## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$S_{\infty}= \dfrac{3}{2}$
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: $S_{\infty}= \dfrac{3}{2}$