## College Algebra (10th Edition)

The geometric series converges. $S_{\infty} = \dfrac{18}{5}$
RECALL: (1) In the infinite geometric series: $$\sum_{k=1}^{\infty}c \cdot r^{n-1}$$ $r$ is the common ratio. (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 $\bf\text{Solve for r}:$ Note that when a geometric series is summation notation, the expression being raised to a power is the common ratio. Thus, the common ratio of the given series is $-\dfrac{2}{3}$. Since $|-\frac{2}{3}|=\frac{2}{3} \lt 1$, the series converges. $\bf\text{Find the sum:}$ The first term of the series can be evaluated by substituting $1$ for $k$: $a_1 = 6\left(-\dfrac{2}{3}\right)^{1-1} = 6\left(-\dfrac{2}{3}\right)^0=6(1) = 6$ With $a_1=6$ and $r=-\dfrac{2}{3}$, solve for the sum using the formula in part (2) above to obtain: $S_{\infty} = \dfrac{a_1}{a-r} \\S_{\infty}=\dfrac{6}{1-(-\frac{2}{3})} \\S_{\infty}=\dfrac{6}{\frac{3}{3}+\frac{2}{3}} \\S_{\infty}=\dfrac{6}{\frac{5}{3}} \\S_{\infty}=6 \cdot \dfrac{3}{5} \\S_{\infty}=\dfrac{18}{5}$