Answer
convergent;
$S_{\infty}=2$
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 $|r|\ge1$.
(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:
$\require{cancel}
r = \dfrac{-\frac{3}{2}}{3}
\\r=-\dfrac{3}{2} \cdot \dfrac{1}{3}
\\r=-\dfrac{\cancel{3}}{2} \cdot \dfrac{1}{\cancel{3}}
\\r=-\dfrac{1}{2}$
Since $|-\frac{1}{2}|<1$, then the series is convergent.
Solve for the sum using the formula above, with $a=3$ and $r=-\frac{1}{2}$, to obtain:
$\require{cancel}
S_{\infty}=\dfrac{3}{1-(-\frac{1}{2})}
\\S_{\infty}=\dfrac{3}{\frac{2}{2}+\frac{1}{2}}
\\S_{\infty}=\dfrac{3}{\frac{3}{2}}
\\S_{\infty} = 3 \cdot \frac{2}{3}
\\S_{\infty} = \cancel{3} \cdot \dfrac{2}{\cancel{3}}
\\S_{\infty}=2$