## College Algebra (10th Edition)

The series diverges because $r=\dfrac{4}{3}$.
RECALL: (1) The common ratio $r$ of a geometric sequence is equal to the quotient of any term and the term before it. $r=\dfrac{a_n}{a_{n-1}}$ (2) A geometric series converges if $|r| \lt 1$. The sum of a convergent infinite geometric series is given by the formula: $S_{\infty}=\dfrac{a_1}{1-r}$ where $r$ = common ratio $a_1$ = first term $\bf\text{Solve for r}:$ $r=\dfrac{a_2}{a_1} = \dfrac{12}{9}=\dfrac{4}{3}$ Since $|\frac{4}{3}|=\frac{4}{3} \not\lt 1$, the infinite geometric diverges.