Answer
Series Converges.
and
$S_{\infty}=12$
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 $r \ =common \ ratio \ =\dfrac{1}{3}$
Since $r=|\dfrac{1}{3}| \lt 1$, so the infinite geometric Series Converges.
First term: $a_1=8(\dfrac{1}{3})^{1-1}=(8)(1)=8$
Next, $S_{\infty}=\dfrac{8}{1-\dfrac{1}{3}}=\dfrac{8}{2/3}$
Therefore, the sum of the convergent infinite geometric series is: $S_{\infty}=12$
