Answer
series converges;
$S_{\infty}=\dfrac{8}{5}$
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
$\bf\text{Solve for r}:$
$r=\dfrac{a_2}{a_1} = \dfrac{-\frac{1}{2}}{2}=-\dfrac{1}{2} \cdot \dfrac{1}{2}=-\dfrac{1}{4}$
Since $|-\frac{1}{4}|=\frac{1}{4} \lt 1$, the infinite geometric series converges.
$\bf\text{Find the sum of the infinite geometric series}:$
With $a_1 = 6$ and $r=-\frac{1}{4}$,
$S_{\infty} = \dfrac{a_1}{1-r}
\\S_{\infty}= \dfrac{2}{1-(-\frac{1}{4})}
\\S_{\infty}=\dfrac{2}{\frac{4}{4} + \frac{1}{4}}
\\S_{\infty}=\dfrac{2}{\frac{5}{4}}
\\S_{\infty}= 2 \cdot \dfrac{4}{5}
\\S_{\infty}= \dfrac{8}{5}$
