Answer
divergent
Work Step by Step
RECALL:
(1) The sum of an infinite geometric series is convergent if $|r| < 1$. The sum is given by the formula:
$S_{\infty}=\dfrac{a}{1-r}$
(2) An infinite geometric series is divergent if $|r|>1$.
(3) The common ratio $r$ of a geometric series can be found by dividing any term by the term before it:
$r = \dfrac{a_n}{a_{n-1}}$
Solve for $r$ to obtain:
$\require{cancel}
r = \dfrac{\frac{3}{2}}{1}
\\r=\dfrac{3}{2}$
Since $|\frac{3}{2}|\lt1$, then the series is divergent.