Answer
Series Converges.
and
$S_{\infty}= 6$
Work Step by Step
The common ratio of a geometric sequence is equal to the quotient of any term and the term before it:
$ \ r = \dfrac{a_n}{a_{n-1}}$ or, $r=\dfrac{a_2}{a_1}$
The sum of a convergent infinite geometric series is given by the formula:
$S_{\infty}=\dfrac{a_1}{1-r}$ and a geometric series converges if $|r| \lt 1$.
where $r \ =common \ ratio \ =\dfrac{2}{3}$
Since $r=|\dfrac{2}{3}|=\dfrac{2}{3} \lt 1$, so the infinite geometric series Converges.
First term: $a_1=3(\dfrac{2}{3})^{1}=\dfrac{6}{3}=2$
Next, $S_{\infty}=\dfrac{2}{1-(\dfrac{2}{3})}=\dfrac{2}{1/3}$
Therefore, the sum of the convergent infinite geometric series is: $S_{\infty}= 6$
