Answer
$r=\frac{1}{3}$
$S_\infty=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}$
Work Step by Step
In this given geometric series we have to find the ratio, in order to calculate the infinite sum:
$r=\frac{a_2}{a_1}=\frac{\frac{1}{3}}{1}=\frac{1}{3}$
The sum of an infinite geometric series exists only if $\mid r\mid<1$, here it is true. Therefore the sum converges.
The sum of an infinite geometric series is:
$S_\infty=\frac{a_1}{1-r}$
Here: $S_\infty=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}$
