Answer
Series Converges.
and
$S_{\infty}=\dfrac{8}{3}$
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 $r \ =common \ ratio \ =\dfrac{-1}{2}$
Since $r=|\dfrac{-1}{2}|=\dfrac{1}{2} \lt 1$, so the infinite geometric series Converges.
First term: $a_1=4(\dfrac{-1}{2})^{1-1}=(4)(1)=4$
Next, $S_{\infty}=\dfrac{4}{1-(\dfrac{-1}{2})}=\dfrac{4}{3/2}$
Therefore, the sum of the convergent infinite geometric series is: $S_{\infty}=\dfrac{8}{3}$
