Find the Sum of the Infinite Geometric Series -6 , -3 , -3/2 , -3/4

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by ${\displaystyle \frac{1}{2}}$ gives the next term. In other words, ${\displaystyle a_n=a_1{r}^{n-1}}$.

The sum of a series ${\displaystyle S_n}$ is calculated using the formula ${\displaystyle S_n=\frac{a(1-r^n)}{1-r}}$. For the sum of an infinite geometric series ${\displaystyle S_{\infty }}$, as ${\displaystyle n}$ approaches ${\displaystyle \infty }$, ${\displaystyle 1-r^n}$ approaches ${\displaystyle 1}$. Thus, ${\displaystyle \frac{a(1-r^n)}{1-r}}$ approaches ${\displaystyle \frac{a}{1-r}}$.

The values ${\displaystyle a=-6}$ and ${\displaystyle r=\frac{1}{2}}$ can be put in the equation ${\displaystyle S_{\infty }}$.