# How do you find the sum of the infinite geometric series 3-1+1/3...?

Well, it seems as if we are multiplying by $- \frac{1}{3}$ because from 3 to -1 would be $3 \cdot - \frac{1}{3}$ and from -1 to 1/3 would be $- 1 \cdot - \frac{1}{3}$
From here, we can just apply the infinite series formula, $\frac{1}{1 - r}$ because the r is between -1 and 1.
Thus, the sum is $\frac{1}{1 - \left(- \frac{1}{3}\right)}$ or $\frac{1}{1 + \frac{1}{3}}$ or $\frac{1}{\frac{4}{3}}$
or $\frac{3}{4}$