Answer
series converges;
$S_{\infty} = 16$
Work Step by Step
RECALL:
(1)
The common ratio $r$ of a geometric sequence is equal to the quotient of any term and the term before it.
$r=\dfrac{a_n}{a_{n-1}}$
(2)
A geometric series converges if $|r| \lt 1$. The sum of a convergent infinite geometric series is given by the formula:
$S_{\infty}=\dfrac{a_1}{1-r}$
where
$r$ = common ratio
$a_1$ = first term
$\bf\text{Solve for r}:$
$r=\dfrac{a_2}{a_1} = \dfrac{4}{8}=\dfrac{1}{2}$
Since $|\frac{1}{2}|\lt 1$, the infinite geometric series converges.
$\bf\text{Find the sum of the infinite geometric series}:$
With $a_1 = 8$ and $r=\frac{1}{2}$,
$S_{\infty} = \dfrac{a_1}{1-r}
\\S_{\infty}= \dfrac{8}{1-\frac{1}{2}}
\\S_{\infty}=\dfrac{8}{\frac{1}{2}}
\\S_{\infty}= 8 \cdot \dfrac{2}{1}
\\S_{\infty}= 16$
