Answer
$S_{\infty}= \dfrac{3}{2}$
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 and $a_1$= the first term
Now, $r=\dfrac{a_2}{a_1} = \dfrac{(1/3)}{1}=\dfrac{1}{3}$
Since $\dfrac{1}{3}|\lt 1$, so the infinite geometric series converges.
Next, we will find the sum of the infinite geometric series when $a_1 = 1$ and $r=\dfrac{1}{3}$,
$S_{\infty} = \dfrac{a_1}{1-r} = \dfrac{1}{1-\dfrac{1}{3}}
\\ =\dfrac{1}{\dfrac{2}{3}} \\ = \dfrac{3}{2} $
Therefore, the sum of the convergent infinite geometric series is: $S_{\infty}= \dfrac{3}{2}$
