Answer
$\frac{r~\alpha}{c-r~\alpha} = \frac{r\alpha}{c}+ (\frac{r\alpha}{c})^2+(\frac{r\alpha}{c})^3+(\frac{r\alpha}{c})^4+...$
Work Step by Step
$\frac{\Delta \lambda}{\lambda} = \frac{r~\alpha}{c-r~\alpha}$
We can expand the right side of the equation:
$\frac{r~\alpha}{c-r~\alpha} = \frac{r\alpha}{c}+ (\frac{r\alpha}{c})^2+(\frac{r\alpha}{c})^3+(\frac{r\alpha}{c})^4+...$
