## College Algebra (10th Edition)

series converges; $S_{\infty}= 6$
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 $\bf\text{Solve for r}:$ $r=\dfrac{a_2}{a_1} = \dfrac{\frac{4}{3}}{2}=\dfrac{4}{3} \cdot \dfrac{1}{2} = \dfrac{4}{6}=\dfrac{2}{3}$ Since $|\frac{2}{3}|\lt 1$, the infinite geometric series converges. $\bf\text{Find the sum of the infinite geometric series}:$ With $a_1 = 2$ and $r=\frac{2}{3}$, $S_{\infty} = \dfrac{a_1}{1-r} \\S_{\infty}= \dfrac{2}{1-\frac{2}{3}} \\S_{\infty}=\dfrac{2}{\frac{1}{3}} \\S_{\infty}= 2 \cdot \dfrac{3}{1} \\S_{\infty}= 6$