## College Algebra (10th Edition)

The infinite geometric series converges. $S_{\infty} = \dfrac{8}{3}$
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 $-\dfrac{1}{2}$. Since $|-\frac{1}{2}|=\frac{1}{2} \lt 1$, the series converges. $\bf\text{Find the sum:}$ The first term of the series can be evaluated by substituting $1$ for $k$: $a_1 = 4\left(-\dfrac{1}{2}\right)^{1-1} = 4\left(-\dfrac{1}{2}\right)^0=4(1) = 4$ With $a_1=4$ and $r=-\dfrac{1}{2}$, solve for the sum using the formula in part (2) above to obtain: $S_{\infty} = \dfrac{a_1}{a-r} \\S_{\infty}=\dfrac{4}{1-(-\frac{1}{2})} \\S_{\infty}=\dfrac{4}{\frac{2}{2}+\frac{1}{2}} \\S_{\infty}=\dfrac{4}{\frac{3}{2}} \\S_{\infty}=4 \cdot \dfrac{2}{3} \\S_{\infty}=\dfrac{8}{3}$