Answer
$S_\infty=-\frac{8}{15}\approx -0.5333$
The sum converges.
Work Step by Step
In a geometric series the ratio between consecutive terms is constant. In order to find that ratio we simply divide any two consecutive terms of the series.
The first term of the series is when $i=1$, the second is when $i=2$. $r=\frac{a_2}{a_1}=\frac{(-\frac{2}{3})\times (-\frac{1}{4})^{2-1}}{(-\frac{2}{3})\times (-\frac{1}{4})^{1-1}}=\frac{\frac{2}{12}}{-\frac{2}{3}}=-\frac{1}{4}$
The sum of an infinite geometric series converges if $\mid r\mid<1$, Here, it is true, therefore the sum converges.
The sum of an infinite series is:
$S_\infty=\frac{a_1}{1-r}$
Here:
$S_\infty=\frac{-\frac{2}{3}}{1-(-\frac{1}{4})}=\frac{-\frac{2}{3}}{\frac{5}{4}}=-\frac{8}{15}\approx -0.5333$
