If x is positive then the sum to infinity of the series (1)/(1+3...

The series is a G.P. with common ratio $=\left(\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right)$ and |\mathrm{r}|=\left|\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right|$is less than 1 since$\mathrm{x}$is positive$\mathrm{S}_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\frac{1}{1+3 \mathrm{x}}}{1-\left\{-\left(\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right)\right\}}=\frac{1}{2}