## College Algebra 7th Edition

convergent; $S_{\infty}=\dfrac{3}{4}$
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: $S_{\infty}=\dfrac{1}{1-(-\frac{1}{3})} \\S_{\infty}=\dfrac{1}{\frac{3}{3}+\frac{1}{3}} \\S_{\infty}=\dfrac{1}{\frac{4}{3}} \\S_{\infty} = 1 \cdot \frac{3}{4} \\S_{\infty}=\dfrac{3}{4}$