=\frac{100}{19} $$