Answer
convergent;
$S_{\infty}=\dfrac{3}{2}$
Work Step by Step
RECALL:
(1) The sum of an infinite geometric series is convergent if $|r| < 1$. The sum is given by the formula:
$S_{\infty}=\dfrac{a}{1-r}$
(2) An infinite geometric series is divergent if $|r|>1$.
(3) The common ratio $r$ of a geometric series can be found by dividing any term by the term before it:
$r = \dfrac{a_n}{a_{n-1}}$
Solve for $r$ to obtain:
$r = \dfrac{\frac{1}{3}}{1}
\\r=\dfrac{1}{3}$
Since $|\frac{1}{3}|<1$, then the series converges.
Solve for the sum using the formula above, with $a=1$ and $r=\frac{1}{3}$, to obtain:
$S_{\infty}=\dfrac{1}{1-\frac{1}{3}}
\\S_{\infty}=\dfrac{1}{\frac{2}{3}}
\\S_{\infty} = 1 \cdot \frac{3}{2}
\\S_{\infty}=\dfrac{3}{2}$
