Answer
series converges;
$S_{\infty}= \dfrac{3}{2}$
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
$\bf\text{Solve for r}:$
$r=\dfrac{a_2}{a_1} = \dfrac{\frac{1}{3}}{1}=\dfrac{1}{3}$
Since $\frac{1}{3}|\lt 1$, the infinite geometric series converges.
$\bf\text{Find the sum of the infinite geometric series}:$
With $a_1 = 1$ and $r=\frac{1}{3}$,
$S_{\infty} = \dfrac{a_1}{1-r}
\\S_{\infty}= \dfrac{1}{1-\frac{1}{3}}
\\S_{\infty}=\dfrac{1}{\frac{2}{3}}
\\S_{\infty}= 1 \cdot \dfrac{3}{2}
\\S_{\infty}= \dfrac{3}{2}$