## Precalculus (10th Edition)

Converges, sum: $\frac{4}{7}$
An infinite geometric series converges if and only if $|r|\lt1$, where $r$ is the common ratio. If it converges, then it equals $\frac{a_1}{1-r}$ where $a_1$ is the first term. The common ratio is the quotient of two consecutive terms: $r=\frac{a_2}{a_1}=\dfrac{-\frac{3}{4}}{1}=-\frac{3}{4}$. $|-\frac{3}{4}|=\frac{3}{4}\lt1$, thus it converges. Hence the sum (since $a_1=1$): $\dfrac{1}{1-(-\frac{3}{4})}=\frac{4}{7}$